COMPULSORY
1
READINGS
1
According to the author of the module, the compulsory readings do not infringe known copyright.
COMPULSORY READINGS
Reading #1
PROPERTIES OF GASES
Complete reference :
Chem1 VirtualText Book
URL: http://www.chem1.com/acad/pdf/c1gas.pdf
Date Accessed 26 August 2006
ABSTRACT
This text introduces the properties of gases as exhibited in the gas laws. It then uses the
kinetic model to explain and justify the behaviour of gases from a molecular viewpoint.
RATIONALE
If gases do exist as particles then they must be explainable on this basis – kinetic molecular
model. This reading is essential for the understanding of learning content in Unit I of this
module. It introduces the gas laws followed by a kinetic explanation and justification for them.
This topic is covered in Sections 5 – 7 of the text.
Reading #2
CHEMICAL ENERGETICS
Complete reference:
http://www.chem1.com/acad/webtext/virtualtextbook.html
Date Accessed 26 August 2006
ABSTRACT
This document introduces the learner to the fundamental concepts of thermodynamics. It
begins by defining terms such as energy, the various types of energy (kinetic and potential
energy) and units of energy. It then goes on to explain the conceptual knowledge for
understanding thermodynamics. It goes further to look at the interconversion of energy as it
tackles the first law of thermodynamics and its applications: in thermochemistry (energy
changes in chemical reactions), thermochemical reactions, bond energies, energy content of
foods and fuels.
RATIONALE
Energy changes are critical to an understanding of many of the concepts encountered in this
module. This document therefore is critical reading for Unit II and Unit III of the module.
Concepts initially developed in Unit II are applied in Unit III in studying the energy changes
involved in chemical reactions.
Reading #3
Complete reference:
THERMODYNAMIC EQUILIBRIUM
http://www.chem1.com/acad/webtext/virtualtextbook.html
Date Accessed 26 August 2006
ABSTRACT
The reference text covers aspects of the drive towards chemical change in terms of
spreading of energy through entropy and Gibb’s free energy. Concepts are developed from a
molecular point of view to explain physical phenomenon. Factors that favour spontaneity of
processes and the condition for equilibrium are discussed.
RATIONALE
Thermodynamic equilibrium is essential to understanding why reactions are spontaneous
and how far they go. Spontaneity is the subject of Unit IV and is well covered by this
reference text. Concepts developed earlier in Unit II are used in studying factors that promote
spontaneity of chemical change.
Reading # 4
Complete reference:
GAS LAWS
ABSTRACT
This text covers the basics of the properties of gases. It begins by looking at the three states
of matter before focusing on the gases. It reviews the gas laws of Boyle, Charles and
Avogadro giving examples of how they are used in solving gas problems. The ideal gas
equation is also discussed.
RATIONALE
Sound knowledge of the basic properties and the laws governing the behaviour of
gases is necessary before tackling the learning material in Unit I. The kinetic
molecular theory, which is the major focus of Unit I explains the behaviour of gases at
a molecular level and justifies the gas laws.
Reading(s) #1
Properties of Gases
Introduction to the ideal gas laws, kinetic-molecular theory and real gases
A Chem1 Virtual Textbook chapter
Stephen K. Lower1 • Simon Fraser University
Table of contents
1
2
3
4
5
6
7
Introduction: observable properties of gases . . . . . . . . . . . . . . . . . . . . . . . . 3
What’s special about gases? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
The pressure of a gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
The temperature of a gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
The basic gas laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Pressure-volume relations: Boyle's law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
How the temperature affects the volume: Charles’ law . . . . . . . . . . . . . . . . . 11
Volume and the number of molecules: Avogadro’s law . . . . . . . . . . . . . . . . . 12
The ideal gas equation of state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Molar volume and density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
Molar volume of a gas: standard temperature and pressure . . . . . . . . . . . . . . 14
Molecular weight and density of a gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
Mixtures of gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Mole fractions and volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Dalton's law of partial pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
Some applications of Dalton’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
Molecules in motion: introduction to kinetic molecular theory . . . . . . . . . 21
The kinetic-molecular model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Kinetic molecular interpretation of gas properties . . . . . . . . . . . . . . . . . . . . . 22
Diffusion: random motion with direction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
Density fluctuations: Why is the sky blue? . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
Light bulbs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
Viscosity of gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
Distribution of gas molecules in a gravitational field . . . . . . . . . . . . . . . . . . . 26
The ionosphere and radio communication . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
More on the kinetic-molecular model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
The velocities of gas molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
The Boltzmann distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
Derivation of the ideal gas equation of state . . . . . . . . . . . . . . . . . . . . . . . . . . 32
How far does a molecule travel between collisions? . . . . . . . . . . . . . . . . . . . . 34
Real gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
Effects of intermolecular forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Equations of state for real gases: the van der Waals equation . . . . . . . . . . . . 38
Condensation and the critical point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
1. To contact the author, please use the Web form at http://www.chem1.com/VT_mail.html
Chem1 Properties of Gases
About this document:
The Chem1 Virtual Textbook is a collection of reference textbook chapters and tutorial units
providing in-depth coverage of topics in college-level General
This document covers its topic at a level appropriate for first-year college chemistry.
It was last modified on 13 April 2005.
It can be downloaded from http://www.chem1.com/acad/pdf/c1gas.pdf
A Web-based version is available at http://www.chem1.com/acad/webtext/gas/.
Page 2 of 41
Chem1 Properties of Gases
1 • Introduction: observable properties of gases
1 • Introduction: observable properties of gases
Throughout much of human history, “airs” or gases were not believed to be matter at all;
their apparently weightless nature and their ability to move about freely and fill all available space, while carrying with them definite physical properties such as odor and sometimes color, conferred upon them a somewhat mysterious nature. Even the scientist Robert
Boyle wrote about “The Strange Subtility, Great Efficacy and Determinate Nature of Effluviums".
It's interesting, however, that around 550 BCE the Greek philospher Anaximenes maintained that
all matter consists of air: "It is from air that all the things that exist , have existed, or will exist come
into being."
The invention of the sensitive balance in the early seventeenth century showed once and for
all that gases have weight and are therefore matter. Guericke's invention of the air pump
(which led directly to his discovery of the vacuum) launched the “pneumatic era" of chemistry long before the existence of atoms and molecules had been accepted. Indeed, the behavior
of gases was soon to prove an invaluable tool in the development of the atomic theory of matter.
The study of gases allows us to understand the behavior of matter at its simplest: individual
particles, acting as individuals, almost completely uncomplicated by interactions and interferences between each other. Later on, our knowledge of gases will serve as the pathway to
our understanding of the far more complicated condensed phases (liquids and solids) in
which the theory of gases will no longer give us correct answers, but it will still provide us
with a useful model that will at least help us to rationalize the behavior of these more complicated states of matter.
1.1 What’s special about gases?
Let us start with what we can observe experimentally about gases.
• First, we know that a gas has no definite volume or shape; a gas will fill whatever volume is
available to it. Contrast this to the behavior of a liquid, which always has a distinct upper surface
when its volume is less than that of the space it occupies.
• The other outstanding characteristic of gases is their low densities, compared with those of liquids and solids. One mole of liquid water at 298 K and 1 atm pressure occupies a volume of 18.8
cm3, whereas the same quantity of water vapor at the same temperature and pressure has a volume of 30200 cm3, more than 1000 times greater.
• The most remarkable property of gases, however, is that to a very good approximation, they all
behave the same way in response to changes in temperature and pressure, expanding or contracting by predictable amounts. This is very different from the behavior of liquids or solids, in
which the properties of each particular substance must be determined individually.
We will see later that each of these three macroscopic characteristics of gases follows directly
from the microscopic view— that is, from the atomic nature of matter.
What’s special about gases?
Page 3 of 41
Chem1 Properties of Gases
1 • Introduction: observable properties of gases
1.2 The pressure of a gas
The molecules of a gas, being in continuous motion, frequently strike the inner walls of their
container. As they do so, they immediately bounce off without loss of kinetic energy, but the
reversal of direction (acceleration) imparts a force to the container walls. This force, divided
by the total surface area on which it acts, is the pressure of the gas.
The pressure of a gas is observed by measuring the pressure that must be applied externally
in order to keep the gas from expanding or contracting. To visualize this, imagine some gas
trapped in a cylinder having one end enclosed by a freely moving piston. In order to keep the
gas in the container, a certain amount of weight (more precisely, a force, f ) must be placed on
the piston so as to exactly balance the force exerted by the gas on the bottom of the piston,
and tending to push it up. The pressure of the gas is simply the quotient f/A, where A is the
cross-section area of the piston.
Pressure units
The unit of pressure in the SI system is the pascal (Pa), defined as a force of one newton per
square metre (1 Nm–2 = 1 kg m–1 s–2.)
At the Earth's surface, the force of gravity acting on a 1 kg mass is 9.81 N. Thus if, in the above
Figure, the weight is 1 kg and the surface area of the piston is 1 M, the pressure of the gas would
be 9.81 Pa. A 1-gram weight acting on a piston of 1 cm2 cross-section would exert a pressure of
98.1 pA. (If you wonder why the pressure is higher in the second example, consider the number of
cm2 contained in 1 m2.)
In chemistry, it is more common to express pressures in units of atmospheres or torr:
1 atm = 101325 Pa = 760 torr.
The older unit millimetre of mercury (mm Hg) is almost the same as the torr; it is defined as
one mm of level difference in a mercury barometer at 0°C. In meteorology, the pressure unit
most commonly used is the bar:
1 bar = 106 N m–2 = 0.987 atm.
In engineering work the pound per square inch is still widely used; standard atmospheric
pressure is 14.7 psi.
The pressure of a gas
Page 4 of 41
Chem1 Properties of Gases
1 • Introduction: observable properties of gases
How is pressure measured?
Atmospheric pressure and the barometer The column of air above us exerts a force on each
1-cm2 of surface equivalent to a weight of about 1034 g.
This figure is obtained by solving Newton's law f = ma for m, using the acceleration of gravity for a:
f
101375 kg m-1 s–2
m = --- = ---------------------------------------------- = 10340 = 1034gcm –2
a
9.8m –2
So, if several kilos of air are constantly pressing down on your
body, why do you not feel it?
Answer: because every other part of your body (including within
your lungs and insides) also experiences the same pressure, so
there is no net force (other than gravity) acting on you.
In the early 17th century the Italian EVANGELISTA
TORRICELLI invented a device to measure this pressure. The barometer consists of a vertical glass
tube closed at the top and evacuated, and open at
the bottom, where it is immersed in a dish of a liquid. The atmospheric pressure acting on this liquid
will force it up into the evacuated tube until the
weight of the liquid column exactly balances the
atmospheric pressure. If the liquid is mercury, the
height supported will be about 760 cm; this height
corresponds to standard atmospheric pressure.
How is the air pressure of 1034 g cm–3 related to the 760-mm
height of the mercury column in the barometer? What if water
were used in place of mercury?
Answer: The density of Hg is 13.6 g cm–3, so in a column of 1cm2 cross-section, the height needed to counter the atmospheric
pressure would be (1034 g / 1 cm2) / (13.6 g cm–3) = 76 cm.
The density of water is only 1/13.6 that of mercury, so standard atmospheric pressure would
support a water column whose height is 13.6 x 76 cm = 1034 cm, or 10.3 m. You would have
to read a water barometer from a fourth-story window!
Torricelli's invention overturned the then-common belief that air (and by extension, all
gases) are weightless.
The fact that we live at the bottom of a sea of air was most spectacularly demonstrated in 1654,
when two teams of eight horses were unable to pull apart two 14-inch copper hemispheres (the
"Magdeburg hemispheres") which had been joined together and then evacuated with Guericke's
newly-invented vacuum pump.
The pressure of a gas
Page 5 of 41
Chem1 Properties of Gases
1 • Introduction: observable properties of gases
The manometer
A modification of the barometer, the U-tube manometer,
provides a simple device for measuring the pressure of
any gas in a container. The U-tube is partially filled with
mercury, one end is connected to container, while the
other end is left open to the atmosphere. The pressure
inside the container is found from the difference in height
between the mercury in the two sides of the U-tube.
The pressure of a gas
Page 6 of 41
Chem1 Properties of Gases
1 • Introduction: observable properties of gases
1.3 The temperature of a gas
If two bodies are at different temperatures, heat will flow from the warmer to the cooler one
until their temperatures are the same. This is the principle on which thermometry is based;
the temperature of an object is measured indirectly by placing a calibrated device known as
a thermometer in contact with it. When thermal equilibrium is obtained, the temperature
of the thermometer is the same as the temperature of the object.
Temperature scales
A thermometer makes use of some temperature-dependent quantity, such as the density of a
liquid, to allow the temperature to be found indirectly through some easily measured quantity such as the length of a mercury column. The resulting scale of temperature is entirely
arbitrary; it is defined by locating its zero point, and the size of the degree unit.
At one point in the 18th century, 35 different temperature scales were in use!
The Celsius temperature scale locates the zero point at the freezing temperature of water;
the Celsius degree (C °)1 is defined as 1/100 of the difference between the freezing and boiling temperatures of water at 1 atm pressure.
The older Fahrenheit scale placed the zero point at the coldest temperature it was possible to
obtain at the time (by mixing salt and ice.) The 100° point was set with body temperature
(later found to be 98.6°F.) On this scale, water freezes at 32°F and boils at 212°F. The Fahrenheit scale is a finer one than the Celsius scale; there are 180 Fahrenheit degrees in the
same temperature interval that contains 100 Celsius degrees, so 1F° = 9/5 C . Since the zero
points are also different by 32F, conversion between temperatures expressed on the two
scales requires the addition or subtraction of this offset, as well as multiplication by the
ratio of the degree size.
You should be able to derive the formula for this conversion.
Absolute temperature
In 1787 the French mathematician and physicist JACQUES CHARLES discovered that for each
Celsius degree that the temperature of a gas is lowered, the volume of the gas will diminish
by 1/273 of its volume at 0°C. The obvious implication of this is that if the temperature could
be reduced to –273°C, the volume of the gas would contract to zero. Of course, all real gases
condense to liquids before this happens, but at sufficiently low pressures their volumes are
linear functions of the temperature (Charles' Law), and extrapolation of a plot of volume as a
function of temperature predicts zero volume at -273°C. This temperature, known as absolute zero, corresponds to the total absence of thermal energy.
The temperature scale on which the zero point is –273.15°C was suggested by LORD KELVIN,
and is usually known as the Kelvin scale. Since the sizes of the Kelvin and Celsius degrees
are the same, conversion between the two scales is a simple matter of adding or subtracting
273.15; thus room temperature, 20°, is about 293 K.
1. Notice that temperature is expressed by placing the degree symbol in front of the scale
abbreviation (37°C), whereas a temperature interval is written with the degree sign following the sumbol (2 C°).
The temperature of a gas
Page 7 of 41
Chem1 Properties of Gases
2 • The basic gas laws
A rather fine point to note: the degree symbol is not used with the "K", which should always be separated from the preceding number by a space. See here for an explanation.
Because the Kelvin scale is based on an absolute, rather than on an arbitrary zero of temperature, it plays a special significance in scientific calculations; most fundamental physical
relations involving temperature are expressed mathematically in terms of absolute temperature. In engineering work, an absolute scale based on the Fahrenheit degree is sometimes
used; this is known as the Rankine scale.
2 • The basic gas laws
The "pneumatic" era of chemistry began with the discovery of the vacuum around 1650
which clearly established that gases are a form of matter. The ease with which gases could
be studied soon led to the discovery of numerous emprical (experimental) laws that proved
fundamental to the later development of chemistry and led indirectly to the atomic view of
matter.
2.1 Pressure-volume relations: Boyle's law
ROBERT BOYLE (1627-91) showed that the volume of air trapped by a liquid in the closed
short limb of a J-shaped tube decreased in exact proportion to the pressure produced by the
liquid in the long part of the tube. The trapped air acted much like a spring, exerting a force
opposing its compression. Boyle called this effect “the spring of the air", and published his
results in a pamphlet of that title.
Some of Boyle’s actual data showing the
constancy of the PV product
The effect can be seen in a simple J-shaped tube in which air is trapped in the short limb as
mercury is poured into the right side. The difference between the heights of the two mercury
columns gives the pressure (76 cm = 1 atm), and the volume of the air is calculated from the
length of the air column and the tubing diameter. In Boyle's experiment he used a simple air
pump invented by his friend Robert Hooke.
Boyle's law can be expressed as
PV = constant
(1)
or, equivalently,
Pressure-volume relations: Boyle's law
Page 8 of 41
Chem1 Properties of Gases
2 • The basic gas laws
P 1V 1 = P 2V 2
(2)
These relations hold true only if the number of molecules n and the temperature are constant. This is a relation of inverse proportionality; any change in the pressure is exactly compensated by an opposing change in the volume. As the pressure decreases toward zero, the
volume will increase without limit. Conversely, as the pressure is increased, the volume
decreases, but can never reach zero. There will be a separate P-V plot for each temperature;
a single P-V plot is therefore called an isotherm.
Shown here are some isotherms for one mole of an
ideal gas at several different temperatures. Each
plot has the shape of a hyperbola— the locus of all
points having the property x y = a, where a is a constant. You will see later how the value of this constant (PV=25 for the 300K isotherm shown here) is
determined.
It is very important that you understand this kind
of plot which governs any relationship of inverse
proportionality. You should be able to sketch out
such a plot when given the value of any one (x,y)
pair.
A related type of plot with which you should
be familiar shows the product PV as a function of the pressure. You should understand
why this yields a straight line, and how this
set of plots relates to the one immediately
above.
Pressure-volume relations: Boyle's law
Page 9 of 41
Chem1 Properties of Gases
2 • The basic gas laws
Problem Example 1
In an industrial process, a gas confined to a volume of 1 L at a pressure of 20 atm is allowed
to flow into a 12-L container by opening the valve that connects the two containers. What
will be the final pressure of the gas?
Solution: The final volume of the gas is (1 + 12)L = 13 L. The gas expands in inverse
proportion to the two volumes:
P2 = (20 atm) × (1 L / 13 L) = 1.5 atm
Note that there is no need to make explicit use of any “formula”
in problems of this kind!
Pressure-volume relations: Boyle's law
Page 10 of 41
Chem1 Properties of Gases
2 • The basic gas laws
2.2 How the temperature affects the volume: Charles’ law
All matter expands when heated, but gases are special in that their degree of expansion is
independent of their composition. The French scientists JACQUES CHARLES (1746-1823) and
JOSEPH GAY-LUSSAC (1778-1850) independently found that if the pressure is held constant,
the volume of any gas changes by the same fractional amount (1/273 of its value) for each C°
change in temperature.
The volume of a gas confined against a constant pressure is directly
proportional to the absolute temperature.
A graphical expression of the law of Charles and Gay-Lussac can be seen in these plots of the
volume of one mole of an ideal gas as a function of its temperature at various constant pressures.
• What do these plots show? The straight-line plots show that the ratio V/T (and thus dV/dT) is a
constant at any given pressure. Thus we can express the law algebraically as
V/T = constant or V1/T1 = V2/T2. (Don’t memorize this formula!)
• What is the significance of the extrapolation to zero volume? If a gas contracts by 1/273 of its volume
for each degree of cooling, it should contract to zero volume at a temperature of –273°C. This, of
course, is the absolute zero of temperature, and this extrapolation of Charles' law is the first evidence of the special significance of this temperature.
• Why do the plots for different pressures have different slopes? The lower the pressure, the greater
the volume (Boyle's law), so at low pressures the fraction (V/273) will have a larger value. You
might say that the gas must "contract faster" to reach zero volume when its starting volume is
larger.
Problem Example 2
The air pressure in a car tire is 30 psi (pounds per square inch) at 10°C. What will the pressure be after driving has raised its temperature to 45°C ? (Assume that the volume remains
unchanged.)
Solution: The gas expands in direct proportion to the ratio of the absolute temperatures:
P2 = (30 psi) × (318K / 283K) = 33.7 psi
How the temperature affects the volume: Charles’ law
Page 11 of 41
Chem1 Properties of Gases
2 • The basic gas laws
2.3 Volume and the number of molecules: Avogadro’s law
Gay-Lussac noticed that when two gases react, they do so in volume ratios that can always
be expressed as small whole numbers. Thus when hydrogen burns in oxygen, the volume of
hydrogen consumed is always exactly twice the volume of oxygen. The Italian scientist
AMADEO AVOGADRO (1776-1856) drew the crucial conclusion: these volume ratios must be
related to the relative numbers of molecules that react, and thus the famous "E.V.E.N principle":
Equal volumes of gases measured at the same temperature and pressure
contain equal numbers of molecules
Avogadro's law thus predicts a directly proportional relation between the number of moles of
a gas and its volume.
This relationship, originally known as Avogadro's Hypothesis, was crucial in establishing the formulas of simple molecules at a time (around 1811) when the distinction between atoms and molecules was not clearly understood. In particular, the existence of diatomic molecules of elements
such as H2, O2, and Cl2 was not recognized until the results of combining-volume experiments
such as those depicted below could be interpreted in terms of the E.V.E.N. principle.
Once it was shown that equal volumes of hydrogen and oxygen do not combine in the manner depicted in (1), it became clear that these elements exist as diatomic molecules and that
the formula of water must be H2O rather than HO as previously thought.
Volume and the number of molecules: Avogadro’s law
Page 12 of 41
Chem1 Properties of Gases
2 • The basic gas laws
2.4 The ideal gas equation of state
If the variables P, V, T and n (the number of moles) have known values, then a gas is said to
be in a definite state, meaning that all other physical properties of the gas are also defined.
The relation between these state variables is known as an equation of state. By combining
the expressions of Boyle's, Charles', and Avogadro's laws (you should be able to do this!) we
can write the very important ideal gas equation of state
in which the proportionality constant R is known as the gas constant. This is one of the few
equations you must commit to memory in this course; you should also know the common
value and units of R.
An ideal gas is defined as a hypothetical substance that obeys the ideal gas
equation of state.
We will see later that all real gases behave more and more like an ideal gas as the pressure
approaches zero. A pressure of only 1 atm is sufficiently close to zero to make this relation
useful for most gases at this pressure.
In order to depict the relations between the three variables P, V and T we need a threedimensional graph.
Fig. 1: PVT surface for an ideal gas
Each point on the curved surface represents a
possible combination of (P,V,T) for an arbitrary
quantity of an ideal gas. The three sets of lines
inscribed on the surface correspond to states in
which one of these three variables is held constant.
The red curved lines, being lines of constant
temperature, are isotherms, and are plots of
Boyle's law. These isotherms are also seen projected onto the x-y plane at the top right.The yellow lines are isobars and represent Charles' law
plots; they are projected onto the x-z plane at the
bottom.
The green lines, known as isochors, show all values of (P,V,T) with various fixed volumes.
The ideal gas equation of state
Page 13 of 41
Chem1 Properties of Gases
3 • Molar volume and density
Problem Example 3
A biscuit made with baking powder has a volume of 20 mL, of which one-fourth consists of
empty space created by gas bubbles produced when the baking powder decomposed to CO2.
What weight of NaHCO3 was present in the baking powder in the biscuit? Assume that the
gas reached its final volume during the baking process when the temperature was 400°C.
(Baking powder consists of sodium bicarbonate mixed with some other solid that produces an
acidic solution on addition of water, initiating the reaction NaHCO3(s) + H+ → Na+ + H2O +
CO2
Solution: Use the ideal gas equation to find the number of moles of CO2 gas; this will be the
same as the number of moles of NaHCO3 (84 g mol–1) consumed :
9.1E–6 mol × 84 g mol–1 = 0.0076 g
3 • Molar volume and density
Although all gases closely follow the ideal gas law PV = nRT under appropriate conditions,
each gas is also a unique chemical substance consisting of molecular units that have definite
masses. In this lesson we will see how these molecular masses affect the properties of gases
that conform to the ideal gas law.
Following this, we will look at gases that contain more than one kind of molecule— in other
words, mixtures of gases.
3.1 Molar volume of a gas: standard temperature and pressure
You will recall that the molar mass of a pure substance is the mass of 6.02×1023 (Avogadro's
number) of particles or molecular units of that substance. Molar masses are commonly
expressed in units of grams per mole (g mol–1) and are often referred to as molecular
weights.
As was explained in the preceding lesson, equal volumes of gases, measured at the same
temperature and pressure, contain equal numbers of molecules (this is the "EVEN" principle, more formally known as Avogadro's law.)
Molar volume of a gas: standard temperature and pressure
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Chem1 Properties of Gases
3 • Molar volume and density
This means that one mole of any gas will occupy the same volume as one mole of any other
gas at a given temperature and pressure. The magnitude of this volume will of course
depend on the temperature and pressure, so as a means of convenient comparison it is customary to define a set of conditions T†=†273K and P†=†1†atm as standard temperature
and pressure, usually denoted STP. Substituting these values into the ideal gas equation of
state and solving for V yields a volume of 22.414†litres for 1 mole
The standard molar volume 22.4†L†mol–1 is a value worth memorizing, but remember
that it is valid only at STP. The molar volume at other temperatures and pressures can easily be found by simple proportion.
Problem Example 4
What would the volume of one mole of air be at 20°C on top of Muana Kea, Haw'aii
(altitude 4.2 km) where the air pressure is approximately 600 kPa?
Solution: Apply Boyle's and Charles' laws as successive correction factors to the
standard sea-level pressure of 1013†kPa:
The molar volume of a substance can tell us something about how much space each molecule
occupies, as the following example shows.
Problem Example 5
Estimate the average distance between the molecules in a gas at 1†atm pressure
and 0°C.
Solution. Consider a 1-cm3 volume of the gas, which will contain
(6.02E23 mol–1)/(22400 cm3 mol–1) = 2.69E19 cm–3.
The volume per molecule (not the same as the volume of a molecule, which for an
ideal gas is zero!) is just the reciprocal of this, or 3.72E–20 cm3. Assume that the
molecules are evenly distributed so that each occupies an imaginary box having
this volume. The average distance between the centers of the molecules will be
defined by the length of this box, which is the cube root of the volume per molecule:
(3.72 × 10–20)1/3 = 3.38 × 10–7 cm = 3.4 nm
Molar volume of a gas: standard temperature and pressure
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Chem1 Properties of Gases
3 • Molar volume and density
3.2 Molecular weight and density of a gas
The molar volumes of all gases are the same when measured at the same temperature and pressure. But the molar
masses of different gases will vary. This means that different gases will have different densities (different masses per
unit volume). If we know the molecular weight of a gas, we can calculate its density.
Problem Example 6
Uranium hexafluoride UF6 gas is used in the isotopic enrichment of natural uranium.
Calculate its density at STP.
Solution: The molecular weight of UF6 is 352.
(353 g molñ1) /(22.4 mol L–1) = 15.7 g L–1
Note: there is no need to look up a "formula" for this calculation; simply combine the molar
mass and molar volume in such a way as to make the units come out correctly.
More importantly, if we can measure the density of an unknown gas, we have a convenient
means of estimating its molecular weight. This is one of many important examples of how a
macroscopic measurement (one made on bulk matter) can yield microscopic information
(that is, about molecular-scale objects.)
Determination of the molecular weight of a gas from its density is known as the Dumas
method, after the French chemist JEAN DUMAS (1800-1840) who developed it. One simply
measures the weight of a known volume of gas and converts this volume to its STP equivalent, using Boyle's and Charles' laws. The weight of the gas divided by its STP volume yields
the density of the gas, and the density multiplied by 22.4†mol–1 gives the molecular weight.
Pay careful attention to the examples of gas density calculations shown in your textbook.
You will be expected to carry out calculations of this kind, converting between molecular
weight and gas density.
Gas densities are now measured in industry by electro-mechanical devices such as vibrating
reeds which can provide continuous, on-line records at specific locations, as within pipelines.
Problem Example 7
Calculate the approximate molar mass of a gas whose measured density is 3.33 g/L at 30∞C
and 780 torr.
Solution. From the ideal gas equation, the number of moles contained in one litre of the gas
is
The molecular weight is therefore (33 g L–1)/(.0413 mol L–1) = 80.6 g mol–1
Gas density measurements can be a useful means of estimating the composition of a mixture
of two different gases:
Molecular weight and density of a gas
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Chem1 Properties of Gases
4 • Mixtures of gases
4 • Mixtures of gases
Because most of the volume occupied by a gas consists of empty space, there is nothing to
prevent two or more kinds of gases from occupying the same volume. Homogeneous mixtures of this kind are generally known as solutions, but it is customary to refer to them simply as gaseous mixtures.
4.1 Mole fractions and volumes
We can specify the composition of gaseous mixtures in many different ways, but the most
common ones are by volumes and by mole fractions. From Avogadro's Law we know that
"equal volumes contains equal numbers of molecules", so when we say that air, for example,
is 21 percent oxygen and 78 percent nitrogen by volume, this is the same as saying that
these same percentages of the molecules in air consist of O2 and N2. Similarly, in 1.0 mole of
air, there is 0.21 mol of O2 and 0.78 mol of N2 (the other 0.1 mole consists of various of trace
gases, but is mostly neon.)
Note that you could never assume a similar equivalance with mixtures of liquids or solids, to which
the E.V.E.N. principle does not apply.
These last two numbers (.21 and .78) also express the mole fractions of oxygen and nitrogen
in air. Mole fraction means exactly what it says: the fraction of the molecules that consist of
a specific substance. This is expressed algebraically by
so in the case of oxygen, its mole fraction is
Don't let this type of notation put you off! The summation sign Σni (Greek Sigma)
simply means to add up the n's (number of moles) of every substance present. Thus
if O2 is the "i-th" substance as in the expression immediately above, the summation
runs from i=1 through i=3.
Mole fractions and volumes
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Chem1 Properties of Gases
4 • Mixtures of gases
Problem Example 8
A mixture of O2 and nitrous oxide, N2O, is sometimes used as a mild anesthetic in dental
surgery. A certain mixture of these gases has a density of 1.482 g L–1 at 25 and 0.980†atm.
What was the mole-percent of N2O in this mixture?
Solution: First, find the density the gas would have at STP:
The molar mass of the mixture is (5 g L–)(22.4 L mol–1) = 37.0 g mol–1. The molecular
weights of O2 and N2 are 32 and 44, respectively. Thus the fraction of the heavier gas in the
mixture is
4.2 Dalton's law of partial pressures
The ideal gas equation of state applies to mixtures just as to pure gases. It was in fact with a
gas mixture, ordinary air, that Boyle, Gay-Lussac and Charles did their early experiments.
The only new concept we need in order to deal with gas mixtures is the partial pressure, a
concept invented by the famous English chemist JOHN DALTON (1766-1844). Dalton reasoned that the low density and high compressibility of gases indicates that they consist
mostly of empty space; from this it follows that when two or more different gases occupy the
same volume, they behave entirely independently.
"Every gas is a vacuum to every other gas."
John Dalton
The contribution that each component of a gaseous mixture makes to the total pressure of
the gas is known as the partial pressure of that gas. Dalton himself stated this law in the
simple and vivid way quoted above. The usual way of stating Dalton's Law of Partial Pressures is
The total pressure of a gas is the sum of the partial pressures
of its components
This is expressed algebraically as
or, equivalently,
Dalton's law of partial pressures
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Chem1 Properties of Gases
4 • Mixtures of gases
(If you feel a need to memorize these formulas, you probably don't really understand Dalton's
Law!)
Problem Example 9
What is the mole fraction of carbon dioxide in a mixture consisting of equal masses of CO2
(MW=44) and neon (MW=20.2)?
Solution: Assume any arbitrary mass, such as 100 g, find the equivalent numbers of moles
of each gas, and then substitute into the definition of mole fraction:
nCO2 = (100 g) / (44 g mol–1) = 2.3 mol
nNe= (100 g) / (20.2 g mol–1) = 4.9 mol
XNe = (2.3 mol) / (2.3 mol + 4.9 mol) = 0.43
Problem Example 10
Calculate the mass of each component present in a mixture of fluorine (MW 19.0) and xenon
(MW†131.3) contained in a 2.0-L flask. The partial pressure of Xe is 350†torr and the total
pressure is 724†torr at 25°C.
Solution: From Dalton's law, the partial pressure of F2 is (724 – 350)†=†374 torr:
The mole fractions are XXe = 350/724 = .48 and XF2 = 374/724 = .52 . The total number of
moles of gas is
The mass of Xe is (131.3 g mol–1) ◊ (.48 ◊× .078 mol) = 4.9 g
Dalton's law of partial pressures
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Chem1 Properties of Gases
4 • Mixtures of gases
Molar mass of a gas mixture
A gas made up of more than one kind of molecule will have a molar mass that is a weighted
average of the molar masses of its components. (By the way, the older term "molecular
weight" implies a certain mass possessed by individual molecules, so it is commonly modified by the prefix ìaverageî in the context of a gas mixture.) The molar mass of a mixture of
gases is just the sum of the mole fractions of each gas, multiplied by the molar mass of that
substance.
Problem Example 11
Find the average molar mass of dry air whose volume-composition is O2 (21%), N2 (78%)
and Ar (1%).
Solution: The average molecular weight is the mole-fraction-weighted sum of the molecular
weights of its components. The mole fractions, of course, are the same as the volume-fractions (E.V.E.N. principle.)
m = (.21 x 32) + (.78 x 28) + (.01 x 20) = 28
4.3 Some applications of Dalton’s Law
Collecting gases over water
A common laboratory method of collecting the gaseous product
of a chemical reaction is to conduct it into an inverted tube or
bottle filled with water, the opening of which is immersed in a
larger container of water. This arrangement is called a pneumatic trough, and was widely used in the early days of chemistry. As the gas enters the bottle it displaces the water and
becomes trapped in the upper part.
The volume of the gas can be observed by means of a calibrated
scale on the bottle, but what about its pressure? The total pressure confining the gas is just that of the atmosphere transmitting its force through the water. But liquid water itself is always in equilibrium with its
vapor, so the space in the top of the tube is a mixture of two gases: the gas being collected,
and gaseous H2O. The partial pressure of H2O is known as the vapor pressure of water and
it depends on the temperature. In order to determine the quantity of gas we have collected,
we must use Dalton's Law to find the partial pressure of that gas.
Some applications of Dalton’s Law
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Chem1 Properties of Gases
5 • Molecules in motion: introduction to kinetic molec-
Problem Example 12
Oxygen gas was collected over water as shown above. The atmospheric pressure was
754†torr, the temperature was 22°C, and the volume of the gas was 155†mL. The vapor
pressure of water at 22°C is 19.8†torr. Use this information to estimate the number of moles
of O2 produced.
Solution. From Dalton's law, PO2= Ptotal – PH2O = 754 — 19.8 = 734 torr.
Scuba Diving
Our respiratory systems are designed to maintain the proper oxygen concentration in
the blood when the partial pressure of O2 is 0.21 atm, its normal sea-level value.
Below the water surface, the pressure increases by 1 atm for each 10.3-m increase in
depth; thus a scuba diver at 10.3 m experiences a total of 2 atm pressure pressing on
the body. In order to prevent the lungs from collapsing, the air the diver breathes
should also be at about the same pressure.
But at a total pressure of 2 atm, the partial pressure of O2 in ordinary air would be
0.42 atm; at a depth of 100†ft (about 30 m), the O2 pressure of .8†atm would be far
too high for health. For this reason, the air mixture in the pressurized tanks that
scuba divers wear must contain a smaller fraction of O2. This can be achieved most
simply by raising the nitrogen content, but high partial pressures of N2 can also be
dangerous owing to its greater tendency to dissolve in the blood, which can create a
condition known as nitrogen narcosis. The preferred diluting agent for sustained
deep diving is helium, which has very little tendency to dissolve in the blood even at high pressures.
5 • Molecules in motion: introduction to kinetic
molecular theory
The properties such as temperature, pressure, and volume, together with other properties
related to them (density, thermal conductivity, etc.) are known as macroscopic properties of
matter; these are properties that can be observed in bulk matter, without reference to its
underlying structure or molecular nature.
By the late 19th century the atomic theory of matter was sufficiently well accepted that scientists began to relate these macroscopic properties to the behavior of the individual molecules, which are described by the microscopic properties of matter. The outcome of this effort
was the kinetic molecular theory of gases. This theory applies strictly only to a hypothetical
substance known as an ideal gas; we will see, however, that under many conditions it
describes the behavior of real gases at ordinary temperatures and pressures quite accurately, and serves as the starting point for dealing with more complicated states of matter.
Some applications of Dalton’s Law
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Chem1 Properties of Gases
5 • Molecules in motion: introduction to kinetic molec-
5.1 The kinetic-molecular model
The basic tenets of the kinetic-molecular theory are as follows:
1.
2.
3.
4.
5.
A gas is composed of molecules that are separated by average distances that are much greater than the sizes of the
molecules themselves. The volume occupied by the molecules of the gas is negligible compared to the volume
of the gas itself.
The molecules of an ideal gas exert no attractive forces on each other, or on the walls of the container.
The molecules are in constant random motion, and as material bodies, they obey Newton's laws of motion. This
means that the molecules move in straight lines until they collide with each other or with the walls of the container. (image source)
Collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but
the total kinetic energy is conserved. Collisions are not “sticky”.
The average kinetic energy of the gas molecules is directly proportional to the absolute temperature.
(Notice that the term ìaverageî is very important here; the velocities and kinetic energies of individual molecules
will span a wide range of values, and some will even have zero velocity at a given instant.) This implies that all
molecular motion would cease if the temperature were reduced to absolute zero.
According to this model, most of the volume occupied by a gas is empty space; this is the
main feature that distinguishes gases from condensed states of matter (liquids and solids) in
which neighboring molecules are constantly in contact. Gas molecules are in rapid and continuous motion; at ordinary temperatures and pressures their velocities are of the order of
0.1-1†km/sec and each molecule experiences approximately 1010 collisions with other molecules every second.
5.2 Kinetic molecular interpretation of gas properties
If gases do in fact consist of widely-separated particles, then the observable properties of
gases must be explainable in terms of the simple mechanics that govern the motions of the
individual molecules.
Kinetic interpretation of gas pressure The kinetic molecular theory
makes it easy to see why a gas should exert a pressure on the walls of
a container. Any surface in contact with the gas is constantly bombarded by the molecules. At each collision, a molecule moving with
momentum mv strikes the surface. Since the collisions are elastic, the
molecule bounces back with the same velocity in the opposite direction. This change in velocity ∆V is equivalent to an acceleration a;
According to Newton's second law, a force f†=†ma is thus exerted on
the surface of area A exerting a pressure P†=†f/A.
Kinetic interpretation of temperature According to the kinetic molecular theory, the average
kinetic energy of an ideal gas is directly proportional to the absolue temperature. Kinetic
energy is the energy a body has by virtue of its motion:
As the temperature of a gas rises, the average velocity of the molecules will increase; a doubling of the temperature will increase this velocity by a factor of four. Collisions with the
walls of the container will transfer more momentum, and thus more kinetic energy, to the
walls. If the walls are cooler than the gas, they will get warmer, returning less kinetic
energy to the gas, and causing it to cool until thermal equilibrium is reached. Because temThe kinetic-molecular model
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Chem1 Properties of Gases
5 • Molecules in motion: introduction to kinetic molec-
perature depends on the average kinetic energy, the concept of temperature only applies to a
statistically meaningful sample of molecules. We will have more to say about molecular
velocities and kinetic energies farther on.
Kinetic explanation of Boyle's law Boyle's law is easily explained by the kinetic molecular
theory. The pressure of a gas depends on the number of times per second that the molecules
strike the surface of the container. If we compress the gas to a smaller volume, the same
number of molecules are now acting against a smaller surface area, so the number striking
per unit of area, and thus the pressure, is now greater.
Kinetic explanation of Charles' law Kinetic molecular theory states that an increase in temperature raises the average kinetic energy of the molecules. If the molecules are moving
more rapidly but the pressure remains the same, then the molecules must stay farther
apart, so that the increase in the rate at which molecules collide with the surface of the container is compensated for by a corresponding increase in the area of this surface as the gas
expands.
Kinetic explanation of Avogadro's law If we increase the number of gas molecules in a
closed container, more of them will collide with the walls per unit time. If the pressure is to
remain constant, the volume must increase in proportion, so that the molecules strike the
walls less frequently, and over a larger surface area.
5.3 Diffusion: random motion with direction
Diffusion refers to the transport of matter through a concentration gradient; the rule is
that substances move (or tend to move) from regions of higher concentration to those of
lower concentration. The diffusion of tea out of a teabag into water, or of perfume from a person, are common examples; we would not expect to see either process happening in reverse!
When the stopcock is opened, random motions
cause each gas to diffuse into the other container.
After diffusion is complete (right), individual molecules of both kinds continue to pass between the
flasks in both directions.
It might at first seem strange that the random motions of molecules can lead to a completely
predictable drift in their ultimate distribution. The key to this apparent paradox is the distinction between an individual and the population. Although we can say nothing about the
fate of an individual molecule, the behavior of a large collection ("population") of molecules is
subject to the laws of statistics. We won't go into the details here, but one simple example
should suffice: consider what happens when the stopcock is opened that separates two containers of different gases under identical condtions.
5.4 Density fluctuations: Why is the sky blue?
Diffusion ensures that molecules will quickly distribute themselves throughout the volume
occupied by the gas in a thoroughly uniform manner. The chances are virtually zero that sufficiently more molecules might momentarily find themselves near one side of a container
than the other to result in an observable temporary density or pressure difference. This is a
result of simple statistics. But statistical predictions are only valid when the sample population is large.
Diffusion: random motion with direction
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Chem1 Properties of Gases
5 • Molecules in motion: introduction to kinetic molec-
Consider what would happen if we consider extremely small volumes of space: cubes that
are about 10–7†cm on each side, for example. Such a cell would contain only a few molecules,
and at any one instant we would expect to find some containing more or less than others,
although in time they would average out to the same value. The effect of this statistical
behavior is to give rise to random fluctuations in the density of a gas over distances comparable to the dimensions of visible light waves. When light passes through a medium whose
density is non-uniform, some of the light is scattered. The kind of scattering due to random
density fluctuations is called Rayleigh scattering, and it has the property of affecting (scattering) shorter wavelengths more effectively than longer wavelengths. The clear sky appears
blue in color because the blue (shorter wavelength) component of sunlight is scattered more.
The longer wavelengths remain in the path of the sunlight, available to delight us at sunrise
or sunset.
From http://www.uccs.edu/~tchriste/
What we have been discussing is a form of what is known as fluctuation phenomena. If a
massless partition were inserted into the middle of a container of a gas, the random fluctuations in pressure of a gas on either side would not always completely cancel out, especially
when the pressures are quite small. These pressure fluctuations would cause the partition to
vibrate.
5.5 Light bulbs
An interesting application involving several aspects of the kinetic molecular behavior of
gases is the use of a gas, usually argon, to extend the lifetime of incandescent lamp bulbs. As
a light bulb is used, tungsten atoms evaporate from the filament and condense on the cooler
inner wall of the bulb, blackening it and reducing light output. As the filament gets thinner
in certain spots, the increased electrical resistance results in a higher local power dissipation, more rapid evaporation, and eventually the filament breaks.
The pressure inside a lamp bulb must be sufficiently low for the mean free path of the gas
molecules to be fairly long; otherwise heat would be conducted from the filament too rapidly,
Light bulbs
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Chem1 Properties of Gases
5 • Molecules in motion: introduction to kinetic molec-
and the bulb would melt. (Thermal conduction depends on intermolecular collisions, and a
longer mean free path means a lower collision frequency). A complete vacuum would minimize heat conduction, but this would result in such a long mean free path that the tungsten
atoms would rapidly migrate to the walls, resulting in a very short filament life and extensive bulb blackening.
Around 1910, the General Electric Company hired IRVING LANGMUIR as one of the first
chemists to be employed as an industrial scientist in North America. Langmuir quickly saw
that bulb blackening was a consequence of the long mean free path of vaporized tungsten
molecules, and showed that the addition of a small amount of argon will reduce the mean
free path, increasing the probability that an outward-moving tungsten atom will collide with
an argon atom. A certain proportion of these will eventually find their way back to the filament, partially reconstituting it. Krypton would be a better choice of gas than argon, since
its greater mass would be more effective in changing the direction of the rather heavy tungsten atom. Unfortunately, krypton, being a rarer gas, is around 50 times as expensive as
argon, so it is used only in ìpremiumî light bulbs. The more recently-developed halogen-cycle
lamp is an interesting chemistry-based method of prolonging the life of a tungsten-filament
lamp.
5.6 Viscosity of gases
Gases, like all fluids, exhibit a resistance to flow, a property known as viscosity. The basic
cause of viscosity is the random nature of thermally-induced molecular motion. In order to
force a fluid through a pipe or tube, an additional non-random translational motion must be
superimposed on the thermal motion.
There is a slight problem, however. Molecules flowing near the center of the pipe collide
mostly with molecules moving in the same direction at about the same velocity, but those
that happen to find themselves near the wall will experience frequent collisions with the
wall. Since the molecules in the wall of the pipe are not moving in the direction of the flow,
they will tend to absorb more kinetic energy than they return, with the result that the gas
molecules closest to the wall of the pipe lose some of their forward momentum. Their random
thermal motion will eventually take them deeper into the stream, where they will collide
with other flowing molecules and slow them down. This gives rise to a resistance to flow
known as viscosity; this is the reason why long gas transmission pipelines need to have
pumping stations every 100 km or so.
Viscosity of gases
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Chem1 Properties of Gases
5 • Molecules in motion: introduction to kinetic molec-
Origin of gas viscosity in a pipeline. The magnified view on the right shows the boundary region
where random movements of molecules in directions other than the one of the flow (1) move toward
the confining surface and temporarily adsorb to it (2). After a short time, thermal energy causes the
molecule to be released (4) with most of its velocity not in the flow direction. A rapidly-flowing molecule
(3) collides with it (5) and loses some of its flow velocity.
As you know, liquids such as syrup or honey exhibit smaller viscosities at higher temperatures as the increased thermal energy reduces the influence of intermolecular attractions,
thus allowing the molecules to slip around each other more easily. Gases, however, behave in
just the opposite way; gas viscosity arises from collisiion-induced transfer of momentum
from rapidly-moving molecules to slow ones that have been released from the boundary
layer. The higher the temperature, the more rapidly the molecules move and collide with
each other, so the higher the viscosity.
5.7 Distribution of gas molecules in a gravitational field
Everyone knows that the air pressure decreases with altitude. This effect is easily understood qualitatively through the kinetic molecular theory. Random thermal motion tends to
move gas molecules in all directions equally. In the presence of a gravitational field, however, motions in a downward direction are slightly favored. This causes the concentration,
and thus the pressure of a gas to be greater at lower elevations and to decrease without limit
at higher elevations.
Distribution of gas molecules in a gravitational field
Page 26 of 41
Chem1 Properties of Gases
The pressure at any elevation in a vertical
column of a fluid is due to the weight of the
fluid above it. This build-up in the pressure
follows an exponential law.
5 • Molecules in motion: introduction to kinetic molec-
Notice how much more slowly the pressure falls off
at higher altitudes; this is because 90 percent of the
molecules in the air are below 10 km.
Graphic: http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/
pw/prs/hght.rxml
Decrease of pressure with
altitude for air at 25°C.
Note the constant increment of
altitude (6.04 km, the "half
height") required to reduce the
pressure by half its value. This
reflects the special property of
an exponential function a†=†ey,
namely that the derivitave da/dy
is just ey itself.
The exact functional relationship between pressure and altitude is known as the barometric distribution law. It is easily derived using first-year calculus. For air at 25°C the pressure Ph at any altitude is given by Ph = Po e–.11h in which Po is the pressure at sea level. This
is a form of the very common exponential decay law which we will encounter in several different contexts in this course. An exponential decay (or growth) law describes any quantity
whose rate of change is directly proportional to its current value, such as the amount of
money in a compound-interest savings account or the density of a column of gas at any altitude. The most important feature of any quantity described by this law is that the fractional
rate of change of the quantity in question (in this case, ∆P/P or in calculus, dP/P, is a constant.) This means that the increase in altitude required to reduce the pressure by half is
also a constant, about 6†km in the Earth’s atmosphere.
Because heavier molecules will be more strongly affected by gravity, their concentrations will
fall off more rapidly with elevation. For this reason the partial pressures of the various components of the atmosphere will tend to vary with altitude. The difference in pressure is also
affected by the temperature; at higher temperatures there is more thermal motion, and hence
Distribution of gas molecules in a gravitational field
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Chem1 Properties of Gases
6 • More on the kinetic-molecular model
a less rapid fall-off of pressure with altitude. Owing to atmospheric convection and turbulence, these effects are not observed in the lower part of the atmosphere, but in the uppermost
parts of the atmosphere the heavier molecules do tend to drift downward.
5.8 The ionosphere and radio communication
At very low pressures, mean free paths are sufficiently great that collisions between molecules become rather infrequent. Under these conditions, highly reactive species such as ions,
atoms, and molecular fragments that would ordinarily be destroyed on every collision can
persist for appreciable periods of time. The most important example of this occurs at the top
of the Earth's atmosphere, at an altitude of 200†km, where the pressure is about 10–7 atm.
Here the mean free path will be 107 times its value at 1†atm, or about 1†m. In this part of
the atmosphere, known as the thermosphere, the chemistry is dominated by species such as
O, O2+ and HO which are formed by the action of intense solar ultraviolet light on the normal atmospheric gases near the top of the stratosphere. The high concentrations of electrically charged species in these regions (sometimes also called the ionosphere) reflect radio
waves and are responsible for around-the-world transmission of mid-frequency radio signals.
http://apollo.lsc.vsc.edu/classes/met130/notes/chapter1/ion2.html
The ion density in the lower part of the ionosphere (about 80†km altitude) is so high that the
radiation from broadcast-band radio stations is absorbed in this region before these waves
can reach the reflective high-altitude layers. However, the pressure in this region (known as
the D-layer) is great enough that the ions recombine soon after local sunset, causing the Dlayer to disappear and allowing the waves to reflect off of the upper (F-layer) part of the ionosphere. This is the reason that distant broadcast stations can only be heard at night.
6 • More on the kinetic-molecular model
In this section, we look in more detail at some aspects of the kinetic-molecular model and
how it relates to our empirical knowledge of gases. For most students, this will be the first
application of algebra to the development of a chemical model; this should be educational in
itself, and may help bring that subject back to life. As before, your emphasis should on
understanding these models and the ideas behind them, there is no need to memorize any of
the formulas.
The ionosphere and radio communication
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Chem1 Properties of Gases
6 • More on the kinetic-molecular model
6.1 The velocities of gas molecules
At temperatures above absolute zero, all molecules are in motion. In the case of a gas, this
motion consists of straight-line jumps whose lengths are quite great compared to the dimensions of the molecule. Although we can never predict the velocity of a particular individual
molecule, the fact that we are usually dealing with a huge number of them allows us to know
what fraction of the molecules have kinetic energies (and hence velocities) that lie within
any given range.
Many molecules, many velocities
The trajectory of an individual gas molecule consists of a series of straight-line paths interrupted by collisions. What happens when two molecules collide depends on their relative
kinetic energies; in general, a faster or heavier molecule will impart some of its kinetic
energy to a slower or lighter one. Two molecules having identical masses and moving in
opposite directions at the same speed will momentarily remain motionless after their collision.
If we could measure the instantaneous velocities of all the molecules in a sample of a gas at
some fixed temperature, we would obtain a wide range of values. A few would be zero, and a
few would be very high velocities, but the majority would fall into a more or less well defined
range. We might be tempted to define an average velocity for a collection of molecules, but
here we would need to be careful: molecules moving in opposite directions have velocities of
opposite signs. Because the molecules are in a gas are in random thermal motion, there will
be just about as many molecules moving in one direction as in the opposite direction, so the
velocity vectors of opposite signs would all cancel and the average velocity would come out to
zero. Since this answer is not very useful, we need to do our averaging in a slightly different
way.
The proper treatment is to average the squares of the velocities, and then take the square
root of this value. The resulting quantity is known as the root mean square, or RMS velocity
which we will denote simply by v.
The formula relating the RMS velocity to the temperature and molar mass is surprisingly
simple, considering the great complexity of the events it represents:
in which m is the molar mass in kg mol–1, and k†=†R/6.02E23, the “gas constant per molecule”, is known as the Boltzmann constant.
The velocities of gas molecules
Page 29 of 41
Chem1 Properties of Gases
6 • More on the kinetic-molecular model
Problem Example 13
What is the average velocity of a nitrogen molecules at 300K?
Solution: The molar mass of N2 is 28.01 g. Substituting in the above equation and expressing R in energy units, we obtain
Recalling the definition of the joule (1 J†=†1†kg†m2†s–2) and taking the square root,
or
A simpler formula for estimating average molecular velocities is
in which V is in units of meters/sec, T is the absolute temperature and m the molar mass in
grams.
Note that the N2 molecule velocity calculated above is comparable
to that of a rifle bullet (typically 300-500 m s–1.)
6.2 The Boltzmann distribution
If we were to plot the number of molecules whose velocities fall
within a series of narrow ranges, we would obtain a slightly
asymmetric curve known as a velocity distribution. The peak of
this curve would correspond to the most probable velocity. This
velocity distribution curve is known as the Maxwell-Boltzmann
distribution. This distribution law was first worked out around
1850 by the great Scottish physicist, James Clerk Maxwell, who
is better known for disovering the laws of electromagnetic radiation. Later, LUDWIG BOLTZMANN (1844-1906) put the relation on
a sounder theoretical basis and simplified the mathematics
somewhat.
The Boltzmann distribution
Page 30 of 41
Chem1 Properties of Gases
6 • More on the kinetic-molecular model
The derivation of the Boltzmann curve is a bit too
complicated to go into here, but its physical basis is
easy to understand. Consider a large population of
molecules having some fixed amount of kinetic
energy. As long as the temperature remains constant, this total energy will remain unchanged, but
it can be distributed among the molecules in many
different ways, and this distribution will change
continually as the molecules collide with each other
and with the walls of the container.
It turns out, however, that kinetic energy is
acquired and handed around only in discrete
amounts which are known as quanta. Once the molecule has a given number of kinetic energy quanta,
these can be apportioned amongst the three directions of motion in many different ways, each resulting in a distinct total velocity state for the molecule.
The greater the number of quanta, (that is, the
greater the total kinetic energy of the molecule) the greater the number of possible velocity
states. If we assume that all velocity states are equally probable, then simple statistics predicts that higher velocities will be more favored because so many more of hese are available
(2).
Although the number of possible higher-energy states is greater, the lower-energy states are
more likely to be occupied (1). This is because only so much kinetic energy available to the
gas as a whole; every molecule that acquires kinetic energy in a collision leaves behind
another molecule having less. This tends to even out the kinetic energies in a collection of
molecules, and ensures that there are always some molecules whose instantaneous velocity
is near zero. The net effect of these two opposing tendencies, one favoring high kinetic energies and the other favoring low ones, is the peaked curve 3 seen above. Notice that because
of the assymetry of this curve, the mean (rms†average) velocity is not the same as the most
probable velocity, which is defined by the peak of the curve.
At higher temperatures (or with lighter molecules) the latter constraint becomes less important, and the mean velocity increases, but with a wider velocity distribution, the number of
molecules having any one velocity diminishes, so the curve tends to flatten out.
How the temperature affects the distribution of molecular velocities.
Notice how the left ends of the plots are
anchored at zero velocity (there will
always be a few molecules that happen
to be at rest.) As a consequence, the
curves flatten out as the higher temperatures make additional higher-velocity
states of motion more accessible. The
area under each plot is the same for a
constant number of molecules.
The Boltzmann distribution
Page 31 of 41
Chem1 Properties of Gases
6 • More on the kinetic-molecular model
How the molecular weight
affects the velocity
distribution
All molecules have the same
kinetic energy (mv2/2) at the
same temperature, so the fraction
of molecules with higher velocities
will increase as the molecular
weight decreases.
This plot was taken froma U of Florida
Chemistry page which has a good discussion of kinetic-molecular theory,
and includes some very nice graphics:
http://itl.chem.ufl.edu/2041_f97/lectures/lec_d.html
Boltzmann distribution and planetary atmospheres
The ability of a planet to retain an atmospheric gas depends on the average velocity (and
thus on the temperature and mass) of the gas molecules and on the planet's mass, which
determines its gravity and thus the escape velocity. In order to retain a gas for the age of the
solar system, the average velocity of the gas molecules should not exceed about one-sixth of
the escape velocity. The escape velocity from the Earth is 11.2 km/s, and 1/6 of this is about 2
km/s. Examination of the above plot reveals that hydrogen molecules can easily achieve this
velocity, and this is the reason that hydrogen, the most abundant element in the universe, is
almost absent from Earth's atmosphere.
Although hydrogen is not a significant atmospheric component, water vapor is. A very small
amount of this diffuses to the upper part of the atmosphere, where intense solar radiation breaks
down the H2O into H2. Escape of this hydrogen from the upper atmosphere amounts to about
2.5†×†1010†g/year.
6.3 Derivation of the ideal gas equation of state
The ideal gas equation of state came about by combining the empirically determined laws of
Boyle, Charles, and Avogadro, but one of the triumphs of the kinetic molecular theory was
the derivation of this equation from simple mechanics in the late nineteenth century. This is
a beautiful example of how the principles of elementary mechanics can be applied to a simple model to develop a useful description of the behavior of macroscopic matter, and it will be
worth your effort to follow and understand the derivation.
We begin by recalling that the pressure of a gas arises from the force exerted when molecules collide
with the walls of the container. This force can be found from Newton's law
(3)
in which v is the velocity component of the molecule in the direction perpendicular to the
wall.
To evaluate the derivative, which is the velocity change per unit time, consider a single molecule of a gas contained in a cubic box of length l. For simplicity, assume that the molecule is
moving along the x-axis which is perpendicular to a pair of walls, so that it is continually
bouncing back and forth between the same pair of walls. When the molecule of mass m
Derivation of the ideal gas equation of state
Page 32 of 41
Chem1 Properties of Gases
6 • More on the kinetic-molecular model
strikes the wall at velocity vx (and thus with a momentum mvx) it will rebound elastically
and end up moving in the opposite direction with momentum –mvx. The total change in
momentum per collision is thus 2mvx.
After the collision the molecule must travel a distance l to the opposite wall, and then back
across this same distance before colliding again with the wall in question. This determines
the time between successive collisions with a given wall; the number of collisions per second
will be vx/2l. The force exerted on the wall is the rate of change of the momentum, given by
the product of the momentum change per collision and the collision frequency:
(4)
Pressure is force per unit area, so the pressure exerted by the molecule on the wall of crosssection l2 becomes
(5)
in which V is the volume of the box.
We have calculated the pressure due to a single molecule moving at a constant velocity in a
direction perpendicular to a wall. If we now introduce more molecules, we must interpret v2
as an average value which we will denote by
.
Also, since the molecules are moving randomly in all directions, only one-third of their total
velocity will be directed along any one cartesian axis, so the total pressure exerted by N molecules becomes
(6)
in which V is the volume of the container.
Recalling that mv2/2 is the average translational kinetic energy we can rewrite the above as
(7)
We know that the average kinetic energy is directly proportional to the temperature, so that
for a single molecule,
(8)
Derivation of the ideal gas equation of state
Page 33 of 41
Chem1 Properties of Gases
6 • More on the kinetic-molecular model
in which the proportionality constant k is known as the Boltzmann constant. For one mole of
molecules we express this proportionality constant as (6.02E23)k = R in which is R is the
familiar gas constant.
Substituting into (8) yields the ideal gas equation
PV = RT
Since the product PV has the dimensions of energy, so does RT, and this quantity in fact represents the average translational kinetic energy per mole. The relationship between these
two energy units can be obtained by recalling that 1†atm is 1.013E5 N m–2, so that
The gas constant R is one of the most important fundamental constants relating to the macroscopic behavior of matter. It is commonly expressed in both pressure-volume and in energy
units:
R = 0.082057 L atm mol–1 K–1 = 8.314 J mol–1 K–1
Notice that the Boltzmann constant k, which appears in many expressions relating to the
statistical treatment of molecules, is just R/6.02E23, the "gas constant per molecule."
6.4 How far does a molecule travel between collisions?
Molecular velocities tend to be very high by our everyday standards (typically around
500†metres per sec), but even in gases, they bump into each other so frequently that their
paths are continually being deflected in a random manner, so that the net movement
(diffusion) of a molecule from one location to another occurs rather slowly.
How close can two molecules get?
Each molecule is surrounded by an imaginary
sphere (gray circle) whose radius σ is equal to
the sum of the radii of the colliding molecules.
This sphere defines the excluded volume,
within which the center of another molecule
cannot enter.
The average distance a molecule moves between such collisions is called the mean free
path. This distance, denoted by λ (lambda), depends on the number of molecules per unit
volume and on their size. To avoid collision, a molecule of diameter σ must trace out a path
corresponding to the axis of an imaginary cylinder whose cross-section is πσ2. Eventually it
will encounter another molecule (extreme right in the diagram below) that has intruded into
this cylinder and defines the terminus of its free motion.
How far does a molecule travel between collisions?
Page 34 of 41
Chem1 Properties of Gases
7 • Real gases
The volume of the cylinder is πσ2/λ. At each collision the molecule is diverted to a new path
and traces out a new exclusion cylinder. After colliding with all n molecules in one cubic centimetre of the gas it will have traced out a total exclusion volume of Ɍɖ2Ʌ. Solving for
É… and applying a correction factor Å„2 to take into account exchange of momentum
between the colliding molecules (the detailed argument for this is too complicated to go into
here), we obtain
(9)
Small molecules such as He, H2 and CH4 typically have diameters of around 30-50†nm. At
STP the value of n, the number of molecules per cubic metre, is
(10)
Substitution into (10) yields a value of around 10–7†m (100†nm) for the mean free path of
most molecules under these conditions. Although this may seem like a very small distance, it
typically amounts to 100 molecular diameters, and more importantly, about 30 times the
average distance between molecules. This explains why such gases conform very closely to
the ideal gas law at ordinary temperatures and pressures.
On the other hand, at each collision the molecule can be expected to change direction.
Because these changes are random, the net change in location a molecule experiences during
a period of one second is typically rather small. Thus in spite of the high molecular velocities, the speed of molecular diffusion in a gas is usually quite small.
7 • Real gases
The "ideal gas laws" as we know them do a remarkably good job of describing the behavior of
a huge number chemically diverse substances as they exist in the gaseous state under ordinary environmental conditions, roughly around 1†atm pressure and a temperature of
300†K. This is especially so when we consider that some of the basic tenets of the ideal gas
model have to be abandoned in order to explain such properties as
• the average distance between collisions (the molecules really do take up space)
• the viscosity of a gas flowing through a pipe (the molecules do get temporarily "stuck" on the pipe
surface, and are therefore affected by intermolecular attractive forces.)
Even so, many of the common laws such as Boyle's and Charles' continue to describe these
gases quite well even under conditions where these phenomena are evident.
How far does a molecule travel between collisions?
Page 35 of 41
Chem1 Properties of Gases
7 • Real gases
Under ordinary environmental condtions (moderate pressures and
above 0°C), the isotherms of substances we normally think of as
gases don't appear to differ very greatly from the hyperbolic form
(PV/RT) = constant.
... but over a wider range of
conditions, things begin to get more complicated.
Thus Isopentane, shown here, behaves in a reasonably ideal manner above 210 K, but below this temperature the isotherms become somewhat distorted,
and at 185K and below they cease to be continuous,
showing peculiar horizontal segments in which
reducing the volume does not change the pressure.
It turns out that real gases eventually begin to follow their own unique equations of state, and ultimately even cease to be gases. In this unit we will
see why this occurs, what the consequences are, and
how we might modify the ideal gas equation of state
to extend its usefulness over a greater range of temperatures and pressures.
7.1 Effects of intermolecular forces
According to Boyle's law, the product PV is a constant at any given temperature, so a plot of
PV as a function of the pressure of an ideal gas yields a horizontal straight line. This implies
that any increase in the pressure of the gas is exactly counteracted by a decrease in the volume as the molecules are crowded closer together. But we know that the molecules themselves are finite objects having volumes of their own, and this must place a lower limit on
the volume into which they can be squeezed. So we must reformulate the ideal gas equation
of state as a relation that is true only in the limiting case of zero pressure:
So what happens when a real gas is subjected to a very high pressure? The outcome varies
with both the molar mass of the gas and its temperature, but in general we can see the the
effects of both repulsive and attractive intermolecular forces:
Repulsive forces: As a gas is compressed, the individual molecules begin to get in each
other's way, giving rise to a very strong repulsive force acts to oppose any further volume
decrease. We would therefore expect the PV vs P line to curve upward at high pressures,
a nd this is in fact what is observed for all gases at sufficiently high pressures.
Attractive forces: At very close distances, all molecules repel each other as their electron clouds come into contact. At greater distances, however, brief statistical fluctuations in the distribution these electron clouds give rise to a universal attractive force
Effects of intermolecular forces
Page 36 of 41
Chem1 Properties of Gases
7 • Real gases
between all molecules. The more electrons in the molecule (and thus the greater the
molecular weight), the greater is this attractive force. As long as the energy of thermal
motion dominates this attractive force, the substance remains in the gaseous state, but
at sufficiently low temperatures the attractions dominate and the substance condenses
to a liquid or solid.
The universal attractive force described above is known as the dispersion, or London force.
There may also be additional (and usually stronger) attractive forces related to charge
imbalance in the molecule or to hydrogen bonding. These various attractive forces are often
referred to collectively as van†der†Waals forces.
A plot of PV/RT as a function of pressure is a very sensitive indicator of deviations from
ideal behavior, since such a plot is just a horizonal line for an ideal gas. The two illustrations
below show how these plots vary with the nature of the gas, and with temperature.
Effects of intermolecular attractions
Intermolecular attractions, which generally increase with molecular weight,
cause the PV product to decrease as
higher pressures bring the molecules
closer together and thus within the range
of these attractive forces; the effect is to
cause the volume to decrease more rapidly than it otherwise would. But as the
molecules begin to intrude on each others' territory, repulsive forces always
eventually win out.
Temperature reduces the effects of intermolecular
attractions
At higher temperatures, increased thermal motions overcome the effects of intermolecular attractions which normally dominate at lower pressures. So all gases behave
more ideally at higher temperatures.
For any gas, there is a special temperature (the Boyle
temperature) at which attractive and repulsive forces
exactly balance each other at zero pressure. As you can
see in this plot for methane, this balance does remain as
the pressure is incrased.
The effect of intermolecular attractions on a PV-vs.-P plot would be to hold the molecules
slightly closer together, so that the volume would decrease more rapidly than the pressure
increases. The resulting curve would dip downward as the pressure increases, and this dip
would be greater at lower temperatures and for heavier molecules. At higher pressures, however, the stronger repulsive forces would begin to dominate, and the curve will eventually
bend upward as before. The effects of intermolecular interactions are most evident at low
temperatures and high pressures; that is, at high densities.
Effects of intermolecular forces
Page 37 of 41
Chem1 Properties of Gases
7 • Real gases
7.2 Equations of state for real gases: the van der Waals equation
How might we modify the ideal gas equation of state to take into account the effects of intermolecular interactions? The first and most well known answer to this question was offered
by the Dutch scientist J.D.†VAN†DER†WAALS (1837-1923) in 1873.
van†der†Waals recognized that the molecules themselves take up space that subtracts from the volume of
the container, so that the “volume of the gas” V in the
ideal gas equation should be replaced by the term
(V – b) which is known as the excluded volume, typically of the order of 20-100†cm3†mol–1. The excluded
volume (indicated by the gray circle at the right) surrounding any molecule defines the closest possible
approach of any two molecules during collision.
The intermolecular attractive forces act to slightly diminish the frequency and intensity of
encounters between the molecules and the walls of the container; the effect is the same as if
the pressure of the gas were slightly higher than it actually is. This imaginary increase is
called the internal pressure, and we can write
Peffective = Pideal + Pinternal
Thus we should replace the P in the ideal gas equation by
Pideal = Peffective – Pinternal
Since the attractions are between pairs of molecules, the total attractive force is proportional
to the square of the number of molecules per volume of space, and thus for a fixed number of
molecules such as one mole, the force is inversely proportional to the square of the volume of
the gas; the smaller the volume, the closer are the molecules and the greater the attractions
between pairs (hence the square term) of molecules. The pressure that goes into the corrected ideal gas equation is
in which the constants a and b depend respectively on the magnitudes of the attractive and
repulsive forces in a particular gas. The complete van†der†Waals equation of state thus
becomes
Although you do not have to memorize this equation, you are expected to understand it and to
explain the significance of the terms it contains. You should also understand that the
van†der†Waals constants a and b must be determined empirically for every gas. This can be done
by plotting the P-V behavior of the gas and adjusting the values of a and b until the van†der†Waals
equation results in an identical plot. The constant a is related in a simple way to the molecular
radius; thus the determination of a constitutes an indirect measurment of an important microscopic
quantity.
Equations of state for real gases: the van der Waals equation
Page 38 of 41
Chem1 Properties of Gases
7 • Real gases
molar
mass/g
substance
a
b
(L2-atm mol–2)
(L mol–1)
hydrogen H2
2
0.244
0.0266
helium He
4
0.034
0.0237
methane CH4
16
2.25
0.0428
water H2O
18
5.46
0.0305
nitrogen N2
28
1.39
0.0391
carbon dioxide CO2
44
3.59
0.0427
carbon tetrachloride CCl4
110
20.4
0.1383
Table 1: van der Waals constants for some gases
The van†der†Waals equation is only one of many equations of state for real gases. More elaborate equations are required to describe the behavior of gases over wider pressure ranges.
These generally take account of higher-order nonlinear attractive forces, and require the use
of more empirical constants. Although we will make no use of them in this course, they are
widely employed in chemical engineering work in which the behavior of gases at high pressures must be accurately predicted.
7.3 Condensation and the critical point
The most striking feature of real gases is that they cease to remain gases as the temperature
is lowered and the pressure is increased. The plot below illustrates this behavior; as the volume is decreased, the lower-temperature isotherms suddenly change into straight lines.
Under these conditions, the pressure remains constant as the volume is reduced. This can
only mean that the gas is “disappearing” as we squeeze the system down to a smaller volume. In its place, we obtain a new state of matter, the liquid. In the shaded region, two
phases, liquid, and gas, are simultaneously present. Finally, at very small volume all the gas
has disappeared and only the liquid phase remains. At this point the isotherms bend
strongly upward, reflecting our common experience that a liquid is practically incompressible.
Condensation and the critical point
Page 39 of 41
Chem1 Properties of Gases
7 • Real gases
PVT surface of a real gas
To better understand this plot, look at the isotherm labeled 1. As the gas is compressed
from 1 to 2, the pressure rises in much the
same way as Boyle's law predicts. Compression beyond 2, however, does not cause any
rise in the pressure. What happens instead is
that some of the gas condenses to a liquid. At
3, the substance is entirely in its liquid state.
The very steep rise to 4 corresponds to our
ordinary experience that liquids have very
low compressibilities. The range of volumes
possible for the liquid (indicated by the width
of the green wedge-shaped cross section)
diminishes as the critical temperature is
approached.
Critical constants of some gases
gas
Pc, atm
Vc
cm3 mol–1
He
2.3
58
5.2
Ne
27
42
44
Ar
48
75
150
Xe
58
119
290
H2
13
65
33
O2
50
78
155
N2
34
90
126
CO2
73
94
302
H2O
73
94
302
NH3
111
72
405
CH4
46
99
191
Tc, K
Note especially
• The very low critical pressure and temperature of helium, reflecting the very small
intermolecular attractions of this atom.
• Tc of the noble gas elements increases with
atomic number
• Hydrogen gas cannot be liquified above 33
K; this poses a major difficulty in the use of
hydrogen as an automotive fuel; storage as a
high-pressure gas requires heavy steel containers which add greatly to its effective
weight-per-joule of energy storage.
• The properties of carbon dioxide (particularly its use as a supercritical fluid) are
described above.
• The high Tc of water is another manifestation of its "anomalous" properties relating to
hydrogen-bonding.
Supercritical fluids
The maximum temperature at which the two phases can coexist is called the critical temperature. The set of (P,V,T) corresponding to this condition is known as the critical point. Liquid
Condensation and the critical point
Page 40 of 41
Chem1 Properties of Gases
7 • Real gases
and gas can coexist only within the regions indicated by the wedge-shaped cross section on
the left and the shaded area in the diagram above. An important consequence of this is that
a liquid phase cannot exist at temperatures above the critical point.
The critical point of a gas is defined by its critical temperature, pressure and volume,
denoted by Tc, Pc, and Vc.
The critical temperature of carbon dioxide is 31°C, so you can tell whether the temperature is
higher or lower than this by shaking a CO2 fire extinguisher; on a warm day, you will not hear
any liquid sloshing around inside. The critical temperature of water is 374K, and that of
hydrogen is only 33K.
Critical behavior of carbon dioxide
At temperatures below 31°C (the critical temperature), CO2 acts somewhat like an ideal gas
at higher pressures (1). Below this temperature,
subjecting the gas to a higher pressure eventually causes condensation to begin. Thus at
21°C, at a pressure of about 62†atm, the volume can be reduced from 200†cm3 to about
55†cm3 without any further rise in the pressure.
Instead of the gas being compressed, it is
replaced with the far more compact liquid as the
gas is essentially being "squeezed" into its liquid phase. After all of the gas has disappeared
(2), the pressure rises very rapidly because
now all that remains is an almost incompressible liquid. The isotherm 3 that passes through
the critical point is called the critical isotherm.
Above this isotherm (4), CO2 exists only as a
supercritical fluid.
One intriguing consequence of the very limited bounds of the liquid state is that you could
start with a gas at large volume and low temperature, raise the temperature, reduce the volume, and then reduce the temperature so as to arrive at the liquid region at the lower left,
without ever passing through the two-phase region, and thus without undergoing condensation!
Supercritical fluids
The supercritical state of matter, as the fluid above the critical point is often called, possesses the flow properties of a gas and the solvent properties of a liquid. The density of a
supercritical fluid can be changed over a wide range by adusting the pressure; this, in turn,
changes its solubility, which can thus be optimized for a particular application. The picture
at the right shows a commercial laboratory device used for carrying out chemical reactions
under supercritical conditions.
Supercritical carbon dioxide is widely used to dissolve the caffeine out of coffee beans
and as a dry-cleaning solvent. Supercritical water has recently attracted interest as a
medium for chemically decomposing dangerous environmental pollutants such as PCBs.
Condensation and the critical point
Page 41 of 41
Reading(s) #2
Chemical energetics: heat and work
http://www.chem1.com/acad/webtext/energetics/CE01.html
Chemical Energetics: an introduction to chemical thermodynamics and the First Law
Introduction: the basics of energy, heat and
work
ho me Ener g y Fir s t L a w | C hem i c al en erg y | T he r mo c hem i st r y | Ap pli c at i o ns
On this page:
All chemical changes are
Energy: what is it?
accompanied by the absorption or
Energy units
release of heat. The intimate
System and surroundings
Observable properties and the state of a system
connection between matter and
Heat and work
energy has been a source of
Concept map
wonder and speculation from the
most primitive times; it is no
accident that fire was considered one of the four basic elements (along
with earth, air, and water) as early as the fifth century BCE. In this unit
we will review some of the fundamental concepts of energy and heat and
the relation between them. We will begin the study of thermodynamics,
which treats the energetic aspects of change in general, and we will finally
apply this specifically to chemical change. Our purpose will be to provide
you with the tools to predict the energy changes associated with chemical
processes. This will build the groundwork for a more ambitious goal: to
predict the direction and extent of change itself.
One of the interesting things about thermodynamics is that although it deals with
matter, it makes no assumptions about the microscopic nature of that matter.
Thermodynamics deals with matter in a macroscopic sense; it would be valid even
if the atomic theory of matter were wrong. This is an important quality, because it
means that reasoning based on thermodynamics is unlikely to require alteration as
new facts about atomic structure and atomic interactions come to light.
Energy
Energy is one of the most fundamental and universal concepts of physical science,
but one that is remarkably difficult to define in way that is meaningful to most
people. This perhaps reflects the fact that energy is not a “thing” that exists by
itself, but is rather an attribute of matter (and also of electromagnetic radiation)
that can manifest itself in different ways. It can be observed and measured only
indirectly through its effects on matter that acquires, loses, or possesses it.
1 of 7
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Chemical energetics: heat and work
http://www.chem1.com/acad/webtext/energetics/CE01.html
Kinetic energy and potential energy
Whatever energy may be, there are basically two kinds: kinetic and potential.
Kinetic energy is associated with the motion of an object; a body with a
mass m and moving at a velocity v possesses the kinetic energy mv2 /2.
In the 17th Century, the great mathematician Gottfried Leibniz
(1646-1716) suggested the distinction between vis viva ("live energy")
and vis mortua ("dead energy"), which later became known as kinetic
energy and potential energy.
Potential energy is energy a body has by virtue of its location in a force
field— a gravitational, electrical, or magnetic field. For example, if an object of
mass m is raised off the floor to a height h, its potential energy increased by mgh,
where g is a proportionality constant known as the acceleration of gravity.
Similarly, the potential energy of a particle having an electric charge q depends on
its location in an electrostatic field. Electrostatic potential energy plays a major role
in chemistry; the potential energies of electrons in the force field created by atomic
nuclei play a major role in determining the energies of atoms and molecules.
Energy scales are always arbitrary
You might at first think that a book sitting on the table has zero kinetic energy
since it is not moving. In truth, however, that the earth itself is moving; it is
spinning on its axis, it is orbiting the sun, and the sun itself is moving away from
the other stars in the general expansion of the universe. Since these motions are
normally of no interest to us, we are free to adopt an arbitrary scale in which the
velocity of the book is measured with respect to the table; on this so-called
laboratory coordinate system, the kinetic energy of the book can be considered
zero.
We do the same thing with potential energy. If the book is on the table, its
potential energy with respect to the surface of the table will be zero. If we adopt
this as our zero of potential energy, and then push the book off the table, its
potential energy will be negative after it reaches the floor.
Energy units
Energy is measured in terms of its ability to perform work or to transfer heat.
Mechanical work is done when a force f displaces an object by a distance d: w = f
× d. The basic unit of energy is the joule. One joule is the amount of work done
when a force of 1 newton acts over a distance of 1 m; thus 1 J = 1 N-m. The
newton is the amount of force required to accelerate a 1-kg mass by 1 m/sec2 , so
the basic dimensions of the joule are kg m2 s– 2.The other two units in wide use.
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Chemical energetics: heat and work
http://www.chem1.com/acad/webtext/energetics/CE01.html
the calorie and the BTU (British thermal unit) are defined in terms of the heating
effect on water. Because of the many forms that energy can take, there are a
correspondingly large number of units in which it can be expressed, a few of
which are summarized below.
1 calorie will raise the temperature of 1 g of water by 1
1 cal = 4.184 J
C°. The “dietary” calorie is actually 1 kcal.
1 BTU (British Thermal Unit) will raise the temperature
of 1 lb of water by 1F°.
The erg is the c.g.s. unit of energy and a very small
one; the work done when a 1-dyne force acts over a
distance of 1 cm.
1 BTU = 1055 J
1 J = 107 ergs
1 erg = 1 d-cm = 1 g cm2
s –2
The electron-volt is even tinier: 1 e-v is the work
required to move a unit electric charge (1 C) through a
potential difference of 1 volt.
1 J = 6.24 × 1018 e-v
The watt is a unit of power, which measures the rate of
energy flow in J sec–1. Thus the watt-hour is a unit of
energy. An average human consumes energy at a rate
of about 100 watts; the brain alone runs at about 5
watts.
1 J = 2.78 × 10–4 watt-hr
1 w-h = 3.6 kJ
The liter-atmosphere is a variant of force-displacement
1 L-atm = 101.325 J
work associated with volume changes in gases.
The huge quantities of energy consumed by cities and
1 quad = 1015 Btu = 1.05
countries are expressed in quads; the therm is a similar
× 1018 J
but smaller unit.
If the object is to obliterate cities or countries with
1 ton of TNT = 4.184 GJ
nuclear weapons, the energy unit of choice is the ton of
(by definition)
TNT equivalent.
In terms of fossil fuels, we have barrel-of-oil equivalent, 1 bboe = 6.1 GJ
cubic-meter-of-natural gas equivalent, and ton-of-coal
1 cmge = 37-39 mJ
equivalent.
1 toce = 29 GJ
System and surroundings
The thermodynamic view of the world requires that we
be very precise about our use of certain words. The two
most important of these are system and surroundings. A
thermodynamic system is that part of the world to
which we are directing our attention. Everything that is
not a part of the system constitutes the surroundings.
The system and surroundings are separated by a
boundary. If our system is one mole of a gas in a
container, then the boundary is simply the inner wall of the container itself. The
boundary need not be a physical barrier; for example, if our system is a factory or
a forest, then the boundary can be wherever we wish to define it. We can even
focus our attention on the dissolved ions in an aqueous solution of a salt, leaving
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the water molecules as part of the surroundings. The single property that the
boundary must have is that it be clearly defined, so we can unambiguously say
whether a given part of the world is in our system or in the surroundings.
If matter is not able to pass across the boundary, then the system is said
to be closed; otherwise, it is open. A closed system may still exchange
energy with the surroundings unless the system is an isolated one, in
which case neither matter nor energy can pass across the boundary.
The tea in a closed Thermos bottle approximates a closed system over
a short time interval.
Properties and the state of a system
The properties of a system are those quantities such as the pressure, volume,
temperature, and its composition, which are in principle measurable and capable
of assuming definite values. There are of course many properties other than those
mentioned above; the density and thermal conductivity are two examples.
However, the pressure, volume, and temperature have special significance because
they determine the values of all the other properties; they are therefore known as
state properties because if their values are known then the system is in a
definite state.
Change of state: the meaning of Δ
In dealing with thermodynamics, we must be able to unambiguously define the
change in the state of a system when it undergoes some process. This is done by
specifying changes in the values of the different state properties using the symbol
Δ (delta) as illustrated here for a change in the volume:
ΔV = Vfinal – Vinitial
We can compute similar delta-values for changes in P, V, ni (the number of moles
of component i), and the other state properties we will meet later.
Heat and work
Heat and work are both measured in energy units, so they must both represent
energy. How do they differ from each other, and from just plain “energy” itself?
First, recall that energy can take many forms: mechanical, chemical, electrical,
radiation (light), and thermal, or heat. So heat is a form of energy, but it differs
from all the others in one crucial way. All other forms of energy are
interconvertible: mechanical energy can be completely converted to electrical
energy, and the latter can be completely converted to heat. However, complete
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conversion of heat into other forms of energy is impossible. This is the
essence of the Second Law of Thermodynamics which is covered in another
unit; for the moment, we will simply state that this fact places heat in a category of
its own and justifies its special treatment.
There is another special property of heat that
you already know about: thermal energy can be
transferred from one body (i.e., one system) to
another. We often refer to this as a "flow" of
heat, recalling the 18th-century notion that heat
was an actual substance called “caloric” that
could flow like a liquid. Moreover, you know that heat can only flow from a
system at a higher temperature to one at a lower temperature. This special
characteristic is often used to distinguish heat from other modes of transferring
energy across the boundaries of a system.
Work, like energy, can take various forms: mechanical, electrical, gravitational,
etc. All have in common the fact that they are the product of two factors, an
intensity term and a capacity term. For example, the simplest form of mechanical
work arises when an object moves a certain distance against an opposing force.
Electrical work is done when a body having a certain charge moves through a
potential difference.
type of
work
intensity factor
capacity
factor
formula
change in
distance
f Δx
mechanical
force
gravitational
gravitational potential
mass
(a function of height)
electrical
potential difference
quantity
of charge
mgh
QΔ V
Performance of work involves a transformation of energy; thus when a book
drops to the floor, gravitational work is done (a mass moves through a
gravitational potential difference), and the potential energy the book had before it
was dropped is converted into kinetic energy which is ultimately dispersed as heat.
Mechanical
work is the
product of
the force
exerted on
a body and
the distance
it is moved:
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1 N-m = 1 J
(Illustration
from the Ben
Wiens Energy
site)
Heat and work are best thought of as processes by which energy is exchanged,
rather than as energy itself. That is, heat “exists” only when it is flowing, work
“exists” only when it is being done.
When two bodies are placed in thermal contact and energy
flows from the warmer body to the cooler one,we call the
process “heat”. A transfer of energy to or from a system by
any means other than heat is called “work”.
Interconvertability of heat and work
Work can be completely converted into heat (by friction, for
example), but heat can only be partially converted to work.
Conversion of heat into work is accomplished by means of a heat
engine. The science of thermodynamics developed out of the need
to understand the limitations of steam-driven heat engines at the
beginning of the Industrial Age. A fundamental law of Nature, the
Second Law of Thermodynamics, states that the complete
conversion of heat into work is impossible. Something to think
about when you purchase fuel for your car!
Concept map
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Chemical Energetics: an introduction to chemical thermodynamics and the First Law
The First Law of Thermodynamics
ho me | Ene r g y Fi r st L aw C hem i c al en erg y | T he r mo c hem i st r y | Ap pli c at i o ns
On this page:
"Energy cannot be created or
Internal Energy
destroyed"ó this fundamental law of
The First Law of Thermodynamics
nature, more properly known as
Pressure-volume work
Reversible Processes
conservation of energy, is familiar
Constant-pressure processes: the enthalpy
to anyone who has studied science.
Heat capacity and specific heat
Under its more formal name of the
Concept Map
First Law of Thermodynamics, it
governs all aspects of energy in
science and engineering applications. It's special importance in
Chemistry arises from the fact that virtually all chemical reactions
are accompanied by the uptake or release of energy.
Internal energy
Internal energy is simply the totality of all forms of kinetic and potential
energy of the system. Thermodynamics makes no distinction between these two
forms of energy and it does not assume the existence of atoms and molecules.
But since we are studying thermodynamics in the context of chemistry, we can
allow ourselves to depart from “pure” thermodynamics enough to point out that
the internal energy is the sum of the kinetic energy of motion of the molecules,
and the potential energy represented by the chemical bonds between the atoms
and any other intermolecular forces that may be operative.
How can we know how much internal energy a system possesses? The answer
is that we cannot, at least not on an absolute basis; all scales of energy are
arbitrary. The best we can do is measure changes in energy. However, we are
perfectly free to define zero energy as the energy of the system in some arbitrary
reference state, and then say that the internal energy of the system in any
other state is the difference between the energies of the system in these two
different states.
The First Law
This law is one of the most fundamental principles of the physical world. Also
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known as the Law of Conservation of Energy, it states that energy can not
be created or destroyed; it can only be redistributed or changed from one form to
another.
A way of expressing this law that is generally more useful in Chemistry is that
any change in the internal energy of a system is given by the sum of the heat q
that flows across its boundaries and the work w done on the system by the
surroundings.
ΔU = q + w
(1) must know this!
This says that there are two kinds of processes, heat and work, that can lead to a
change in the internal energy of a system. Since both heat and work can be
measured and quantified, this is the same as saying that any change in the energy
of a system must result in a corresponding change in the energy of the world
outside the system- in other words, energy cannot be created or destroyed.
There is an important sign convention for heat and work that you
are expected to know. If heat flows into a system or the
surroundings to do work on it, the internal energy increases and the
sign of q or w is positive. Conversely, heat flow out of the system or
work done by the system will be at the expense of the internal
energy, and will therefore be negative. (Note that this is the opposite
of the sign convention that was commonly used in much of the
pre-1970 literature.)
The full significance of Eq. 1 cannot be grasped without understanding that U is
a state function. This means that a given change in internal energy ΔU can
follow an infinite variety of pathways corresponding to all the possible
combinations of q and w that can add up to a given value of ΔU.
As a simple example of how this principle can simplify our
understanding of change, consider two identical containers of water
initially at the same temperature. We place a flame under one until its
temperature has risen by 1°C. The water in the other container is
stirred vigorously until its temperature has increased by the same
amount. There is now no physical test by which you could determine
which sample of water was warmed by performing work on it, by
allowing heat to flow into it, or by some combination of the two
processes. In other words, there is no basis for saying that one
sample of water now contains more “work”, and the other more
“heat”. The only thing we can know for certain is that both samples
have undergone identical increases in internal energy, and we can
determine the value of simply by measuring the increase in the
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temperature of the water.
An early statement of energy conservation by René Descartes
(1596-1650) explained that in creating the world, God established
“vortices”, states of motion that could interact and be transformed,
but which in their totality would endure in perpetuity.
There seems to be no end of schemes perpetrated by cranks, kooks
(and perhaps even a few crooks!) to foist off onto the science-naïve
public, crackpot schemes to obtain "free energy" from various
sources, usually in blissful ignorance of the First Law. Schemes for
using water as a fuel are especially popular, as can be seen on these
Web sites promoting "Aquafuel" (see these do-it-yourself
instructions) and "Brown's Gas". Of course, the "real" reason this is
not being done is a vast conspiracy involving the big oil companies
and the scientists who work for them!
Pressure-volume work
The kind of work most frequently associated with chemical change occurs when
the volume of the system changes owing to the disappearance or formation of
gaseous substances. This is sometimes called expansion work or PV-work, and it
can most easily be understood by reference to the simplest form of matter we can
deal with, the hypothetical ideal gas.
The figure shows a quantity of gas confined in a cylinder by means
of a moveable piston. Weights placed on top of the piston exert a
force f over the cross-section area A, producing a pressure P = f / A
which is exactly countered by the pressure of the gas, so that the
piston remains stationary. Now suppose that we heat the gas
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slightly; according to Charles’ law, this will cause the gas to expand,
so the piston will be forced upward by a distance Δx. Since this
motion is opposed by the force x, a quantity of work f Δx will be
done by the gas on the piston. By convention, work done by the
system (in this case, the gas) on the surroundings is negative, so the
work is given by
w = – f Δx
When dealing with a gas, it is convenient to think in terms of the
more relevant quantities pressure and volume rather than force and
distance. We can accomplish this by multiplying the second term by
A/A which of course leaves it unchanged:
By grouping the terms differently, but still not changing anything,
we obtain
Since pressure is force per unit area and the product of the length A
and the area has the dimensions of volume, this expression becomes
w = –P ΔV
(2)
It is important to note that although P and V are state functions, the work is not.
As is shown farther below, the quantity of work done will depend on whether
the same net volume change is realized in a single step (by setting the external
pressure to the final pressure P), or in multiple stages by adjusting the restraining
pressure on the gas to successively smaller values approaching the final value of
P.
Problem Example 1
Find the amount of work done on the surroundings when 1
liter of an ideal gas, initially at a pressure of 10 atm, is
allowed to expand at constant temperature to 10 liters by a)
reducing the external pressure to 1 atm in a single step, b)
reducing P first to 5 atm, and then to 1 atm, c) allowing the
gas to expand into an evacuated space so its total volume is
10 liters.
Solution. First, note that ΔV, which is a state function, is the
same for each path:
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V2 = (10/1) × (1 L) = 10 L, so ΔV = 9 L.
For path (a), w = –(1 atm)× (9 L) = –9 L-atm.
For path (b), the work is calculated for each stage
separately:
w = –(5 atm) × (2–1 L) – (1 atm) × (10–2 L) = –13
L-atm
For path (c) the process would be carried out by removing
all weights from the piston in Fig. 1 so that the gas expands
to 10 L against zero external pressure. In this case w = (0
atm) × 9 L = 0; that is, no work is done because there
is no force to oppose the expansion.
Adiabatic and isothermal processes
When a gas expands, it does work on the surroundings; compression of a gas to
a smaller volume similarly requires that the surroundings perform work on the
gas. If the gas is thermally isolated from the surroundings, then the process is
said to occur adiabatically. In an adiabatic change, q = 0, so the First Law
becomes ΔU = 0 + w. Since the temperature of the gas changes with its internal
energy, it follows that adiabatic compression of a gas will cause it to warm up,
while adiabatic expansion will result in cooling.
In contrast to this, consider a gas that is allowed to slowly excape from a
container immersed in a constant-temperature bath. As the gas expands, it does
work on the surroundings and therefore tends to cool, but the thermal gradient
that results causes heat to pass into the gas from the surroundings to exactly
compensate for this change. This is called an isothermal expansion. In an
isothermal process the internal energy remains constant and we can write the
First Law as 0 = q + w, or q = –w, illustrating that the heat flow and work done
exactly balance each other.
Because no thermal insulation is perfect, truly adiabatic
processes do not occur. However, heat flow does take time, so a
compression or expansion that occurs more rapidly than thermal
equilibration can be considred adiabatic for practical purposes.
If you have ever used a hand pump to inflate a bicycle tire, you
may have noticed that the bottom of the pump barrel can get
quite warm. Although a small part of this warming may be due
to friction, it is mostly a result of the work you (the surroundings) are doing on
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the system (the gas.)
Adiabatic expansion and contractions are especially important in understanding
the behavior of the atmosphere. Although we commonly think of the
atmosphere as homogeneous, it is really not, due largely to uneven heating and
cooling over localized areas. Because mixing and heat transfer between adjoining
parcels of air does not occur rapidly, many common atmospheric phenomena can
be considered at least quasi-adiabatic. A more detailed exposition of this topic is
given in Part 5 of this unit.
Reversible processes
From Problem Example 1 we see that when a gas expands into a vacuum
(Pexternal = 0) the work done is zero. This is the minimum work the gas can do;
what is the maximum work the gas can perform on the surroundings? To
answer this, notice that more work is done when the process is carried out in two
stages than in one stage; a simple calculation will show that even more work can
be obtained by increasing the number of stages— that is, by allowing the gas to
expand against a series of successively lower external pressures. In order to
extract the maximum possible work from the process, the expansion would have
to be carried out in an infinite sequence of infinitessimal steps. Each step yields
an increment of work P ΔV which can be expressed as (RT/V) dV and integrated:
Although such a path (which corresponds to what is called a reversible
process) cannot be realized in practice, it can be approximated as closely as
desired.
Even though no real process can take place reversibly (it would take an infinitely
long time!), reversible processes play an essential role in thermodynamics. The
main reason for this is that qrev and wrev are state functions which are
important and are easily calculated. Moreover, many real processes take place
sufficiently gradually that they can be treated as approximately reversible
processes for easier calculation.
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These plots illustrate
the effects of various
degrees of reversibility
on the amount of work
done when a gas
expands, and the work
that must be done in
order to restore it to its
initial state by
recompressing it. The
work, in each case, is
proportional to the
shaded area on the
plot.
Each
expansion-compression
cycle leaves the gas
unchanged, but in all
but the one in the
bottom row, the
surroundings are
forever altered, having
expended more work in
compressing the gas
than was performed on
it when the gas
expanted.
Only when the processes
are carried out in an
infinite number of steps
will the system and the
surroundings be restored
to their initial states—
this is the meaning of
thermodynamic
reversibility.
Heat changes at constant pressure: the
enthalpy
Most chemical processes are accompanied by changes in the volume of the
system, and therefore involve both heat and work terms. If the process takes
place at a constant pressure, then the work is given by PΔV and the change in
internal energy will be
ΔU = q – PΔV
(3)
Thus the amount of heat that passes between the system and the surroundings is
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given by
q = ΔU +PΔV
(4)
Problem Example 2
Hydrogen chloride gas readily dissolves in water, releasing
75.3 kJ/mol of heat in the process. If one mole of HCl at 298
K and 1 atm pressure occupies 24.5 liters, find ΔU for the
system when one mole of HCl dissolves in water under these
conditions.
Solution: In this process the volume of liquid remains
practically unchanged, so ΔV = –24.5 L. The work done is
w = –PΔV = –(1 atm)(–24.5 L) = 24.6 L-atm
(The work is positive because it is being done on the system
as its volume decreases due to the dissolution of the gas
into the much smaller volume of the solution.) Using the
conversion factor 1 L-atm = 101.33 J mol–1 and
substituting in Eq. 3 (above) we obtain
ΔU= q +PΔV = –(75300 J) + [101.33 J/L-atm) × (24.5
L-atm)] = –72.82 kJ
In other words, if the gaseous HCl could dissolve without
volume change, the heat released by the process (75.3 kJ)
would cause the system’s internal energy to diminish by
75.3 kJ. But the disappearance of the gaseous phase
reduces the volume of the system. This is equivalent to
compression of the system by the pressure of the
atmosphere performing work on it and consuming part of
the energy that would otherwise be liberated, reducing the
net value of ΔU to –72.82 kJ.
Since both ΔU and ΔV in Eq 4 are state functions, then qP, the heat that is
absorbed or released when a process takes place at constant pressure, must also
be a state function and is known as the enthalpy change ΔH.
ΔH ≡ qP = ΔU + PΔV
(5)
Since most processes that occur in the laboratory, on the surface of
the earth, and in organisms do so under a constant pressure of one
atmosphere, Eq 5 is the form of the First Law that is of greatest
interest to most of us most of the time.
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The heat capacity
For systems in which no change in composition (chemical reaction) occurs,
things are even simpler: to a very good approximation, the enthalpy depends only
on the temperature. This means that the temperature of such a system can serve
as a direct measure of its enthalpy. The functional relation between the internal
energy and the temperature is given by the heat capacity measured at constant
pressure:
(6)
(or ΔH/ΔT if you don’t care for calculus!) An analogous quantity relates the heat
capacity at constant volume to the internal energy:
The difference between CP and CV is of importance only when the
volume of the system changes significantly— that is, when different
numbers of moles of gases appear on either side of the chemical
equation. For reactions involving only liquids and solids, Cp and Cv
are for all practical purposes identical.
Heat capacity can be expressed in joules or calories per mole per degree (molar
heat capacity), or in joules or calories per gram per degree; the latter is called the
specific heat capacity or just the specific heat.
The greater the heat capacity of a substance, the
smaller will be the effect of a given absorption or loss
of heat on its temperature.
Heat capacities of some common substances
substance
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C J /g-K
C J/mol-K
Aluminum
0.900
24.3
Copper
0.386
24.5
Lead
0.128
26.4
Mercury
0.140
28.3
Zinc
.387
25.2
Alcohol (ethanol)
2.4
111
Water
4.186
75.2
Ice (–10° C)
2.05
36.9
Gasoline (n-octane)
.53
2.2
Glass
.84
Note especially the
following:
Both specific (per
gram) heat
capacities and molar
heat capacities are
given.
The molar heat
capacities of the
metallic elements are
almost identical. This
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is the basis of the
Law of Dulong and
Petit, which served
as an important tool
for estimating the
atomic weights of
some elements.
Carbon (graphite/diamond)
.157 / .147
7.88 / 7.38
Sodium chloride
0.854
49.9
Rock (granite)
.790
Air
1.006
The intermolecular
hydrogen bonding in
water and alcohols
results in
anomalously high
heat capacities for
these liquids; the
same is true for ice,
compared to other
solids.
The values for
graphite and
diamond are
consistent with the
principle that solids
that are more
“ordered” tend to
have larger heat
capacities.
Concept map
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Chemical Energetics: an introduction to chemical thermodynamics and the First Law
Molecules as energy carriers and converters
hom e | Ene r gy | F i r st La w C hem i ca l e nerg y T he r m oc hem i st r y | Ap pli c a ti o ns
On this page:
All molecules at temperatures above
Chemical energy: what is it?
absolue zero possess thermal
How molecules take up thermal energy
energy— the randomized kinetic
Energetics of chemical reactions
Heat changes at constant pressure: the enthalpy
energy associated with the various
Changes in enthalpy and internal energy
motions the molecules as a whole, and
How the enthalpy depends on the temperature
also the atoms within them, can
Concept Map
undergo. Polyatomic molecules also
possess potential energy in the form
of chemical bonds. Molecules are thus both vehicles for storing and transporting
energy, and the means of converting it from one form to another when the
formation, breaking, or rearrangement of the chemical bonds within them is
accompanied by the uptake or release of heat.
Chemical Energy
When you buy a liter of gasoline for your car, a cubic metre of natural gas to heat
your home, or a small battery for your flashlight, you are purchasing energy in a
chemical form. In each case, some kind of a chemical change will have to occur
before this energy can be released and utilized: the fuel must be burned in the
presence of oxygen, or the two poles of the battery must be connected through an
external circuit (thereby initiating a chemical reaction inside the battery.) And
eventually, when each of these reactions is complete, our source of energy will be
exhausted; the fuel will be used up, or the battery will be “dead”.
Chemical substances are made of atoms, or more generally, of positively charged
nuclei surrounded by negatively charged electrons. A molecule such as
dihydrogen, H2 , is held together by electrostatic attractions mediated by the
electrons shared between the two nuclei. The total potential energy of the molecule
is the sum of the repulsions between like charges and the attractions between
electrons and nuclei:
PEtotal = PEelectron-electron + PEnucleus-nucleus + PEnucleus-electron
In other words, the potential energy of a molecule depends on the time-averaged
relative locations of its constituent nuclei and electrons. This dependence is
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expressed by the familiar potential energy curve which serves as an important
description of the chemical bond between two atoms.
Translation refers to movement of an object as a complete unit.
Translational motions of molecules in solids or liquids are restricted to
very short distances comparable to the dimensions of the molecules
themselves, whereas in gases the molecules typically travel hundreds of
molecular diameters between collisions.
In gaseous hydrogen, for example, the molecules will be moving freely from one
location to another; this is called translational motion, and the molecules
therefore possess translational kinetic energy KEtrans = mv2 /2, in which v
stands for the average velocity of the molecules; you may recall from your study
of gases that v, and therefore KEtrans, depends on the temperature.
In addition to translation, molecules can possess other kinds of motion. Because a
chemical bond acts as a kind of spring, the two atoms in H2 will have a natural
vibrational frequency. In more complicated molecules, many different modes
of vibration become possible, and these all contribute a vibrational term KEvib to
the total kinetic energy. Finally, a molecule can undergo rotational motions
which give rise to a third term KErot. Thus the total kinetic energy of a molecule is
the sum
KEtotal = KEtrans + KEvib + KErot
The total energy of the molecule (its internal energy U) is just the sum
U = KEtotal + PEtotal
(1)
Although this formula is simple and straightforward, it cannot take us very far in
understanding and predicting the behavior of even one molecule, let alone a large
number of them. The reason, of course, is the chaotic and unpredictable nature of
molecular motion. Fortunately, the behavior of a large collection of molecules,
like that of a large population of people, can be described by statistical methods.
How molecules take up thermal energy
As noted above, the heat capacity of a substance is a measure of how sensitively
its temperature is affected by a change in heat content; the greater the heat
capacity, the less effect a given flow of heat q will have on the temperature.
We also pointed out that temperature is a measure of the average kinetic energy
due to translational motions of molecules. If vibrational or rotational motions
are also active, these will also accept thermal energy and reduce the amount that
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goes into translational motions. Because the temperature depends only on the
latter, the effect of the other kinds of motions will be to reduce the dependence of
the internal energy on the temperature, thus raising the heat capacity of a
substance.
Vibrational and rotational motions are not possible for monatomic species such as
the noble gas elements, so these substances have the lowest heat capacities.
Moreover, as you can see in the leftmost column of Table 3, their heat capacities
are all the same. This reflects the fact that translational motions are the same for all
particles; all such motions can be resolved into three directions in space, each
contributing one degree of freedom to the molecule and 1/2 R to its heat capacity.
(R is the gas constant, 8.314 J K– 1).
Molar heat
capacities of some
gaseous substances
at constant
pressure.
Values are in kJ
mol–1.
monatomic diatomic triatomic
He
20.5 CO
29.3 H2O 33.5
Ne
20.5 N2
29.5 D2O 34.3
Ar
20.5 F 2
31.4 CO2 37.2
Kr
20.5 Cl2 33.9 CS2
45.6
Whereas monatomic molecules can only possess translational thermal energy, two
additional kinds of motions become possible in polyatomic molecules. A linear
molecule has an axis that defines two perpendicular directions in which rotations
can occur; each represents an additional degree of freedom, so the two together
contribute a total of 1/2 R to the heat capacity. For a non-linear molecule, rotations
are possible along all three directions of space, so these molecules have a rotational
heat capacity of 3/2 R. Finally, the individual atoms within a molecule can move
relative to each other, producing a vibrational motion. A molecule consisting of N
atoms can vibrate in 3N –6 different ways or modes. Each vibrational mode
contributes R (rather than ½ R) to the total heat capacity. (These results come from
advanced mechanics and will not be proven here.)
monatomic
3/2R
0
0
diatomic
3/2 R
R
R
polyatomic
3/2 R
3/2 R
3N – 6
separation between adjacent
lavels
6.0 × 10–17 J
(O2)
373 J (HCl)
373 J (HCl)
Table 4: Contribution of molecular motions to heat capacity
Now we are in a position to understand why more complicated molecules have
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higher heat capacities. The total kinetic energy of a molecule is the sum of those
due to the various kinds of motions:
KEtotal = KEtrans + KErot + KEvib
When a monatomic gas absorbs heat, all of the energy ends up in translational
motion, and thus goes to increase its temperature. In a polyatomic gas, by contrast,
the absorbed energy is partitioned among the other kinds of motions; since only
the translational motions contribute to the temperature, the temperature rise is
smaller, and thus the heat capacity is larger.
There is one very significant complication, however: classical mechanics predicts
that the energy is always partitioned equally between all degrees of freedom.
Experiments, however, show that this is observed only at quite high temperatures.
The reason is that these motions are all quantized. This means that only certain
increments of energy are possible for each mode of motion, and unless a certain
minimum amount of energy is available, a given mode will not be active at all and
will contribute nothing to the heat capacity.
Distribution of
thermal energy in a
typical diatomic
molecule at 300K
The shading indicates
the average thermal
energy available at this
temperature; only those
levels within this rage
will have significant
occupancy as indicated
by the thickness of the
lines in the two
rightmost columns. At
300K, only the lowest
vibrational state and the
first few rotational states
will be active. Most of
the thermal energy will
be confined to the
translational levels
whose minute spacing
(10–17 J) causes them
to appear as a
continuum.
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Heat
capacity of
dihydrogen
as a
function of
temperature
This plot is
typical of those
for other
substances,
and shows the
practical
consequences
of the spacings
of the various
forms of
hermal energy.
Thus
translational
motions are
available at
virtually all
temperatures,
but
contributions to
heat acapacity
by rotational or
vibrational
motions can
only develop
at
temperatures
sufficiently
large to excite
these motion.
It turns out that translational energy levels are spaced so closely that they these
motions are active almost down to absolute zero, so all gases possess a heat
capacity of at least 3/2 R at all temperatures. Rotational motions do not get started
until intermediate temperatures, typically 300-500K, so within this range heat
capacities begin to increase with temperature. Finally, at very high temperatures,
vibrations begin to make a significant contribution to the heat capacity
The strong intermolecular forces of liquids and many solids allow heat to be
channeled into vibrational motions involving more than a single molecule, further
increasing heat capacities. One of the well known “anomalous” properties of liquid
water is its high heat capacity (75 J mol– 1 K– 1) due to intermolecular hydrogen
bonding, which is directly responsible for the moderating influence of large
bodies of water on coastal climates.
Heat capacities of metals
Metallic solids are a rather special case. In metals, the atoms oscillate about their
equilibrium positions in a rather uniform way which is essentially the same for all
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metals, so they should all have about the same heat capacity. That this is indeed the
case is embodied in the Law of Dulong and Petit. In the 19th century these
workers discovered that the molar heat capacities of all the metallic elements they
studied were around to 25 J mol– 1 K– 1, which is close to what classical physics
predicts for crystalline metals. This observation played an important role in
characterizing new elements, for it provided a means of estimating their molar
masses by a simple heat capacity measurement.
Heat changes at constant pressure: the
enthalpy
Most chemical processes are accompanied by changes in the volume of the system,
and therefore involve both heat and work terms. The work term P ΔV becomes
especially significant when the number of moles of gaseous components changes.
If the process takes place at a constant pressure, then the work is given by PΔV
and the change in internal energy will be
ΔU = qP – PΔV
(2)
Thus the amount of heat qP that passes between the system and the surroundings is
given by
qP = ΔU +PΔV
(3)
Problem Example
Hydrogen chloride gas readily dissolves in water, releasing
75.3 kJ/mol of heat in the process. If one mole of HCl at 298
K and 1 atm pressure occupies 24.5 liters, find the ΔU for
the system when one mole of HCl dissolves in water under
these conditions.
Solution: In this process the volume of liquid remains
practically unchanged, so ΔV = –24.5 L. The work done is
w = –PΔV = –(1 atm)(–24.5 L) = 24.6 L-atm
(The work is positive because it is being done on the system as its volume
decreases due to the dissolution of the gas into the much smaller volume of the
solution.)
Using the conversion factor 1 L-atm = 101.33 J mol–1 and
substituting in Eq. 3 we obtain
ΔU= q +PΔV = –(75300 J) + [101.33 J/L-atm) × (24.5
L-atm)] = –72.82 kJ
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In other words, if the gaseous HCl simply dissolved without
volume change, the heat released by the process (75.3 kJ)
would cause the system’s internal energy to diminish by
75.3 kJ. But the volume decrease due to the disappearance
of the gas is equivalent to compression of the system by the
pressure of the atmosphere; the resulting work done on the
system acts to increase its internal energy, so the net value
of ΔU is –72.82 kJ instead of –75.3 kJ.
Since both ɢU and ɢV in Eq 3 are state functions, then qP, the heat that is
absorbed or released when a process takes place at constant pressure, must also be
a state function and is known as the enthalpy change H.
ΔH ≡ qP = ΔU +PΔV
(4) must know this!
The enthalpy change associated with a chemical reaction
corresponds to the quantity of heat exchanged with the
surroundings when the reactants are converted to products
at constant pressure.
Under the special conditions in which the pressure is 1 atm and the reactants and
products are at a temperature of 298 K, ΔH becomes the standard enthalpy
change ΔH °.
Since most changes that occur in the laboratory, on the surface of the earth, and in
organisms are subjected to a constant pressure of one atmosphere, this is the form
of the First Law that is of greatest interest to most scientists.
Energetics of chemical reactions
The rearrangement of atoms that occurs in a chemical reaction is virtually always
accompanied by the liberation or absorption of heat. If the purpose of the reaction
is to serve as a source of heat, such as in the combustion of a fuel, then these heat
effects are of direct and obvious interest. We will soon see, however, that a study
of the energetics of chemical reactions in general can lead us to a deeper
understanding of chemical equilibrium and the basis of chemical change itself.
In chemical thermodynamics, we define the zero of the enthalpy and internal
energy as that of the elements as they exist in their stable forms at 298K and 1 atm
pressure. Thus the enthalpies H of Xe(g), O2 (g) and C(diamond) are all zero, as
are those of H2 and Cl2 in the reaction
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H 2(g) + Cl2(g) → 2 HCl(g)
The enthalpy of two moles of HCl is smaller than that of the reactants, so the
difference is released as heat. Such a reaction is said to be exothermic. The
reverse of this reaction would absorb the same quantity of heat from the
surroundings and be endothermic.
In comparing the internal energies and enthalpies of different substances as we
have been doing here, it is important to compare equal numbers of moles, because
energy is an extensive property of matter. However, heats of reaction are
commonly expressed on a molar basis and treated as intensive properties.
Changes in enthalpy and internal energy
We can characterize any chemical reaction by the change in the internal energy or
enthalpy:
ΔH = H final – H initial
The significance of this can hardly be exaggerated because ΔH, being a state
function, is entirely independent of how the system gets from the initial state to the
final state. In other words, the value of ΔH or ΔU for a given change in state is
independent of the pathway of the process.
Consider, for example, the oxidation of a lump of sugar to carbon dioxide and
water:
C12H 22O11 + 12 O2(g) → 12 CO2(g) + 11 H2O(l)
This process can be carried out in many ways, for example by burning the sugar
in air, or by eating the sugar and letting your body carry out the oxidation.
Although the mechanisms of the transformation are completely different for these
two pathways, the overall change in the enthalpy of the system (the atoms of
carbon, hydrogen and oxygen that were originally in the sugar) will be identical,
and can be calculated simply by looking up the standard enthalpies of the reactants
and products and calculating the difference
ΔH = [12 × H(CO2)] + [11 × H(H2O)] – H(C12H 22O11) = –5606 kJ
The same quantity of heat is released whether the sugar is burnt in the air or
oxidized in a series of enzyme-catalyzed steps in your body.
Variation of the enthalpy with temperature
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The enthalpy of a system increases with the temperature by the amount ΔH =
CPΔT. The defining relation
ΔH = ΔU + PΔV
tells us that this change is dominated by the internal energy, subject to a slight
correction for the work associated with volume change. Heating a substance causes
it to expand, making ΔV positive and causing the enthalpy to increase slightly
more than the internal energy. Physically, what this means is that if the
temperature is increased while holding the pressure constant, some extra energy
must be expended to push back the external atmosphere while the system expands.
The difference between the dependence of U and H on temperature is only really
significant for gases, since the coefficients of thermal expansion of liquids and
solids are very small.
Phase changes
A plot of the enthalpy of a system as a function of its temperature is called an
enthalpy diagram. The slope of the line is given by Cp . The enthalpy diagram
of a pure substance such as water shows that this plot is not uniform, but is
interrupted by sharp breaks at which the value of C p is apparently infinite,
meaning that the substance can absorb or lose heat without undergoing any
change in temperature at all. This, of course, is exactly what happens when a
substance undergoes a phase change; you already know that the temperature the
water boiling in a kettle can never exceed 100 until all the liquid has evaporated,
at which point the temperature of the steam will rise as more heat flows into the
system.
Enthalpy of carbon
tetrachloride as a
function of
temperature at 1
atm
A plot of the enthalpy of a
system as a function of its
temperature provides a
concise view of its thermal
behavior. The slope of the
line is given by the heat
capacity Cp. All H-vs.-C
plots show sharp breaks
at which the value of Cp is
apparently infinite,
meaning that the
substance can absorb or
lose heat without
undergoing any change in
temperature at all. This,
of course, is exactly what
happens when a
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substance undergoes a
phase change; you
already know that the
temperature of the water
boiling in a kettle can
never exceed 100°C until
all the liquid has
evaporated, at which
point the temperature (of
the steam) will rise as
more heat flows into the
system.
The lowest-temperature
discontinuity on the CCl4
diagram corresponds to a
solid-solid phase transition
associated with a
rearrangement of molecules in
the crystalline solid.
Fusion and boiling are not the only kinds of phase changes that matter can
undergo. Most solids can exist in different structural modifications at different
temperatures, and the resulting solid-solid phase changes produce similar
discontinuities in the heat capacity. Enthalpy diagrams are easily determined by
following the temperature of a sample as heat flows into our out of the substance
at a constant rate. The resulting diagrams are widely used in materials science and
forensic investigations to characterize complex and unknown substances.
Concept map
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Chemical Energetics: an introduction to chemical thermodynamics and the First Law
Thermochemistry and calorimetry
h om e | E n ergy| F i rst Law | Ch em i cal e ne rgy Th erm o che m is t ry Ap pli cat i o ns
On this page:
The heat that flows across the Thermochemical equations and standard states
Enthalpy of formation
boundaries of a system
Hess' Law and thermochemical calculations
undergoing a change is a
Calorimetry
fundamental property that
Concept Map
characterizes the process. It is
easily measured, and if the
process is a chemical reaction carried out at constant pressure, it can also
be predicted from the difference between the enthalpies of the products
and reactants. The quantitative study and measurement of heat and
enthalpy changes is known as thermochemistry.
Thermochemical equations and
standard states
In order to define the thermochemical properties of a process, it is first
necessary to write a thermochemical equation that defines the actual
change taking place, both in terms of the formulas of the substances
involved and their physical states (temperature, pressure, and whether
solid, liquid, or gaseous.
To take a very simple example, here is the complete thermochemical
equation for the vaporization of water at its normal boiling point:
H2O(l, 373 K, 1 atm) → H2O(g, 373 K, 1 atm)
Δ H = 40.7 kJ mol–1
The quantity 40.7 is known as the enthalpy of vaporization (“heat of
vaporization”) of liquid water.
The following points should be kept in mind when writing
thermochemical equations:
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• Thermochemical equations for reactions taking place in solution
must also specify the concentrations of the dissolved species. For
example, the enthalpy of neutralization of a strong acid by a strong
base is
H+(aq, 1M, 298 K, 1 atm) + OH–(aq, 1M, 298 K, 1 atm) →
H2O(l, 373 K, 1 atm) Δ H =56.9 kJ mol–1
in which the abbreviation aq refers to the hydrated ions as they exist
in aqueous solution. Since most thermochemical equations are
written for the standard conditions of 298 K and 1 atm pressure, we
can leave these quantities out if these conditions apply both before
and after the reaction. If, under these same conditions, the substance
is in its preferred (most stable) physical state, then the substance is
said to be in its standard state. Thus the standard state of water at
1 atm is the solid below 0°C, and the gas above 100°C. A
thermochemical quantity such as Δ H that refers to reactants and
products in their standard states is denoted by ΔH° .
• In the case of dissolved substances, the standard state of a solute
is that in which the “effective concentration”, known as the
activity, is unity. For non-ionic solutes the activity and molarity are
usually about the same for concentrations up to about 1M, but for an
ionic solute this approximation is generally valid only for solutions
more dilute than 0.001-0.01M, depending on electric charge and size
of the particular ion.
• Any thermodynamic quantity such as ΔH that is associated with
a thermochemical equation always refers to the number of moles
of substances explicitly shown in the equation. Thus for the
synthesis of water we can write
2 H2(g) + O2(g) → 2 H2O(l)
Δ H = –572 kJ
or
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H2(g) +
½ O2(g) → H2O(l)
Δ H = –286 kJ
Enthalpy of formation
The enthalpy change for a chemical reaction is the difference
ΔH = Hproducts – Hreactants
If the reaction in question represents the formation of one mole of the
compound from its elements in their standard states, as in
H2(g) +
½ O2(g) → H2O(l)
Δ H = –286 kJ
then we can arbitrarily set the enthalpy of the elements to zero and write
Hf ° = ΣHf °products – ΣHf °reactants = –286 kJ – 0 = –268 kJ mol–1
which defines the standard enthalpy of formation of water at 298K.
The standard enthalpy of formation of a compound
is defined as the heat associated with the
formation of one mole of the compound from its
elements in their standard states.
In general, the standard enthalpy change for a reaction is given by
the expression
Δ ΣHf °products – ΣHf °reactants
(1) must know this!
in which the ΣHf ° terms indicate the sums of the standard enthalpies of
formations of all products and reactants. The above definition is one of
the most important in chemistry because it allows us to predict the
enthalpy change of any reaction without knowing any more than the
standard enthalpies of formation of the products and reactants, which are
widely available in tables.
The following examples illustrate some important aspects of the standard
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enthalpy of formation of substances.
• The thermochemical equation defining Hf ° is always written in
terms of one mole of the substance in question:
½ N2(g) + 3/2 H2(g) → NH3(g)
Δ H° = –46.1 kJ (per mole of
NH3)
• A number of elements, of which sulfur and carbon are common
examples, can exist in more then one solid crystalline form. The
standard heat of formation of a compound is always taken in
reference to the forms of the elements that are most stable at 25°C
and 1 atm pressure. In the case of carbon, this is the graphite, rather
than the diamond form:
C(graphite) + O2(g) → CO2(g)
Δ H° = –393.5 kJ mol–1
C(diamond) + O2(g) → CO2(g)
Δ H° = –395.8 kJ mol–1
• The physical state of the product of the formation reaction must
be indicated explicitly if it is not the most stable one at 25°C and 1
atm pressure:
H2(g) +
½ O2(g) → H2O(l)
Δ H° = –285.8 kJ mol–1
H2(g) +
½ O2(g) → H2O(g)
Δ H° = –241.8 kJ mol–1
Notice that the difference between these two
heat of vaporization of water.
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• Although the formation of most molecules from their elements is
an exothermic process, the formation of some compounds is mildly
endothermic:
½ N2(g) + O2(g) → NO2(g)
Δ H° = +33.2 kJ mol–1
A positive heat of formation is frequently associated with
instability— the tendency of a molecule to decompose into its
elements, although it is not in itself a sufficient cause. In many cases,
however, the rate of this decomposition is essentially zero, so it is
still possible for the substance to exist. In this connection, it is worth
noting that all molecules will become unstable at higher
temperatures.
• The thermochemical reactions that define the heats of formation
of most compounds cannot actually take place; for example, the
direct synthesis of methane from its elements
C(graphite) + 2 H2(g) → CH4(g)
cannot be observed directly owing to the large number of other
possible reactions between these two elements. However, the
standard enthalpy change for such a reaction be found indirectly
from other data, as explained in the next section.
• The standard enthalpy of formation of gaseous atoms from the
element is known as the heat of atomization. Heats of atomization
are always positive, and are important in the calculation of bond
energies.
Fe(s) → Fe(g)
Δ H° = 417 kJ mol–1
• The standard heat of formation of a dissolved ion such as Cl– (that
is, formation of the ion from the element) cannot be measured
because it is impossible to have a solution containing a single kind
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of ion. For this reason, ionic enthalpies are expressed on a separate
scale on which Hf° of the hydrogen ion at unit activity (1 M
effective concentration) is defined as zero. Thus for Ca2+(aq), Hf° =
–248 kJ mol–1; this means that the reaction
Ca(s) → Ca2+(aq) + 2e–(aq)
½ H2(g) → H+(aq) + e–(aq)
(Of course, neither of these reactions can take place by itself, so ionic
enthalpies must be measured indirectly.)
Hess’ law and thermochemical
calculations
You probably know that two or more chemical equations can be
combined algebraically to give a new equation. Even before the science
of thermodynamics developed in the late nineteenth century, it was
observed that the heats associated with chemical reactions can be
combined in the same way to yield the heat of another reaction. For
example, the standard enthalpy changes for the oxidation of graphite and
diamond can be combined to obtain Δ H° for the transformation between
these two forms of solid carbon, a reaction that cannot be studied
experimentally.
C(graphite) + O2(g) → CO2(g)
Δ H° = –393.51 kJ mol–1
C(diamond) + O2(g) → CO2(g)
Δ H° = –395.40 kJ mol–1
Subtraction of the second reaction from the first (i.e., writing the second
equation in reverse and adding it to the first one) yields
C(graphite) → C(diamond)
Δ H° = 1.89 kJ mol–1
This principle, known as Hess’ law of independent heat summation
is a direct consequence of the enthalpy being a state function. Hess’ law
is one of the most powerful tools of chemistry, for it allows the change in
the enthalpy (and in other thermodynamic functions) of huge numbers of
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chemical reactions to be predicted from a relatively small base of
experimental data.
Germain Henri Hess (1802-1850) was a
Swiss-born professor of chemistry at St.
Petersburg, Russia. He formulated his famous
law, which he discovered empirically, in
1840. Very little appears to be known about
his other work in chemistry.
Because most substances cannot be prepared directly from their
elements, heats of formation of compounds are seldom determined by
direct measurement. Instead, Hess’ law is employed to calculate
enthalpies of formation from more accessible data. The most important of
these are the standard enthalpies of combustion. Most elements and
compounds combine with oxygen, and many of these oxidations are
highly exothermic, making the measurement of their heats relatively
easy.
For example, by combining the heats of combustion of carbon,
hydrogen, and methane, we obtain the standard enthalpy of formation of
methane, which as we noted above, cannot be determined directly:
The standard heat of atomization refers to the transformation of an
element into gaseous atoms:
C(graphite)→ C(g)
Δ H° = 716.7 kJ
... which is always, of course, an endothermic process. Heats of
atomization are most commonly used for calculating bond energies.
Introduction to Hess's Law at the ChemTeam site
Calorimetry
How are enthalpy changes determined experimentally? First, you must
understand that the only thermal quantity that can be observed directly
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is the heat q that flows into or out of a reaction vessel, and that q is
numerically equal to Δ H° only under the special condition of constant
pressure. Moreover, q is equal to the standard enthalpy change only
when the reactants and products are both at the same temperature,
normally 25°C.
The measurement of q is generally known as calorimetry. A very simple
calorimetric determination of the standard enthalpy of the reaction
H+(aq) + OH–(aq) → H2O(l)
could be carried out by combining equal volumes of 0.1M solutions of
HCl and of NaOH initially at 25°C. Since this reaction is exothermic, a
quantity of heat q will be released into the solution. What we actually
measure is the resulting temperature rise; if we multiply Δ T by the
specific heat capacity of the solution (which will be close to that of pure
water, 4.184 J/g-K), we obtain the number of joules of heat released into
each gram of the solution, and q can then be calculated from the mass of
the solution. Since the entire process is carried out at constant pressure,
we have Δ H° = q.
For reactions that cannot be carried out in dilute aqueous
solution, the reaction vessel is commonly placed within a
larger insulated container of water. During the reaction, heat
passes between the inner and outer containers until their
temperatures become identical. Again, the temperature
change of the water is observed, but in this case the value of q
cannot be found just from the mass and the specific heat
capacity of the water, for we now have to allow for the
absorption of some of the heat by the walls of the inner vessel.
Instead, the calorimeter is “calibrated” by measuring the
temperature change that results from the introduction of a
known quantity of heat. The resulting calorimeter
constant, expressed in J K–1, can be regarded as the “heat
capacity of the calorimeter”. The known source of heat is
usually produced by passing an electric current through a
resistor within the calorimeter.
Δ H = Δ U + Δ (PV) = qV + Δ ngRT
(8)
Since the process takes
place at constant volume,
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the reaction vessel must be
constructed to withstand
the high pressure resulting
from the combustion
process, which amounts to
a confined explosion. The
vessel is usually called a
“bomb”, and the
technique is known as
bomb calorimetry. In
order to ensure complete
combustion, the bomb is
initially charged with pure oxygen above atmospheric pressure. The
reaction is initiated by discharging a capacitor through a thin wire which
ignites the mixture.
Since the process takes place at constant volume, the reaction vessel
must be constructed to withstand the high pressure resulting from the
combustion process, which amounts to a confined explosion. The vessel
is usually called a “bomb”, and the technique is known as bomb
calorimetry. In order to ensure complete combustion, the bomb is
initially charged with pure oxygen above atmospheric pressure. The
reaction is initiated by discharging a capacitor through a thin wire which
ignites the mixture.
Problem Example
A sample of biphenyl (C6H5)2 weighing 0.526 g was ignited in a
bomb calorimeter initially at 25°C, producing a temperature rise
of 1.91 K. In a separate calibration experiment, a sample of
benzoic acid C6H5COOH weighing 0.825 g was ignited under
identical conditions and produced a temperature rise of 1.94 K.
For benzoic acid, the heat of combustion at constant pressure is
known to be 3226 kJ mol–1 (that is, Δ U° = –3226 kJ mol–1.)
Use this information to determine the standard enthalpy of
combustion of biphenyl.
Solution.The calorimeter constant is given by
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The heat released by the combustion of the biphenyl at
constant pressure is then Δ U:
(The negative sign indicates that heat is released in this
process.) From the reaction equation
(C6H5)2(s) + 19/2 O2(g) → 12 CO2(g) + 5 H2O(l)
we have Δ n g = 12 - (19/2) = –5/2. Converting to Δ H, we
substitute into
Δ H = qV + Δ ng RT
ΔH° = ΔU° – (5/2)(8.314 J mol –1 K–1) = –6440 J mol–1
This is the amount of heat that is lost by the system if reaction takes
place at constant pressure and the temperature is restored to its initial
value.
Although calorimetry is simple in principle, its practice is a highly
exacting art, especially when applied to processes that take place slowly
or involve very small heat changes, such as the germination of seeds.
Calorimeters can be
as simple as a foam
plastic coffee cup,
which are often used
in student
laboratories.
Research-grade calorimeters are more
likely to occupy entire table tops...
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rooms!
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The ice
calorimeter is an
important tool for
measuring the heat
capacities of liquids
and solids, as well
as the heats of
certain reactions.
This simple yet
ingenious
apparatus is
essentially a device
for measuring the
change in volume
due to melting of
ice. To measure a
heat capacity, a
warm sample is
placed in the inner
compartment,
which is
surrounded by a
mixture of ice and
water. The heat
withdrawn from the
sample as it cools
causes some of the
ice to melt. Since
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ice is less dense
than water, the
volume of water in
the insulated
chamber decreases.
This causes an
equivalent volume
of mercury to be
sucked into the
inner reservoir from
the outside
container. The loss
in weight of this
container gives the
decrease in volume
of the water, and
thus the mass of
ice melted. This,
combined with the
heat of fusion of
ice, gives the
quantity of heat
lost by the sample
as it cools to 0°C.
Concept map
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Chemical Energetics: an introduction to chemical thermodynamics and the First Law
Some applications of enthalpy and the
First Law
hom e | E ne rgy| F ir st Law | Chem i cal en e rgy | Th e rm o chem i st ry Ap p li cati o n s
In this section, we survey some of
the more common chemistry-related
applications of enthalpy and the
First Law.
On this page:
Enthalpy diagrams
Bond enthalpies and bond energies
Energy content of fuels
Energy content of foods
Thermodynamics and the weather
Concept Map
Enthalpy diagrams
Comparison and interpretation of enthalpy changes is materially aided
by a graphical construction in which the relative enthalpies of various
substances are represented by horizontal lines on a vertical energy scale.
The zero of the scale can be placed anywhere, since energies are always
arbitrary; it is generally most useful to locate the elements at zero energy,
which reflects to the convention that their standard enthlapies of
formation are zero.
An enthalpy diagram for carbon and oxygen and its two stable oxides
is shown below. The labeled arrows show the changes in enthalpy
associated with the various reactions this system can undergo. Notice
how Hess’ law is implicit in this diagram; we can calculate the enthalpy
change for the combustion of carbon monoxide to carbon dioxide, for
example, by subtraction of the appropriate arrow lengths without writing
out the thermochemical equations in a formal way.
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Enthalpy
diagram for
the C-O
system
In this diagram,
we follow the
convention of
assigning zero
enthalpy to
graphite, the
most stable
state of
elemental
carbon at 298 K
and 1 atm
pressure. The
labeled arrows
show the
changes in
enthalpy
associated with
the various
reactions this
system can
undergo.
Enthalpy diagrams
are especially
useful for
comparing groups
of substances
having some
common feature.
This Figure shows
the molar
enthalpies of
species relating to
two hydrogen
halides, with
respect to those of
the elements. From this diagram we can see at a glance that the formation
of HF from the elements is considerably more exothermic than the
corresponding formation of HCl. The upper part of this diagram shows
the gaseous atoms at positive enthalpies with respect to the elements.
The endothermic processes in which the H2 and the dihalogen are
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dissociated into atoms can be imagined as taking place in two stages,
also shown. From the enthalpy change associated with the dissociation
of H2 (218 kJ mol–1), the dissociation enthalpies of F2 and Cl2 can be
calculated and placed on the diagram.
Bond enthalpies and bond energies
The enthalpy change associated with the reaction
HI(g) → H(g) + I(g)
is the heat of dissociation of the HI molecule; it is also the bond energy
of the hydrogen-iodine bond in this molecule. Under the usual standard
conditions, it would be expressed either as the bond enthalpy H°(HI,298)
or internal energy U°(HI,298); in this case the two quantities differ from
each other by Δ PV = RT. Since this reaction cannot be studied directly,
the H–I bond enthalpy is calculated from the appropriate standard
enthalpies of formation:
½ H 2(g) → H(g)
+ 218 kJ
½ I 2(g) → I(g)
+107 kJ
½ H 2(g) + 1/2 I2(g) → HI(g)
–36 kJ
HI(g) → H(g) + I(g)
+299 kJ
Bond energies and enthalpies are important properties of chemical
bonds, and it is very important to be able to estimate their values from
other thermochemical data. The total bond enthalpy of a more complex
molecule such as ethane can be found from the following combination of
reactions:
C2H6(g) → 2C(graphite) + 3 H2(g)
84.7 kJ
3 H2(g) → 6 H(g)
1308 kJ
2 C(graphite) → 2 C(g)
1430 kJ
C2H6(g) → 2 C(g) + 6H(g)
2823 kJ
Pauling’s rule and average bond energy
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The total bond energy of a molecule can be thought of as the sum of the
energies of the individual bonds. This principle, known as Pauling’s
Rule, is only an approximation, because the energy of a given type of
bond is not really a constant, but depends somewhat on the particular
chemical environment of the two atoms. In other words, all we can really
talk about is the average energy of a particular kind of bond, such as
C–O, for example, the average being taken over a representative sample
of compounds containing this type of bond, such as CO, CO2 , COCl2 ,
(CH3 )2 CO, CH3 COOH, etc.
Despite the lack of strict additivity of bond energies,
Pauling’s Rule is extremely useful because it allows one to
estimate the heats of formation of compounds that have not
been studied, or have not even been prepared. Thus in the
foregoing example, if we know the enthalpies of the C–C and
C–H bonds from other data, we could estimate the total bond
enthalpy of ethane, and then work back to get some other
quantity of interest, such as ethane’s enthalpy of formation.
By assembling a large amount of experimental information of this kind, a
consistent set of average bond energies can be obtained. The energies of
double bonds are greater than those of single bonds, and those of triple
bonds are higher still.
H
C
N
O
F
H
436
C
415
N
390
O
464
F
569
Cl
432
Br
370
I
295
Si
395
345
290
160
350
200
439
270
330
200
275
270
240
360
140
185
160
205
255
185
160
200
280
370
540
243
220
190
210
180
359
290
150
210
230
Cl
Br
I
Si
Average energies of some single bonds (kJ/mol)
Energy content of fuels
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A fuel is any substance capable of providing useful amounts of energy
through a process that can be carried out in a controlled manner at
economical cost. For most practical fuels, the process is combustion in
air (in which the oxidizing agent O2 is available at zero cost.) The
enthalpy of combustion is obviously an important criterion for a
substance’s suitability as a fuel, but it is not the only one; a useful fuel
must also be easily ignited, and in the case of a fuel intended for
self-powered vehicles, its energy density in terms of both mass (kJ
kg–1) and volume (kJ m–3) must be reasonably large. Thus substances
such as methane and propane which are gases at 1 atm must be stored as
pressurized liquids for transportation and portable applications.
Energy densites of some common fuels
a) Ethanol is being strongly
promoted as a motor fuel by the
U.S. agricultural industry. Note,
however, that according to some
estimates, it takes 46 MJ of energy
to produce 1 kg of ethanol from
corn. Some other analyses, which
take into account optimal farming
practices and the use of
by-products arrive at different
conclusions; see, for example, this
summary with links to several
reports.
fuel
MJ kg–1
wood (dry)
coal (poor)
15
15
coal (premium)
27
a
ethanol
30
petroleum products
methane
45
54
hydrogenb
140
b) Owing to its low molar mass and
high heat of combustion,
hydrogen possesses an
extraordinarily high energy density,
and would be an ideal fuel if its
critical temperature (33 K, the
temperature above which it cannot
exist as a liquid) were not so low.
The potential benefits of using
hydrogen as a fuel have motivated
a great deal of research into other
methods of getting a large amount
of H2 into a small volume of space.
Simply compressing the gas to a
very high pressure is not practical
because the weight of the
heavy-walled steel vessel required
to withstand the pressure would
increase the effective weight of the
fuel to an unacceptably large value.
One scheme that has shown some
promise exploits the ability of H2 to
“dissolve” in certain transition
metals. The hydrogen can be
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recovered from the resulting solid
solution (actually a loosely-bound
compound) by heating.
Energy content of foods
What, exactly, is meant by the statement that a particular food “contains
1200 calories” per serving? This simply refers to the standard enthalpy of
combustion of the foodstuff, as measured in a bomb calorimeter. Note,
however, that in nutritional usage, the calorie is really a kilocalorie
(sometimes called “large calorie”), that is, 4184 J. Although this unit is
still employed in the popular literature, the SI unit is now commonly
used in the scientific and clinical literature, in which energy contents of
foods are usually quoted in kJ per unit of weight.
Although the mechanisms of oxidation of a carbohydrate such as glucose
to carbon dioxide and water in a bomb calorimeter and in the body are
complex and entirely different, the net reaction involves the same initial
and final states, and must be the same for any possible pathway.
C6 H12O6 + 6 O2 → 6 CO2 + 6 H2 O
ΔH° – 20.8 kJ mol–1
Glucose is a sugar, a breakdown product of starch, and is the most
important energy source at the cellular level; fats, proteins, and other
sugars are readily converted to glucose. By writing balanced equations
for the combustion of sugars, fats, and proteins, a comparison of their
relative energy contents can be made. The stoichiometry of each reaction
gives the amounts of oxygen taken up and released when a given amount
of each kind of food is oxidized; these gas volumes are often taken as
indirect measures of energy consumption and metabolic activity; a
commonly-accepted value that seems to apply to a variety of food
sources is 20.1 J (4.8 kcal) per liter of O2 consumed.
For some components of food, particularly proteins, oxidation may not
always be complete in the body, so the energy that is actually available
will be smaller than that given by the heat of combustion. Mammals, for
example, are unable to break down cellulose (a polymer of sugar) at all;
animals that derive a major part of their nutrition from grass and leaves
must rely on the action of symbiotic bacteria which colonize their
digestive tracts. The amount of energy available from a food can be
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found by measuring the heat of combustion of the waste products
excreted by an organism that has been restricted to a controlled diet, and
subtracting this from the heat of combustion of the food.
type of food
ΔH°, kJ g–1
food
percent
availability
Protein
meat
22.4
Fat
egg
butter
23.4
38.2
39.2
17.2
95
Carbohydrate
animal fat
starch
15.5
99
29.7
100
glucose
(sugar)
ethanol
92
Energy content and availability of the major food components
The amount of energy an animal requires depends on the age, sex,
surface area of the body, and of course on the amount of physical
activity. The rate at which energy is expended is expressed in watts:
1 W = 1 J sec–1. For humans, this value varies from about 200-800 W.
This translates into daily food intakes having energy equivalents of
about 10-15 MJ for most working adults. In order to just maintain weight
in the absence of any physical activity, about 6 MJ per day is required.
animal
kJ hr–1 kJ kg–1 hr–1
mouse
82
17
cat
dog
34
78
6.8
3.3
sheep
human
193
300
2.2
2.1
horse
elephant
1430
5380
1.1
0.7
metabolic rates of some mammals
The above table is instructive in that although larger animals consume
more energy, the energy consumption per unit of body weight decreases
with size. This reflects the fact that many metabolic functions, as well as
heat loss, depend directly or indirectly on the surface area of the body.
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The ratio of surface area to mass is of course greater for smaller animals.
Thermodynamics and the weather
Hydrogen bonds at work
It is common knowledge that large bodies of water have a “moderating”
effect on the local weather, reducing the extremes of temperature that
occur in other areas. Water temperatures change much more slowly than
do those of soil, rock, and vegetation, and this effect tends to affect
nearby land masses. This is largely due to the high heat capacity of water
in relation to that of land surfaces— and thus ultimately to the effects of
hydrogen bonding. The lower efficiency of water as an absorber and
radiator of infrared energy also plays a role.
Adiabatic heating and cooling of the atmosphere
[Adiabatic processes were briefly described in Part 2 of this
document]
The specific heat
capacity of
water is about
four times
greater than that
of soil. This has
a direct
consequence to
anyone who
lives near the
ocean and is
familiar with the
daily variations
in the direction
of the winds
between the
land and the
water. Even
large lakes can
exert a
moderating
influence on the
local weather
due to water's
relative
insensitivity to
temperature
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The above images are from the NDBC site. For a more detailed
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change.
During the
daytime the land
and sea receive
approximately
equal amounts
of heat from the
Sun, but the
much smaller
heat capacity of
the land causes
its temperature
to rise more
rapidly. This
causes the air
above the land
to heat, reducing
its density and
causing it to
rise. Cooler
oceanic air is
drawn in to vill
the void, thus
giving rise to the
daytime sea
breeze.
explanation of land/sea breezes and links to some interesting
meteorological images, see this U. Wisconsin page. An animated
movie that correlates time and temperature data with wind
directions can be seen at the Earth Science textbook site.
In the evening,
both land and
ocean lose heat
by radiation to
the sky, but the
temperature of
the water drops
less than that of
the land,
continuing to
supply heat to
the oceanic air
and causing it to
rise, thus
reversing the
direction of air
flow and
producing the
evening land
breeze.
Why it gets colder as you go higher: the adiabatic lapse
rate
The First Law at work
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The air receives its heat by absorbing
far-infrared radiation from the earth,
which of course receives its heat from
the sun. The amount of heat radiated
to the air immediately above the
surface varies with what's on it
(forest, fields, water, buildings) and
of course on the time and season.
When a parcel of air above a
particular location happens to be
warmed more than the air
immediately surrounding it, this air
expands and becomes less dense. It
therefore rises up through the surrounding air and undergoes further
expansion as it encounters lower pressures at greater altitudes. Whenever
a gas expands against an opposing pressure, it does work on the
surroundings. According to the First Law Δ U = q + w, if this work is not
accompanied by a compensating flow of heat into the system, its internal
energy will fall, and so, therefore, will its temperature. It turns out that
heat flow and mixing are rather slow processes in the atmosphere in
comparison to the convective motion we are describing, so the First Law
can be written as Δ U = w (recall that w is negative when a gas expands.)
Thus as air rises above the surface of the earth it undergoes adiabatic
expansion and cools. The actual rate of temperature decrease with
altitude depends on the composition of the air (the main variable being
its moisture content) and on its heat capacity. For dry air, this results in
an adiabatic lapse rate of 9.8 C° per km of altitude.
Santa Anas and chinooks: those warm, wild winds
Just the opposite happens
when winds develop in
high-altitude areas and head
downhill. As the air descends,
it undergoes compression from
the pressure of the air above it.
The surroundings are now
doing work on the system, and
because the process occurs too
rapidly for the increased internal energy to be removed as heat, the
compression is approximately adiabatic. The resulting winds are warm
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(and therefore dry) and are often very irritating to mucous membranes.
These are known generically as Föhn winds (which is the name given to
those that originate in the Alps.) In North America they are often called
chinooks (or, in winter, "snow melters") when they originate along the
Rocky Mountains. Among the most notorious are the Santa Ana winds
of Southern California which pick up extra heat (and dust) as they pass
over the Mohave Desert before plunging down into the Los Angeles
basin.
More on these kinds of winds
Concept map
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Reading(s) #3
Energy spreading and spontaneous change
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Previous page: Index
Page 1 of 6. - Index
Next page: What is entropy?
The direction of spontaneous change
What are natural processes?
Direction through disorder
From coins to molecules: the spreading of energy
How thermal energy is stored in molecules
Quantum states, microstates, and energy spreading
Energy-spreading changes the world
Energy spreading and sharing in chemical reactions
What are natural processes?
Drop a teabag into a pot of hot water, and you will see the tea diffuse into the water until it is uniformly
distributed throughout the water. What you will never see is the reverse of this process, in which the tea
would be sucked up and re-absorbed by the teabag. The making of tea, like all changes that take place in
the world, possesses a “natural” direction. Processes that occur in this way— that is, when left to
themselves and in the absence of any attempt to drive them in reverse— are known as spontaneous
changes. In many cases our everyday life experiences teach us the direction in which spontaneous change
can occur, and anything that runs counter to these expectations is immediately sensed as weird. In other
cases, including that of most chemical change, we often have no obvious guidelines, and must learn how
to apply the laws of thermodynamics which ultimately govern all spontaneous change.
A non-natural process: Steinberg's
famous New Yorker cartoon.
Saul Steinberg (1914-1999) was one
of America's most admired artists
whose unique and imaginative
cartoon style graced the pages of the
New Yorker magazine for more than
60 years.
Let's begin by thinking about the outcomes of the following four experiments, each of which illustrates a
natural process that proceeds spontaneously only in a single direction.
Experiment 1. A stack of one hundred coins is thrown into the air. After they have come to rest on the
floor, the numbers that land “heads up” and “tails up” are noted.
Net change: ordered coins → randomized coins (roughly equal numbers of heads and tails.)
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Energetics: no relevant net change in energy
Why it cannot go in reverse: Simple statistics shows that the probability of creating more
order (reducing disorder) through a random process such as tossing the coins again is
vanishingly small.
Experiment 2. Two identical blocks of copper, one at 200°C and the other at 100°C, are brought into
contact in a thermally-insulated environment. Eventually the temperatures of both blocks reach 150°C.
Net change: block1 (200°) + block2 (100°) → combined blocks (150°)
Energetics: Heat (randomized molecular kinetic energy) flows from the warmer block to
the cooler one until their temperatures are identical.
Why it cannot go in reverse: Dispersal of kinetic energy amongst the copper atoms is a
random process; the chances that such a process would lead to a non-uniform sharing of
the energy are even smaller than in the case of the 100 coins because of the much greater
number (around 1022) of particles involved.
Experiment 3. A book or some other solid object is held above a table top, and is then allowed to fall.
Net change: book in air
book on table top; potential energy
energy
thermal energy.
organized kinetic
Energetics: At the instant just before the end of its fall, the potential energy the object
acquired when it was raised will exist entirely as kinetic energy mv2/2 in which m is the
mass of the object and v is its velocity. Each atom of which the object is composed will of
course possess a proportionate fraction of this energy, again with its principal velocity
component pointing down. Superimposed on this, however, will be minute thermal
displacements that vary randomly in magnitude and direction from one instant to the
next. The sum total of these constitutes the thermal energy contained in the object.
When the object strikes the table top, its motion ceases and we say its kinetic energy is
zero. Energy is supposed to be conserved, so where did it disappear to? The shock of
impact has resulted in its dispersal into greatly augmented thermal motions of the atoms,
both of the object itself and of the area of the table top where the impact occurred. In
other words, the kinetic energy of organized motion the object had just before its motion
stopped has been transformed into kinetic energy of random or disorganized motion
(thermal energy) which spreads rapidly away from the point of impact.
Why it cannot go in reverse: Once the kinetic energy of the book has been dispersed
amongst the molecules of the book and the table top, the probabability of these
randomized motions reappearing at the surface where the two objects are in contact and
then acting in concert to propel the object back into the air) is negligible.
Experiment 4. One mole of gas, initially at 300 and 2 atm pressure, is allowed to expand to double its
volume, keeping the temperature constant.
Net change: Increase in volume of gas.
Energetics: No change in energy if the gas behaves ideally.
Why it cannot go in reverse: Simple statistics: the probabiliy that N randomly moving
objects (flies in a bottle, for example,) will at any time all be located in one half of the
container is (1/2)N. For chemically-significant values of N (1020, say) this probability is
indistinguishable from zero.
All of the changes described above take place spontaneously, meaning that
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Once they are allowed to commence, they will proceed to the finish without any outside
intervention.
It would be inconceivable that any of these changes could occur in the reverse direction (that is, be
undone) without changing the conditions or actively disturbing the system in some way.
What determines the direction in which spontaneous change will occur? It is clearly not a fall in the
energy, since in most cases cited above the energy of the system did not change. Even in the case of the
falling book, in which the potential energy of the system (the book) falls, energy is conserved overall; if
there is no net loss of energy when these processes operate in the forward or natural direction, it would
not require any expenditure of energy for them to operate in reverse. In other words, contrary to what all
too many people appear to believe,
The direction of a spontaneous process is not governed by the energy change...
... and the First Law of Thermodynamics cannot predict the direction of a natural process.
1.2 Direction through disorder
In our examination of the processes described above, we saw that although the total energy of the system
and the surroundings (and thus, of the world) is unchanged, there is something about the world that has
changed, and this is its degree of disorder.
After coins have been tossed or cards shuffled, the final state is invariably one of greater
disorder. Similarly, the molecules of a gas can occupy a larger number of possible positions in
space if the volume is larger, so the expansion of a gas is similarly accompanied by an
increase in randomness.
A closer look at disorder
How can we express disorder quantitatively? From the example of the coins, you can probably see that
simple statistics plays a role: the probability of obtaining three heads and seven tails after tossing ten coins
is just the ratio of the number of ways that ten different coins can be arranged in this way, to the number
of all possible arrangements of ten coins.
Using the language of molecular statistics, we say that a collection of coins in which a given fraction of its
members are heads-up constitutes a macroscopic state of the system. Since we don’t care which coins
are heads-up, there are clearly numerous configurations of the individual coins which can result in this
“macrostate”. Each of these configurations specifies a microscopic state of the system.
The greater the number of microstates that correspond to a given macrostate,
the greater the probability of that macrostate.
To see what this means, consider the possible outcomes of a toss of four coins:
macrostate ways probability
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0 heads
1 head
1
4
2 heads
6
3 heads
4 heads
4
1
microstates
1/16
TTTT
4/16 = 1/4 HTTT THTT TTHT TTTH
HHTT HTHT HTTH THHT TTHH
6/16 = 3/8 THTH
4/16 = 1/4 HHHT HTHH HHTH THHH
1/16
HHHH
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A toss of four coins will yield one of the five outcomes (macrostates) listed in the leftmost column of the
table. The second column gives the number of “ways”— that is, the number of head/tail configurations of
the set of coins (the number of microstates)— that can result in the macrostate. The probability of a toss
resulting in a particular macrostate is proportional to the number of microstates corresponding to the
macrostate, and is equal to this number, divided by the total number of possible microstates (in this
example, 24 =16). An important assumption here is that all microstates are equally probable; that is, the
toss is a “fair” one in which the many factors that determine the trajectory of each coin operate in an
entirely random way.
1.3 From coins to molecules: the spreading of energy
Disorder is more probable than order because there are so many more ways of achieving it. Thus coins
and cards tend to assume random configurations when tossed or shuffled, and socks and books tend to
become more scattered about a teenager’s room during the course of daily living. But there are some
important differences between these large-scale mechanical, or macro systems, and the collections of
sub-microscopic particles that constitute the stuff of chemistry, and which we will refer to here generically
as molecules. Molecules, unlike macro objects, are capable of accepting, storing, and giving up energy in
tiny amounts (quanta), and act as highly efficient carriers and spreaders of thermal energy as they move
around. Thus, in chemical systems,
1. We are dealing with huge numbers of particles. This is important because statistical predictions are
always more accurate for larger samples. Thus although for the four tosses there is a good chance
(62%) that the H/T ratio will fall outside the range of 0.45 - 0.55, this probability becomes almost
zero for 1000 tosses. To express this in a different way, the chances that 1000 gas molecules
moving about randomly in a container would at any instant be distributed in a sufficiently
non-uniform manner to produce a detectable pressure difference between the two halves of a
container will be extremely small. If we increase the number of molecules to a chemically
significant number (around 102 0, say), then the same probability becomes indistinguishable from
zero.
2. Once the change begins, it proceeds spontaneously. That is, no external agent (a tosser, shuffler, or
teen-ager) is needed to keep the process going. Gases will spontaneously expand if they are allowed
to, and reactions, one started, will proceed toward equilibrium.
3. Thermal energy is continually being exchanged between the particles of the system, and between
the system and the surroundings. Collisions between molecules result in exchanges of momentum
(and thus of kinetic energy) amongst the particles of the system, and (through collisions with the
walls of a container, for example) with the surroundings.
4. Thermal energy spreads rapidly and randomly throughout the various energetically accessible
microstates of the system. The direction of spontaneous change is that which results in the
maximum possible spreading and sharing of thermal energy.
The importance of these last two points is far greater than you might at first think, but to fully appreciate
this, you must recall the various ways in which thermal energy is stored in molecules— hence the
following brief review.
How thermal energy is stored in molecules
Thermal energy is kinetic energy, and thus relates to motion at the molecular scale. What kinds of
molecular motions are possible? For monatomic molecules, there is only one: actual movement from one
location to another, which we call translation. Since there are three directions in space, all molecules
possess three modes of translational motion.
For polyatomic molecules, two additional kinds of motions are possible. One of these is rotation; a linear
molecule such as CO2 in which the atoms are all laid out along the x-axis can rotate along the y- and
z-axes, while molecules having less symmetry can rotate about all three axes. Thus linear molecules
possess two modes of rotational motion, while non-linear ones have three rotational modes.
Finally, molecules consisting of two or more atoms can undergo internal vibrations. For freely moving
molecules in a gas, the number of vibrational modes or patterns depends on both the number of atoms
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and the shape of the molecule, and it increases rapidly as the molecule becomes more complicated.
The relative populations of the
translational, rotational and vibrational
energy states of a typical diatomic
molecule are depicted by the thickness of
the lines in this schematic (not-to-scale!)
diagram. The colored shading indicates the
total thermal energy available at a given
temperature. The numbers at the top show
order-of-magnitude spacings between
adjacent levels. It is readily apparent that
virtually all the thermal energy resides in
translational states.
Notice the greatly different spacing of the three kinds of energy levels. This is extremely important
because it determines the number of energy quanta that a molecule can accept, and, as the following
illustration shows, the number of different ways this energy can be distributed amongst the molecules.
Fig. 1: Accessible energy states. The more closely spaced the quantized energy
states of a molecule, the greater will be the number of ways in which a given
quantity of thermal energy can be shared amongst a collection of these
molecules.
The spacing of molecular energy states becomes closer as the mass and number of bonds in
the molecule increases, so we can generally say that the more complex the molecule, the
greater the density of its energy states.
Quantum states, microstates, and energy spreading
At the atomic and molecular level, all energy is quantized; each particle possesses discrete states of
kinetic energy and is able to accept thermal energy only in packets whose values correspond to the
energies of one or more of these states. Polyatomic molecules can store energy in rotational and
vibrational motions, and all molecules (even monatomic ones) will possess tranlational kinetic energy
(thermal energy) at all temperatures above absolute zero. The energy difference between adjacent
translational states is so minute (roughly 10–30 J) that translational kinetic energy can be regarded as
continuous (non-quantized) for most practical purposes.
The number of ways in which thermal energy can be distributed amongst the allowed states within a
collection of molecules is easily calculated from simple statistics, but we will confine ourselves to an
example here. Suppose that we have a system consisting of three molecules and three quanta of energy to
share among them. We can give all the kinetic energy to any one molecule, leaving the others with none,
we can give two units to one molecule and one unit to another, or we can share out the energy equally
and give one unit to each molecule. All told, there are ten possible ways of distributing three units of
energy among three identical molecules as shown here:
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Each of these ten possibilities represents a distinct microstate that will describe the system at any instant
in time. Those microstates that possess identical distributions of energy among the accessible quantum
levels (and differ only in which particular molecules occupy the levels) are known as configurations.
Because all microstates are equally probable, the probability of any one configuration is proportional to
the number of microstates that can produce it. Thus in the system shown above, the configuration labeled
ii will be observed 60% of the time, while iii will occur only 10% of the time.
huge ... Huge ... HUGE !
As the number of molecules and the number of quanta increases, the number of accessible microstates
grows explosively; if 1000 quanta of energy are shared by 1000 molecules, the number of available
microstates will be around 10600— a number that greatly exceeds the number of atoms in the observable
universe! The number of possible configurations (as defined above) also increases, but in such a way as to
greatly reduce the probability of all but the most probable configurations. Thus for a sample of a gas large
enough to be observable under normal conditions, only a single configuration (energy distribution
amongst the quantum states) need be considered; even the second-most-probable configuration can be
neglected.
The bottom line: any collection of molecules large enough in numbers to have chemical significance will
have its therrmal energy distributed over an unimaginably large number of microstates. The number of
microstates increases exponentially as more energy states ("configurations" as defined above) become
accessible owing to
Addition of energy quanta (higher temperature),
Increase in the number of molecules (resulting from dissociation, for example).
the volume of the system increases (which decreases the spacing between energy states, allowing
more of them to be populated at a given temperature.)
1.4 Energy-spreading changes the world
Energy is conserved; if you lift a book off the table, and let it fall, the total amount of energy in the world
remains unchanged. All you have done is transferred it from the form in which it was stored within the
glucose in your body to your muscles, and then to the book (that is, you did work on the book by
moving it up against the earth’s gravitational field). After the book has fallen, this same quantity of
energy exists as thermal energy (heat) in the book and table top.
What has changed, however, is the availability of this energy. Once the energy has spread into the huge
number of thermal microstates in the warmed objects, the probabiliy of its spontaneously (that is, by
chance) becoming un-dispersed is essentially zero. Thus although the energy is still “there”, it is forever
beyond utilization or recovery.
The profundity of this conclusion was recognized around 1900, when it was first described as the “heat
death” of the world. This refers to the fact that every spontaneous process (essentially every change that
occurs) is accompanied by the “dilution” of energy. The obvious implication is that all of the
molecular-level kinetic energy will be spread out completely, and nothing more will ever change. Not a
happy thought!
Why do gases tend to expand but never contract?
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Everybody knows that a gas, if left to itself, will tend to expand so as to fill the volume within which it is
confined completely and uniformly. What “drives” this expansion? At the simplest level it is clear that
with more space available, random motions of the individual molecules will inevitably disperse them
throughout the space. But as we mentioned above, the allowed energy states that molecules can occupy
are spaced more closely in a larger volume than in a smaller one. The larger the volume available to the
gas, the greater the number of energy states (and thus microstates) its thermal energy can occupy. Since all
such microstates within the thermally accessible range of energies are equally probable, the expansion of
the gas can viewed as a consequence of the tendency of energy to be spread and shared as widely as
possible. Once this has happened, the probability that this sharing of energy will reverse itself (that is, that
the gas will spontaneously contract) is so minute as to be unthinkable.
The same can in fact be said for even other highly probable distributions, such as having
49.999% of the molecules in the left half of the container and 50.001% in the right half. Even
though the number of possible configurations that would yield this distribution of molecules
is uncountably great, it is essentially negligible compared to the number that would
correspond to an exact 50-percent distribution.
Fig 2: Expansion of a gas
The illustration at the far right represents the allowed
thermal energy states of an ideal gas. The larger the
volume in which the gas is confined, the more
closely-spaced are these states, resulting in a huge
increase in the number of microstates into which the
available thermal energy can reside; this can be
considered the origin of the thermodynamic "driving
force" for the spontaneous expansion of a gas.
How energy spreading and sharing is related to heat and temperature
Just as gases spontaneously change their volumes from “smaller-to-larger”, the flow of heat from a
warmer body to a cooler always operates in the direction “warmer-to-cooler” because this allows thermal
energy to occupy a larger number of energy states as new ones are made available by bringing the cooler
body into contact with the warmer one; in effect, the thermal energy becomes more “diluted”.
Fig. 3: Heat flow and energy spreading
(a) Schematic depiction of the thermal energy states in two
separated identical bodies at different temperatures (indicated by
shading.)
(b) When the bodies are brought into thermal contact, thermal
energy flows from the higher occupied levels in the warmer object
into the unoccupied levels of the cooler one until equal numbers
are occupied
As you might expect, the increase in the amount of energy spreading and sharing is proportional to
amount of heat transferred q, but there is one other factor involved, and that is the temperature at which
the transfer occurs. When a quantity of heat q passes into a system at temperature T, the degree of dilution
of the thermal energy is given by q /T
To understand why we have to divide by the temperature, consider the effect of very large and very small
values of T in the denominator. If the body receiving the heat is initially at a very low temperature,
relatively few thermal energy states are occupied, so the amount of energy spreading can be very great.
Conversely, if the temperature is initially large, more thermal energy is already spread around within it,
and absorption of the same amount of energy will have a relatively small effect on the degree of thermal
disorder within the body.
Energy spreading and sharing in chemical reactions
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When a chemical reaction takes place, two kinds of changes relating to thermal energy are involved:
1. The ways that thermal energy can be stored within the reactants will generally be different from
those for the products. For example, in the reaction H2 → 2 H, the reactant dihydrogen possesses
vibrational and rotational energy states, while the atomic hydrogen in the product has translational
states only— but the total number of translational states in two moles of H is twice as great as in one
mole of H2 . Because of their extremely close spacing, translational states are the only ones that
really count at ordinary temperatures, so we can say that thermal energy can become twice as
diluted (“spread out”) in the product than in the reactant. If this were the only factor to consider,
then dissociation of dihydrogen would always be spontaneous and H2 would not exist.
2. In order for this dissociation to occur, however, a quantity of thermal energy (heat) q = DU must be
taken up from the surroundings in order to break the H–H bond. In other words, the ground state
(the energy at which the manifold of energy states begins) is higher in H, as indicated by the vertical
displacement of the right half in each of the four panels below.
Fig. 4: Energy levels in H2 and 2 H
The number of thermally accessible energy states (indicated by the shading)
increases with temperature, but because 2 moles of H possess twice as many
translational states as one mole of H2, dissociation becomes increasingly
favored at higher temperatures.
All molecules spontaneousy absorb heat and dissociate at high temperatures.
The ability of energy to spread into the product molecules is constrained by the availability of sufficient
thermal energy to produce these molecules. This is where the temperature comes in. At absolute zero the
situation is very simple; no thermal enegy is available to bring about dissociation, so the only component
present will be dihydrogen.
As the temperature increases, the number of populated energy states rises, as indicated by the shading in the
diagram. At temperature T1, the number of populated states of H2 is greater than that of 2H, so some of the
latter will be present in the equilibrium mixture, but only as the minority component.
At some temperature T2 the numbers of populated states in the two components of the reaction system will
be identical, so the equilibrium mixture will contain H2 and “2H” in equal amounts; that is, the mole ratio of
H2/H will be 1:2.
As the temperature rises to T3 and above, we see that the number of energy states that are thermally
accessible in the product begins to exceed that for the reactant.
The result is exactly what the LeChâtelier Principle predicts: the equilibrium state for an endothermic
reaction is shifted to the right at higher temperatures.
This is all very well for helping you understand the direct connection between energy spreading when a
chemical reaction occurs, but it is of little help in achieving our goal of predicting the direction and extent
of chemical change. For this, we need to incorporate the concept of energy spreading into
thermodynamics.
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Summary: the key concepts developed on this page
(You are expected to be able to define and explain the significance of terms identified in green sans-serif type.)
1. Spontaneous change is that which, once initiated, proceeds on its own until some state of
equilibrium (mechanical, thermal, chemical, etc.) is attained.
2. All natural processes (those that occur in the world) ultimately involve spontaneous change.
3. Thermal energy is kinetic energy associated with the random motions of atoms and molecules.
These states of motion are quantized into energy states which can be realized in huge numbers of
different ways that we call microstates.
4. All microstates are equally probable, but only those whose energy levels are commensurate with the
available thermal energy as determined by the temperature will be thermally accessible and
likely to be populated.
5. Spontaneous change is driven by the tendency of thermal energy to spread into as many microstates
of the system and surroundings as are thermally accessible.
Home page and index
What is entropy?
Stephen Lower is a retired member of the
Dept of Chemistry, Simon Fraser University
Burnaby/Vancouver Canada
contact the author
Steve's personal Web page
© 2003 by Stephen Lower
all rights reserved
Last modifed 10.9.2003
See here for other Chem1 Virtual Textbook topics
A PDF version of this document is available for printing
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Previous page: Direction of natural
processes
Page 2 of 6. - Index
Next page:The Second Law
What is entropy?
Reversible and irreversible changes
The physical meaning of entropy
Absolute entropies
Standard entropies of substances
Effect of temperature, volume, and concentration on the entropy
The preceding page explained how the tendency of thermal energy to disperse as widely as possible is
what drives all spontaneous processes, including, of course chemical reactions. We now need to
understand how the direction and extent of the spreading and sharing of energy can be related to
measurable thermodynamic properties of substances— that is, of reactants and products.
You will recall that when a quantity of heat q flows from a warmer body to a cooler one, permitting the
available thermal energy to spread into and populate more microstates, that the ratio q/T measures the
extent of this energy spreading. It turns out that we can generalize this to other processes as well, but there
is a difficulty with using q because it is not a state function; that is, its value is dependent on the pathway
or manner in which a process is carried out. This means, of course, that the quotient q/T cannot be a state
function either, so we are unable to use it to get differences between reactants and products as we do with
the other state functions. The way around this is to restrict our consideration to a special class of pathways
that are described as reversible.
Reversible and irreversible changes
A change is said to occur reversibly when it can be carried out in a series of infinitessimal steps, each
one of which can be undone by making a similarly minute change to the conditions that bring the
change about.
For example, the reversible expansion of a gas can be achieved by reducing the external
pressure in a series of infinitessimal steps; reversing any step will restore the system and the
surroundings to their previous state. Similarly, heat can be transferred reversibly between two
bodies by changing the temperature difference between them in infinitessimal steps each of
which can be undone by reversing the temperature difference.
The most widely cited example of an irreversible change is the free expansion of a gas into a vacuum.
Although the system can always be restored to its original state by recompressing the gas, this would
require that the surroundings perform work on the gas. Since the gas does no work on the surrounding in
a free expansion (the external pressure is zero, so PΔV = 0,) there will be a permanent change in the
surroundings. Another example of irreversible change is the conversion of mechanical work into
frictional heat; there is no way, by reversing the motion of a weight along a surface, that the heat released
due to friction can be restored to the system.
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Fig. 5. Reversible and irreversible gas
expansion and compression.
The diagrams show the same expansion
and compression ±ΔV carried out in
different numbers of steps ranging from a
single step at the top to an "infinite"
number of steps at the bottom. As the
number of steps increases, the processes
become less irreversible; that is, the
difference between the work done in
expansion and that required to
re-compress the gas diminishes. In the
limit of an ”infinite” number of steps
(bottom), these work terms are identical,
and both the system and surroundings (the
“world”) are unchanged by the
expansion-compression cycle. In all other
cases the system (the gas) is restored to
its initial state, but the surroundings are
forever changed.
In summary: a reversible change is one carried out in such as way that, when undone, both
the system and surroundings (that is, the world) remain unchanged.
Reversible = impossible: so why bother ?
It should go without saying, of course, that any process that proceeds in infinitissimal steps would take
infinitely long to occur, so thermodynamic reversibility is an idealization that is never achieved in real
processes, except when the system is already at equilibrium, in which case no change will occur anyway!
So why is the concept of a reversible process so important?
The answer can be seen by recalling that the change in the internal energy that characterizes any process
can be distributed in an infinity of ways between heat flow across the boundaries of the system and work
done on or by the system, as expressed by the First Law ΔU = q + w. Each combination of q and w
represents a different pathway between the initial and final states. It can be shown that as a process such as
the expansion of a gas is carried out in successively longer series of smaller steps, the absolute value of q
approaches a minimum, and that of w approaches a maximum that is characteristic of the particular
process. Thus when a process is carried out reversibly, the w-term in the First Law expression has its
greatest possible value, and the q-term is at its smallest. These special quantities wmax and qmin (which we
denote as qrev and pronounce “q-reversible”) have unique values for any given process and are therefore
state functions.
Fig. 6: work and reversibility
Note that the reversible condition
implies wmax and qmin. The
impossibility of extracting all of the
internal energy as work is essentially
a statement of the Second Law.
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Since qrev is a state function, so is qrev/T. This quotient is one of the most important quantities in
thermodynamics, because it expresses the change in energy spreading and sharing that accompanies a
process. Note carefully that this change is always related to the limiting (reversible) value even when the
process is carried out irreversibly and the actual value of q/T is different. Being a state function, qrev/T.
deserves a name and symbol of its own; it is called the entropy, designated by S. Since qrev/T describes a
change in state, we write the definition
Must know this!
...but if no real process can take place reversibly, what use is an expression involving qr e v? This is a
rather fine point that you should understand: although transfer of heat between the system and
surroundings is impossible to achieve in a truly reversible manner, this idealized pathway is only crucial
for the definition of ΔS; by virtue of its being a state function, the same value of ΔS will apply when the
system undergoes the same net change via any pathway. For example, the entropy change a gas
undergoes when its volume is doubled at constant temperature will be the same regardless of whether the
expansion is carried out in 1000 tiny steps (as reversible as patience is likely to allow) or by a single-step
(as irreversible a pathway as you can get!) expansion into a vacuum.
2.1 The physical meaning of entropy
Entropy is a measure of the degree of spreading and sharing of thermal energy within a system.
This “spreading and sharing” can be spreading of the thermal energy in space or its sharing amongst
previously inaccessible microstates of the system. The following table shows how this concept applies to a
number of common processes.
system and process
source of entropy increase of system
This has nothing to do with entropy because
macro objects are unable to exchange thermal
energy with the surroundings within the time scale
of the process
The cooler block contains more unoccupied
Two identical blocks of copper, one microstates, so heat flows from the warmer block
at 20°C and the other at 40°C, are
until equal numbers of microstates are populated
placed in contact.
in the two blocks.
A gas expands isothermally to twice A constant amount of thermal energy spreads over
its initial volume.
a larger volume of space
The increased thermal energy makes additional
microstates accessible. (The increase is by a factor
1 mole of water is heated by 1C°.
of about 1020,000,000,000,000, 000,000,000.)
The effect is the same as allowing each gas to
Equal volumes of two gases are
expand to twice its volume; the thermal energy in
allowed to mix.
each is now spread over a larger volume.
Some of the H2 dissociates to H because at this
One mole of dihydrogen, H2, is
temperature there are more thermally accessible
placed in a container and heated to microstates in the 2 moles of H. (See the column
3000K.
labled T4 in Table 4 in the previous section.)
A deck of cards is shuffled, or 100
coins, initially heads up, are
randomly tossed.
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The composition shifts back to virtually all H2
because this molecule contains more thermally
accessible microstates at low temperatures.
(Column T1 in Table 4.)
The above reaction mixture is
cooled to 300K.
Entropy is an extensive quantity; that is, it is proportional to the quantity of matter in a system; thus 100
g of metallic copper has twice the entropy of 50 g at the same temperature. This makes sense because the
larger piece of copper contains twice as many quantized states able to contain the thermal energy.
Entropy and disorder
Entropy is still described, particularly in older textbooks, as a measure of disorder. In a narrow technical
sense this is correct, since the spreading and sharing of thermal energy does have the effect of
randomizing the disposition of thermal energy within a system. But to simply equate entropy with
“disorder” without further qualification is extremely misleading because the qualifications that it relates
only to the ordering of thermal energy and only at the molecular scale are too easily forgotten. And once
this happens one is easily led to any number of false predictions. It is far better to avoid the term
“disorder” altogether in discussing entropy.
See Entropy is simple, qualitatively. Frank Lambert, J. Chem. Education 79(10) 2002: 1241-
As was explained in Part 1, the distribution of thermal energy in a system is characterized by the number
of quantized microstates that are available; the more of these there are, the greater the entropy of the
system. This is the basis of an alternative definition of entropy
S = k ln Ω
(7)
in which k is the Boltzmann constant (the gas constant per molecule, 1.38 10–23 J K–1) and Ω (omega) is
the number of microstates that correspond to a given macrostate of the system. The quantity Ω is an
unimaginably large number, typically around
that make up the earth is about 1050.
for one mole. By comparison, the number of atoms
The reason S depends on the logarithm of Ω is easy to understand. Suppose we have two
systems (containers of gas, say) with S1, Ω1 and S2, Ω2. If we now redefine this as a single
system (without actually mixing the two gases), then the entropy of the new system will be S =
S 1 + S2 but the number of microstates will be the product Ω1Ω2 because for each state of
system 1, system 2 can be in any of Ω2 states. Because ln(Ω1Ω2) = ln Ω1 + ln Ω2, the
additivity of the entropy is preserved.
2.2 Absolute entropies
Energy values, as you know, are all relative, and must be defined on a scale that is completely arbitrary;
there is no such thing as the absolute energy of a substance, so we can arbitrarily define the enthalpy or
internal energy of an element in its most stable form at 298K and 1 atm pressure as zero.
The same is not true of the entropy; since entropy is a measure of the “dilution” of thermal energy, it
follows that the less thermal energy available to spread through a system (that is, the lower the
temperature), the smaller will be its entropy. In other words, as the absolute temperature of a substance
approaches zero, so does its entropy.
This principle is the basis of the Third law of thermodynamics, which states that the entropy of a
perfectly-ordered solid at 0° K is zero.
The absolute entropy of a substance at any temperature above 0° K must be determined by calculating
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the increments of heat q required to bring the substance from 0° K to the temperature of interest, and then
summing the ratios q/T . Two kinds of experimental measurements are needed:
1. The enthalpies associated with any phase changes the substance may undergo within the temperature
range of interest. Melting of a solid and vaporization of a liquid correspond to sizeable increases in the
number of microstates available to accept thermal energy, so as these processes occur, energy will flow
into a system, filling these new microstates to the extent required to maintain a constant temperature (the
freezing or boiling point); these inflows of thermal energy correspond to the heats of fusion and
vaporization. The entropy increase associated with melting, for example, is just ΔHfusion/Tm.
2. The heat capacity C of a phase expresses the quantity of heat required to change the temperature by a
small amount ΔT , or more precisely, by an infinitessimal amount dT . Thus the entropy increase brought
about by warming a substance over a range of temperatures that does not encompass a phase transition is
given by the sum of the quantities C dT/T for each increment of temperature dT . This is of course just the
integral
Because the heat capacity is itself slightly temperature dependent, the most precise determinations of
absolute entropies require that the functional dependence of C on T be used in the above integral in place
of a constant C . For rough determinations, one can simply plot the enthalpy increment C ΔT as a function
of temperature and measure the area under the curve.
Fig. 7: How the entropy of water
changes with temperature
As the temperature rises, more
microstates become accessible,
allowing thermal energy to be more
widely dispersed. The vertical steps
correspond to phase changes in which
the number and spacing of
microstates changes massively and
discontinuously.
2.3 Standard entropies of substances
The standard entropy of a substance is its entropy at 1 atm pressure. The values found in tables are
normally those for 298K, and are expressed in units of J K–1 mol–1. The table below shows some typical
values for gaseous substances.
He
126
H2
131
CH4
186
Ne
146
N2
192
H2O(g)
187
Ar
155
CO
197
CO2
213
Kr
164
F2
203
C2H6
229
Xe
170
O2
205
n -C3H8
270
Cl2
223
n -C4H10
310
Table 1: Standard entropies of some gases at 298K, J K–1 mol–1
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Note especially how the values given in this table illustrate these important points:
Although the standard internal energies and enthalpies of these substances would be zero, the
entropies are not. This is because there is no absolute scale of energy, so we conventionally set the
“energies of formation” of elements in their standard states to zero. Entropy, however, measures not
energy itself, but its dispersal amongst the various quantum states available to accept it, and these
exist even in pure elements.
It is apparent that entropies generally increase with molecular weight. For the noble gases, this is of
course a direct reflection of the principle that translational quantum states are more closely packed
in heavier molecules, allowing of them to be occupied.
The entropies of the diatomic and polyatomic molecules show the additional effects of rotational
quantum levels.
C(diamond)
2.5
C(graphite)
5.7
Fe
Pb
Na
S(rhombic)
27.1 51.0 64.9 32.0
Si
W
18.9 33.5
Table 2: Entropies of some solid elements at 298 K, J K–1 mol–1
The entropies of the solid elements (Table 2) are strongly influenced by the manner in which the atoms
are bound to one another. The contrast between diamond and graphite is particularly striking; graphite,
which is built up of loosely-bound stacks of hexagonal sheets, appears to be more than twice as good at
soaking up thermal energy as diamond, in which the carbon atoms are tightly locked into a
three-dimensional lattice, thus affording them less opportunity to vibrate around their equilibrium
positions. Looking at all the examples in the above table, you will note a general inverse correlation
between the hardness of a solid and its entropy. Thus sodium, which can be cut with a knife, has almost
twice the entropy of iron; the much greater entropy of lead reflects both its high atomic weight and the
relative softness of this metal. These trends are consistent with the oft-expressed principle that the more
“disordered” a substance, the greater its entropy.
Gases, which serve as efficient
vehicles for spreading thermal energy
over a large volume of space, have
much higher entropies than condensed
phases.
solid
41
liquid
70
gas
186
Table 3: Entropy of water at 298K, J K–1
mol–1
Similarly, liquids, in which the molecular units can interact with each
other in a multiplicity of ways (that is, more microstates) have higher
entropies than solids. An especially interesting comparison can be made
by extrapolating the entropy of ice as measured at 0°C to 25°C so that it
can be compared to the values for liquid and gaseous water at the same
temperature.
2.4 Effect of temperature, volume, and concentration on the entropy
As discussed previously, the entropy of a substance always increases with temperature. This is due mainly
to the increased number of ways that the thermal energy can be distributed amongst the allowed energy
levels as the latter become accessible. The rate of increase dS/dT is just the ratio of the heat capacity to the
temperature C/T . When integrated over a range of temperatures, this yields (for an ideal gas confined to a
fixed volume)
In general, a larger volume also leads to increased entropy. For an ideal gas that expands at a constant
temperature (meaning that it absorbs heat from the surroundings to compensate for the work it does
during the expansion), the increase in entropy is given by
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(10)
Because the pressure of a gas is inversely proportional to its volume, we can easily alter the above relation
to express the entropy change associated with a change in the pressure of a perfect gas:
(11)
The pressure of a gas is directly proportional to its concentration in moles per liter, so we can re-cast this
equation in terms of concentrations:
(12)
Although these equations strictly apply only to perfect gases and cannot be used at all for liquids and
solids, it turns out that in a dilute solution, the solute can often be treated as a gas dispersed in the volume
of the solution, so the last equation can actually give a fairly accurate value for the entropy of dilution of
a solution. We will see later that this has important consequences in determining the equilibrium
concentration of a homogeneous reaction mixture.
Summary: the key concepts developed on this page
(You are expected to be able to define and explain the significance of terms identified in green sans-serif
type.)
1. A reversible process is one carried out in infinitessimal steps which, when undone, both the system and
surroundings (that is, the world) remain unchanged. (See the example of gas expansion-compression
given above.) Although true reversible change cannot be realized in practice, it can always be
approximated.
2. The sum of the heat (q) and work (w) associated with a process is a state function defined by the
First Law ΔU = q + w. Heat and work themselves are not state functions, and therefore depend on the
particular pathway in which a process is carried out.
3. As a process is carried out in a more reversible manner, the value of w approaches its maximum possible
value, and q approaches its minimum possible value.
4. Although q is not a state function, the quotient qrev/T is, and is known as the entropy.
5. Entropy is a measure of the degree of the spreading and sharing of thermal energy within a system.
6. The entropy of a substance increases with its molecular weight and complexity and with temperature.
The entropy also increases as the pressure or concentration becomes smaller. Entropies of gases are
much larger than those of condensed phases.
7. The absolute entropy of a pure substance at a given temperature is the sum of all the entropy it would
acquire on warming from absolute zero (where S=0) to the particular temperature.
Energy spreading and the
direction of spontaneous change
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Stephen Lower is a retired member of the
Dept of Chemistry, Simon Fraser University
Burnaby/Vancouver Canada
contact the author
Steve's personal Web page
© 2003 by Stephen Lower
all rights reserved
Last modifed
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Previous page: What is entropy?
Page 3 of 6. - Index
Next page: Gibbs free energy
The second law of thermodynamics
The direction of spontaneous change (again)
What is a heat engine, and why should you care?
The Second Law: what it means
You will recall that the First Law of thermodynamics, expressed as ΔU = q + w, is essentially a
statement of the law of conservation of energy. The significance of this law is that it tells us that any
proposed process that would violate this condition can be dismissed as impossible, without even inquiring
further into the details of the process.
The second law of thermodynamics goes beyond this by saying, in effect, that the extent to which any
natural process can occur is limited by the dilution of thermal energy (increase in entropy) that
accompanies it, and once the change has occurred, it can never be un-done without spreading even more
energy around.
For example, a piece of ice placed in a warm room will quickly melt, spreading the heat of
fusion it absorbs from the surroundings into the much larger number of energy microstates
available in liquid water. At some later time we can always re-freeze the water by placing it in
a refrigerator, but decreasing the entropy of the water by this means will increase the entropy
of the surroundings by a greater amount as the heat removed by the refrigerator is dissipated
into the surroundings.
There are several formal ways of stating the Second Law, but these will not mean much until we get to the
section on heat engines. For the time being, it is best to just express the most important consequence of
this law: just because the energy is “there” does not mean it will be available to do anything useful. For
example, one might propose a scheme to propel a ship by means of a machine that takes in water, extracts
part of its thermal energy which is used to rotate the propellor, and then tosses the resulting ice cubes
overboard. Such a device would be formally called a perpetual motion machine of the second kind
and would violate the Second Law. (A perpetual motion machine of the first kind is one that would
violate the First Law.)
The U.S. Patent Office frequently receives applications to patent devices whose operation
would not be in accord with the Second Law; in the majority of cases the inventor appears to
be unaware of this fact or, for that matter, of the Second Law. For some time, it has been the
practice of the Patent Office to require that a working model of the device be made available
to verify its operation.
2.6 The direction of spontaneous change
The entropy of the world only increases
All natural processes that allow the free exchange of thermal energy amongst chemically-significant
numbers of particles are accompanied by a spreading or “dilution” of energy that leaves the world forever
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changed. In other words, all spontaneous change leads to an increase in the entropy of the world. At
first sight, this might seem to be inconsistent with our observations of very common instances in which
there is a clear decrease in entropy, such as the freezing of a liquid, the formation of a precipitate, or the
growth of an organism.
... but it’s the entropy of the system plus surroundings that counts! It is important to understand that the
criterion for spontaneous change is the entropy change of the system and the surroundings- that is, of the
“world”, which we denote by ΔStotal:
ΔStotal = ΔSsystem + ΔSsurroundings
The only way the entropy of the surroundings can be affected is by exchange of heat with the system; if
the system absorbs (that is, withdraws from the surroundings) a quantity of heat q , then
ΔSsurroundings = (–q) / T
Note that it does not matter whether the change in the system occurs reversibly or irreversibly;
as mentioned previously, it is always possible to define an alternative (irreversible) pathway in
which the amount of heat exchanged with the surroundings is the same as qrev ; because ΔS is
a state function, the entropy change of the surroundings will have the same value as for the
unrealizable reversible pathway.
If there is no flow of heat into or out of the surroundings, the entropy change of the system and that
of the world are identical. Examples of such processes, which are always spontaneous, are the free
expansion of an ideal gas into a vacuum, and the mixing of two ideal gases. In practice, almost all
processes involving mixing and diffusion can be regarded as driven exclusively by the entropy increase
of the system.
Most processes involving chemical and phase changes involve the exchange of heat with the
surroundings, so their tendency to occur cannot always be predicted by focussing attention on the system
alone. Further, owing to the –q/T term in ΔSsurroundings, the spontaneity of all such processes will depend
on the temperature, as we illustrated for the dissociation of H2 previously.
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As a quantitative example, let us consider the freezing of water. We know that liquid
water will spontaneously change into ice when the temperature drops below 0°C at 1
atm pressure. Since the entropy of the solid is less than that of the liquid, we know the
entropy of the water (the system here) will decrease on freezing. The amount of
decrease is found by dividing the heat of fusion of ice by the temperature for the
reversible pathway, which occurs at the normal freezing point:
If the process is actually carried at 0°C, then the heat of fusion is transferred to the
surroundings at the same temperature, and the entropy of the surroundings increases by
so that ΔStotal = 0. Under these conditions the process can proceed in either direction
(freezing or melting) without affecting the entropy of the world; this means that both
ice and liquid water can be present simultaneously without any change occurring; the
system is said to be in equilibrium.
Suppose now that the water is supercooled to –1°C before it freezes. The entropy change
of the water still corresponds to the reversible value qrev/T = (–6000J)/(273K). The
entropy change of the surroundings, however, is now given by
The total entropy change is now
ΔStotal = (–21.978 + 22.059) J K–1 mol–1 = +0.081 J K–1 mol–1
indicating that the process can now occur (“is spontaneous”) only in the one direction.
Why did we use 273 K when evaluating ΔSsystem and 272 K for calculating ΔSsurroundings?
In the latter case it is possible to formulate a reversible pathway by which heat can be
transferred to the surroundings at any temperature. ΔSsystem, however, is a state
function of water, and will vary with temperature only slightly.
Note that in order to actually freeze water, it must be cooled to very slightly below its
normal freezing point, a condition known as supercooling. Freezing of supercooled water
is of course an irreversible process (once it starts, it cannot be stopped except by raising
the temperature by a finite amount), and the positive value of ΔStotal tells us that this
process will occur spontaneously at temperatures below 273K. Under these conditions,
the process is driven by the entropy increase of the surroundings resulting from flow of
the heat of fusion of water into the surroundings.
In any spontaneous macroscopic change, the entropy of the world increases.
If a process is endothermic or exothermic, heat is exchanged with the surroundings, and we have to
consider the entropy change of the surroundings as well as that of the system in order to predict what the
direction of change will be. Failure to take the surroundings into account is what leads to the apparent
paradox that freezing of a liquid, compression of a gas, and growth of an organism are all processes in
which entropy decreases. It is, of course, only the entropy of the system that undergoes a decrease when
these processes occur.
Does the entropy of the world ever decrease?
The principle that thermal energy (and the molecules containing it) tends to spread out is based on simple
statistics. It must be remembered, however, that the laws of probability have meaningful application only
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to systems made up of large numbers of independent actors. If you trap a hundred flies in a bottle, they
will generally distribute themselves more or less uniformly throughout the container; if there are only four
flies, however, it is quite likely that all of them will occasionally be located in one particular half of the
bottle.
Why the sky is blue. Similarly, you can trust with complete certainty that the spontaneous
movement of half the molecules of the air to one side of the room you now occupy will not
occur, even though the molecules are moving randomly and independently. On the other
hand, if we consider a box whose dimensions are only a few molecular diameters, then we
would expect that the random and short-term displacement of the small number of particles it
contains to one side of the box would occur quite frequently. This is, in fact, the cause of the
blueness of the sky: random fluctuations in the air density over tiny volumes of space whose
dimensions are comparable with the wavelength of light results in selective scattering of the
shorter wavelengths, so that blue light is scattered out, leaving the red light for the enjoyment
of sunset-watchers to the east.
Brownian motion. This refers to the irregular zig-zag-like movement of extremely small
particles such as plant pollen when they are suspended in a drop of liquid. Any such particle is
continually being buffeted by the thermal motions of the surrounding liquid molecules. If size
of the particle is very large compared to that the the liquid molecules, the forces that result
from collisions of these molecules with the particle will cancel out and the particle remains
undisturbed. If the particle is very small, however (perhaps only a thousand times larger than
a molecule of the liquid), then the chances that it will undergo sufficiently more hits from one
direction than from another during a brief interval of time become significant.
In these two examples, the entropy of the system decreases without any compensating flow of heat into
the surroundings, leading to a net (but only temporary) decrease in the entropy of the world. This does
not represent a failure of the Second Law, however, because no one has ever devised a way to extract
useful work from these processes.
2.7 What is a heat engine, and why should you care?
The Industrial Revolution of the 19th century
was largely driven by the invention of the steam
engine. The first major use of such engines was
to pump water out of mines, whose flooding
from natural seepage seriously limited the depths
to which they could be driven, and thus the
availablility of the metal ores that were essential
to the expansion of industrial activities. The
steam engine is a type of heat engine, a device
that converts heat, provided by burning a fuel,
into mechanical work, typically delivered
through the motion of a piston in opposition to
an opposing force. An engine is therefore an
energy conversion device in which, ideally,
every joule of heat released by combustion of
the fuel could be extracted as work at the output
shaft; such an engine would operate at 100
percent efficiency. However, engineers of the
time were perplexed to find that the efficiencies of steam engines were rather low (usually around 20%),
with most of the heat being exhausted uselessly to the environment. Everyone understood that an
efficiency exceeding 100% would be impossible (that would violate conservation of energy, and thus the
First Law), but it was not clear why efficiencies could not rise significantly beyond the small values
observed even as mechanical designs improved
The answer was found by a young French engineer, Sadi Carnot, who in 1824 published an
analysis of an idealized heat engine that is generally considered to be the foundation of the
science of thermodynamics— notwithstanding the fact that Carnot still accepted the belief
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that heat is a fluid-like substance called “caloric”. We will not replicate his analysis here (this
is normally done in more advanced courses in physical chemistry), but will simply state his
conclusion in his own [translated] words:
"The production of motive power is then due in steam-engines not to an actual consumption of
caloric, but to its transportation from a warm body to a cold body... the production of heat alone is not sufficient to
give birth to the impelling power: it is necessary that there should also be cold; without it, the heat would be
useless. The ultimate attainable efficiency of any heat engine will depend on the temperatures at which heat is
supplied to and removed from it."
The fall of heat
The left side of the figure represents a
generalized heat engine into which a
quantity of heat qH, extracted from a
source or “reservoir” at temperature TH
is partly converted into work w. The
remainder of the heat qL is exhausted to
a reservoir at a lower temperature TL. In
practice, TH would be the temperature of
the steam in a steam engine, or the
temperature of the combustion mixture
in an internal combustion or turbine
engine. The low temperature reservoir is
ordinarily that of the local environment.
The efficiency ε (epsilon) of a heat
engine is the fraction of the heat
abstracted from the high temperature
reservoir that can be converted into
work:
ε = w/qH
Fig. 8: Efficiency of a heat engine
Left: common schematic representation of a
heat engine. Right: diagramatic
representation of Eq. 14 (below); the
efficiency is the ratio of the temperature
intervals a :b.
Carnot’s crucial finding (for which he would certainly have deserved a Nobel prize if these had existed at
the time) is that the efficiency is proportional to the ``distance'' in temperature that the heat can “fall” as it
passes through the engine:
(14)
This is illustrated graphically in the right half of Fig. 8, in which the efficiency is simply the
fraction of the “complete” fall (in temperature) to absolute zero (arrow b) that the heat
undergoes in the engine (arrow a.) Clearly, the only way to attain 100% efficiency would be
to set the temperature of the exhaust reservoir to 0°K, which would be impossible. For most
terrestrial heat engines, TL is just the temperature of the environment, normally around 300K,
so the only practical way to improve the efficieny is to make TH as high as possible. This is
the reason that high pressure (superheated) steam is favored in commercial thermal power
plants. The highest temperatures (and the greatest operating efficiencies) are obtained in gas
turbine engines. However, as operating temperatures rise, the costs of dealing with higher
steam pressures and the ability of materials such as turbine blades to withstand high
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temperatures become significant factors, placing an upper limit of around 600K on TH, thus
imposing a maximum of around 50 percent efficiency on thermal power generation.
For nuclear plants, in which safety considerations require lower steam pressures, the
efficiency is lower. One consequence of this is that a larger fraction of the heat is exhausted to
the environment, which may result in greater harm to aquatic organisms when the cooling
water is returned to a stream or estuary.
Problem Example:
Several proposals have been made for building a heat engine that makes use of
the temperature differential between the surface waters of the ocean and cooler
waters that, being more dense, reside at greater depth. If the exhaust temperature
is 5°C, what is the maximum amount of work that could be extracted from 1000
L of surface water at 10°C? (The specific heat capacity of water is 4.184 J
g – 1K– 1.)
Solution: The amount of heat (qH) that must be extracted to cool the water by 5
K is (4.184 J g– 1K– 1)(106 g)(5 K) = 2.09 ¥ 107 J. The ideal thermodynamic
efficiency is given by
The amount of work that could be done would be (.018)(2.09 ¥ 107 J) = 3.7 ¥
106 J
Comment: It may be only 1.8% efficient, but it’s free!
Heat pumps
If a heat engine is run “in reverse” by performing work on it (that is, changing “work out” to “work in” in
Fig 8), it becomes a device for transporting heat against a thermal gradient. Refrigerators and air
conditioners are the most commonly-encountered heat pumps. A heat pump can also be used to heat the
interior of a building. In this application, the low temperature reservoir can be a heat exchanger buried in
the earth or immersed in a well. In this application heat pumps are more efficient than furnaces or electric
heating, but the capital cost is rather high.
Pumping heat uphill: how heat pumps work
The Second Law: what it means
It was the above observation by Carnot that eventually led to the formulation of the Second Law of
Thermodynamics near the end of the 19th Century. One statement of this law (by Kelvin and Planck) is
as follows:
It is impossible for a cyclic process connected to a reservoir at one temperature to produce a positive
amount of work in the surroundings.
To help you understand this statement and how it applies to heat engines, consider the schematic heat
engine in the figure in which a working fluid (combustion gases or steam) expands aginst the restraining
force of a weight that is mechanically linked to the piston. From a thermodynamic perspective, the
working fluid is the system and everything else is surroundings. Expansion of the fluid occurs when it
absorbs heat from the surroundings; return of the system to its initial state requires that the surrounding do
work on the system. Now re-read the above statement of the Second Law, paying special attention to the
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italicized phrases which are explained below:
A cyclic process is one in which the system returns to its inital state. A simple steam engine
undergoes an expansion step (the power stroke), followed by a compression (exhaust stroke) in
which the piston, and thus the engine, returns to its initial state before the process repeats.
“At one temperature” means that the expansion and compression steps operate isothermally. This
means that ΔU = 0; just enough heat is absorbed by the system to perform the work required to raise
the weight, so for this step q = –w.
“A positive amount of work in the surroundings” means that the engine does more work on the
surroundings than the surroundings do on the engine. Without this condition the engine would be
useless.
Note carefully that the Second Law applies only to a cyclic process— isothermal expansion of a gas
against a non-zero pressure always does work on the surroundings, but an engine must repeat this process
continually; to do so it must be returned to its initial state at the end of every cycle. When operating
isothermally, the work –w it does on the surroundings in the expansion step (power stroke) is nullified by
the work +w the surroundings must do on the system in order to complete the cycle.
(Review the graphs of Fig. 5 comparing the work associated with expansion and compression for
single-and multistep processes.
The Second Law can also be stated in an alternative way:
It is impossible to construct a machine operating in cycles that will convert heat into work without
producing any other changes. (Max Planck)
Thus the Second Law does allow an engine to convert heat into work, but only if “other changes”
(transfer of a portion of the heat directly to the surroundings) are allowed. And since heat can only flow
spontaneously from a source at a higher temperature to a sink at a lower temperature, the impossibility of
isothermal conversion of heat into work is implied.
A device that accomplishes the isothermal conversion of heat into work (essentially Fig. 8
with the low-temperature sink missing) is known as a perpetual motion machine of the
second kind. (A perpetual motion machine of the first kind violates the First Law.) The U.S.
Patent and Trademark office is said to receive about 1500 applications for patents on such
devices every year; many of the inventors have no idea that their proposals would violate the
laws of thermodynamics.
The search for perpetual motion machines is a rich history of human folly that goes back to
the 13th century and continues to this day in the form of goofy schemes and not a few scams.
Visit Eric's History of Perpetual Motion and Free Energy Machines for a nicely organized
overview.
Summary: the key concepts developed on this page
(You are expected to be able to define and explain the significance of terms identified in green sans-serif type.)
1. In any macroscopic change, the entropy of the world (that is, system + surroundings) always increases; it
never decreases.
2. Processes that do not exchange heat with the surroundings (such as the free expansion of a gas into a
vacuum) involve entropy change of the system alone, and are always spontaneous.
3. A heat engine is a device that converts heat into work. The fraction of heat that can be converted into work is
limited by the fall in temperature between the input to the engine and the exhaust.
4. According to the Second Law of Thermodynamics, complete conversion of heat into work by a spontaneous
cyclic process is impossible.
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What is entropy?
What is free energy? The Gibbs
function
Stephen Lower is a retired member of the
Dept of Chemistry, Simon Fraser University
Burnaby/Vancouver Canada
contact the author
Steve's personal Web page
© 2003 by Stephen Lower
all rights reserved
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Previous page: The Second Law
Page 4 of 6. - Index
Next page: Free energy and equilibrium
Free energy: the Gibbs function
Physical meaning of the Gibbs function
The standard Gibbs free energy
Why are chemical reactions affected by the temperature?
Free energy, concentrations and escaping tendency
In the previous section we saw that it is the sum of the entropy changes of the system and surroundings
that determines whether a process will occur spontaneously. In chemical thermodynamics we prefer to
focus our attention on the system rather than the surroundings, and would like to avoid having to
calculate the entropy change of the surroundings explicity. The key to doing this is to define a new state
function known as the Gibbs free energy
G = H – TS
(15)
Since H, T and S are all state functions, G is a function of state. This relation is more usefully expressed in
terms of the change
ΔG = ΔH – TΔS
(16)
Must know this!
The Δ-quantities refer to changes of the system; in particular, ΔS is the entropy change of the system and
(being a state function) has a fixed value for a given change. Note, however, that the ΔH term represents
the heat qp that is gained or lost by the system, and that this same quantity of heat (but with opposite sign)
is lost or gained by the surroundings.
For a process that takes place at a temperature T, we can write the total entropy change as the sum of those
for the surroundings and for the system, expressed in terms of the heat that flows into
(16)
Multiplying through by –T , we obtain
which expresses the entropy change of the world in terms of thermodynamic properties of the system
exclusively. If –TΔStotal is denoted by ΔG, then we have Eq. 16 which defines the Gibbs free energy
change for the process.
From the foregoing, you should convince yourself that G (now often referred to as the Gibbs function
rather than as free energy) will decrease in any process occurring at constant temperature and pressure
which is accompanied by an overall increase in the entropy. (The constant temperature and pressure are a
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consequence of the temperature and the enthalpy appearing in the preceding equation.) Since most
chemical and phase changes of interest to chemists take place under such conditions, the Gibbs function is
the most useful of all the thermodynamic properties of a substance, and it is closely linked to the
equilibrium constant.
Problem Example 2
One mole of an ideal gas at 300 K is allowed to expand slowly and isothermally to twice
its initial volume. Calculate q, w, ΔS°sys, ΔS°surr, and ΔG° for this process.
Solution: Assume that the process occurs slowly enough that the it can be considered to
take place reversibly.
a) The work done in a reversible expansion is
so w = (1 mol)(8.314 J mol–1 K–1)(300K)(ln 2) = 1730 J.
b) In order to maintain a constant temperature, an equivalent quantity of
heat must be absorbed by the system: q = 1730 J.
c) The entropy change of the system is given by Eq. 10 (Part 2):
ΔS°sys = R ln (V2/V1) = (8.314 J mol–1 K–1)(ln 2) = 5.76 J mol–1 K–1
d) As was explained in Part 2, the entropy change of the surroundings,
formally defined as qrev/T, is
ΔS°surr = q/T = (–1730 J) / (300 K) = – 5.76 J mol–1 K–1.
e) The free energy change is
ΔG° = ΔH° – T ΔS°sys = 0 – (300 K)(5.76 J K–1) = –1730 J
Comment: Recall that for the expansion of an ideal gas, ΔH° = 0. Note also that
because the expansion is assumed to occur reversibly, the entropy changes in (c) and (d)
cancel out, so that ΔS°world = 0.
Physical meaning of the Gibbs function
Its physical meaning: the maximum work.
As we explained in Part 2, the two quantities qrev (qmin) and wmax associated with reversible processes are
state functions. We gave the quotient qrev/T a new name, the entropy. What we did not say is that wmax
also has its own name, the free energy.
The Gibbs free energy is the maximum useful work (excluding PV work associated with volume changes
of the system) that a system can do on the surroundings when the process occurs reversibly at constant
temperature and pressure.
This work is done at the expense of the internal energy of the system, and whatever part of that is not
extracted as work is exchanged with the surroundings as heat; this latter quantity will have the value –TΔS
.
The Gibbs function: is it free? Is it energy?
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The appellation “free energy” for G has led to so much confusion that many scientists now refer to it
simply as the Gibbs function. The “free” part of the name reflects the steam-engine origins of
thermodynamics with its interest in converting heat into work: ΔG = wmax, the maximum amount of
energy which can be “freed” from the system to perform useful work. A much more serious difficulty,
particularly in the context of chemistry, is that although G has the units of energy (joules, or in its
intensive form, J mol– 1), it lacks one of the most important attributes of energy in that it is not conserved.
Thus although the free energy always falls when a gas expands or a chemical reaction takes place, there
need be no compensating increase in energy anywhere else. Referring to G as an energy also reinforces
the false but widespread notion that a fall in energy must accompany any change. But if we accept that
energy is conserved, it is apparent that the only necessary condition for change (whether the dropping of
a weight, expansion of a gas, or a chemical reaction) is the redistribution of energy. The quantity –ΔG
associated with a process represents the quantity of energy that is “shared and spread”, which as we have
already explained is the meaning of the increase in the entropy. The quotient –ΔG/T is in fact identical
with ΔStotal, the entropy change of the world, whose increase is the primary criterion for any kind of
change.
Who was Gibbs, anyway?
J. Willard Gibbs is considered the father of modern thermodynamics and the
most brilliant American-born scientist of the 19th century. He did most of his
work in obscurity, publishing his difficult-to-understand papers in the
Proceedings of the Connecticut Academy of Sciences, a little-read journal that
was unknown to most of the world.
Some Gibbs links: brief bio - scientific biography
A remarkably rich account of Gibbs' seemingly gray life and his unusual powers of visualization was
written by a noted American poet: Rukeyser, M., Willard Gibbs. Garden City, N.J.: Doubleday Duran and
Co., Inc., 1942.
3.1 The standard Gibbs free energy
In order to make use of free energies to predict chemical changes, we need to know the free energies of
the individual components of the reaction. For this purpose we can combine the standard enthalpy of
formation and the standard entropy of a substance to get its standard free energy of formation
ΔGf ° =ΔHf ° – TΔSf °
and then determine the standard Gibbs free energy of the reaction according to
ΔG° = ΔGf °(products) – ΔGf °(reactants)
As with standard heats of formation, the standard free energy of a substance represents the free energy
change associated with the formation of the substance from the elements in their most stable forms as
they exist under the standard conditions of 1 atm pressure and 298K.tandard Gibbs free energies of
formation are normally found directly from tables. Once the values for all the reactants and products are
known, the standard Gibbs free energy change for the reaction is found in the normal way.
The interpretation of ΔG° for a chemical change is very simple. For a reaction A Æ B, one of the
following three situations will always apply:
ΔG° < 0 net change to the right: A
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The interpretation of ΔG° for
a chemical change is very
simple. For a reaction A Æ B,
one of the three situations
illustrated here will always
apply:
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Table 4: ΔG° as a criterion for spontaneous change
Some textbooks and teachers still say that the ΔG° and thus the criterion for chemical change depends on the two
factors ΔH° and ΔS°, and they sometimes even refer to reactions in which one of these terms dominates as "energy
driven" or "entropy driven" processes. This can be extremely misleading! As explained at the top of this page, all
processes are entropy driven; when –ΔH° exceeds the –TΔS° term, this merely means that the entropy change of the
surroundings is the greater contributor to the entropy change of the system.
Why are chemical reactions affected by the temperature?
The TΔS° term in the equation ΔG° =ΔH° – TΔS° tells us that the temperature dependence of ΔG°
depends almost entirely on the entropy change associated with the process. (We say "almost" because the
values of ΔH° and ΔS° are themselves slightly temperature dependent; both gradually increase with
temperature). In particular, notice that the sign of the entropy change determines whether the reaction
becomes more or less spontaneous as the temperature is raised. Since the signs of both ΔH° and ΔS°
can be positive or negative, we can identify four possibilities as outlined in the table below.
Exothermic reaction, ΔS > 0
C(graphite) + O2 (g)
CO2 (g)
ΔH° = –393 kJ
ΔS° = +2.9 J K–1
ΔG° = –394 kJ at 298 K
This combustion reaction, like most such reactions,
is spontaneous at all temperatures. The positive
entropy change is due mainly to the greater mass of
CO2 compared to O2.
Exothermic reaction, ΔS° < 0
3 H2 + N2
2 NH3 (g)
ΔH° = –46.2 kJ
ΔS° = –389 J K–1
ΔG° = –16.4 kJ at 298 K
The decrease in moles of gas in the Haber ammonia
synthesis drives the entropy change negative. Thus
higher T, which speeds up the reaction, also reduces
its extent.
Endothermic reaction, ΔS° > 0
N2 O4 (g)
2 NO2 (g)
ΔH° = –57.1 kJ
ΔS° = +176 J K–1
ΔG° = +4.8 kJ at 298 K
Dissociation reactions are typically endothermic
with positive entropy change. Ultimately, all
molecules decompose to their atoms at sufficiently
high temperatures.
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Endothermic reaction, ΔS° < 0
1/2 N2 + O2
NO2 (g)
ΔH° = 33.2 kJ
ΔS° = –249 J K–1
ΔG° = +51.3 kJ at 298 K
Although NO2 is thermodynamically unstable, the
reverse of this reaction is kinetically hindered, so
this oxide can exist indefinitely at ordinary
temperatures.
Table 5: Effects of temperature on reaction spontaniety
Must know this!
The plots on the left show how ΔH° and TΔS° combine according to determine the
range of temperatures over which a reaction can take place spontaneously. Be sure
you understand these graphs and can reproduce them for each of the four sign
combinations of ΔH° and ΔS°. These relations are of course governed by the relative
entropies of the reactants and products, illustrated schematically by the spacing of
the quantized energy states; never forget that it is the ability of thermal energy to
spread into as many of these states as possible that ultimately determines the
tendency of a process to take place. The column on the right of the table shows an
example of each class of reaction.
Finding the equilibrium temperature
Notice that the reactions which are spontaneous only above or below a certain temperature have identical
signs for ΔH° and ΔS°. In these cases there will be a unique temperature at which ΔH° = TΔS° and thus
ΔG° = 0 , corresponding to chemical equilibrium. This temperature is given by
(19)
and corresponds to the point at which the lines representing ΔH° and TΔS° cross in a plot of these two
quantities as a function of the temperature.
The values of both ΔH° and ΔS° are themselves somewhat temperature dependent, both
increasing with the temperature. This means that substitution of 298 K values of these
quantities into Eq. 19 will give only an estimate of the equilibrium temperature if this differs
greatly from 298 K.Nevertheless, the general conclusions regarding the temperature
dependence of reactions such as those shown in Table 5 are always correct, even if the
equilibrium temperature cannot be predicted accurately.
Why do most substances have definite melting and boiling points
As a crystalline substance is heated it eventually melts to a liquid, and finally vaporizes. The transitions
between these phases occur abruptly and at definite temperatures; except at the melting and boiling
temperatures, only one of these three phases will be stable— that is, its free energy will be lower than that
of either of the other two phases. To understand the reason for this, recall that as the temperature rises,
TΔS° increases faster than ΔS°, so the free energy (which of course depends on –TΔS°) always falls with
temperature. Furthermore, the temperature dependence of G depends on the entropy (we won’t try to
prove this here), so G falls off fastest for the high-entropy gas phase, and for liquids faster than for solids.
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Fig. 9: Free energy of solid, liquid
and gaseous water as a function of
temperature.
These plots of G° as a function of the
temperature explain why substances
have definite melting and boiling
points. The stable phase is always
the one that has the lowest Gibbs
free energy. The temperatures at
which one phase becomes more
stable than the others are the
melting and boiling points.
In Fig. 9 the free energies of each of the three phases of water are plotted as a function of temperature. At
any given temperature, one of these phases will have a lower free energy than either of the others; this will
of course be the stable phase at that temperature. The cross-over points where two phases have identical
free energies are the temperatures at which both phases can coexist— in other words, the melting and
boiling temperatures. At any other temperature, there is only a single stable phase.
3.2 Free energy, concentrations and escaping tendency
The free energy of a pure liquid or solid at 1 atm pressure is just its molar free energy of formation ΔG°
multiplied by the number of moles present. For gases and substances in solution, we have to take into
account the concentration (which, in the case of gases, is normally expressed in terms of the pressure). We
know that the lower the concentration, the greater the entropy (Eq. 12 in Part 2), and thus the smaller the
free energy.
The free energy of a gas: Standard states
The free energy of a gas depends on its pressure; the higher the pressure, the higher the free energy. Thus
the free expansion of a gas, a spontaneous process, is accompanied by a fall in the free energy. Using Eq.
11 in Part 2 we can express the change in free energy when a gas undergoes a change in pressure from P1
to P2 as
(20)
How can we evaluate the free energy of a specific sample of a gas at some arbitrary pressure? First, recall
that the standard molar free energy G° that you would look up in a table refers to a pressure of 1 atm. The
free energy per mole of our sample is just the sum of this value and any change in free energy that would
occur if the pressure were changed from 1 atm to the pressure of interest
(21)
which we normally write in abbreviated form
G = G° + RT ln P (22)
Escaping tendency
The higher the pressure of a gas, the greater will be the tendency of its molecules to leave the confines of
the container; we will call this the escaping tendency. Equation (22) above tells us that the pressure of a
gas is a directly observable measure of its free energy (G, not G°). Combining these two ideas, we can say
that the free energy of a gas is also a measure of its escaping tendency. The latter term is not used in
traditional thermdynamics because it is essentially synonymous with the free energy, but it is worth
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knowing because it helps us appreciate the physical significance of free energy in certain contexts.
Thermodynamics of mixing and dilution
All substances, given the opportunity to form a homogeneous mixture with other substances, will tend to
become more dilute. This can be rationalized simply from elementary statistics; there are more equally
probable ways of arranging one hundred black marbles and one hundred white marbles, than two
hundred marbles of a single color. For massive objects like marbles this has nothing to do with entropy, of
course. But when we are dealing with huge numbers of molecules capable of storing, exchanging and
spreading thermal energy, mixing and expansion are definitely entropy-driven processes. It can be
argued, in fact, that mixing and expansion are really very similar; after all, when we mix two gases, each
is expanding into the space formerly occupied exclusively by the other.
Fig. 10: Entropy and free energy
of mixing
The two ideal gases in (c) are
separated by a barrier. When the
barrier is removed (d), the two gases
spontaneously mix. ΔS and ΔG for
this process are just twice what they
would be for the expansion of a
single gas (a) to twice its volume (b).
In terms of the spreading of thermal energy the situation is particularly dramatic; the addition of even a
single molecule of B to one mole of a gas A results in a huge increase in the number of
energically-identical (degenerate) microstates that correspond to the interchange of every molecule in the
gas with the new molecule. When actual molar quantities of two gases mix, the number of new microstates
created is beyond comprehension.
Fig. 11: Energy spreading
in expansion and mixing
The tendency of a gas to expand is
due to the more closely-spaced
energy states in the larger volume
(b). When one molecule of a
different kind is introduced into
the gas (c), each microstate in (b)
splits into a huge number of
energetically-identical new states,
denoted (inadequately) by the
dashed lines in (c).
The entropy increase that occurs when the concentration of a substance is reduced through dilution or
mixing is given by Eq. 12 in Part 2. We can similarly define the Gibbs free energy of dilution or
mixing by substituting this equation into the definition of ΔG°:
(23)
If the substance in question forms an ideal solution with the other components, then ΔHdil is by definition
zero, and we can write
(24)
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These relations tell us that the dilution of a substance from an initial concentration C1 to a more dilute
concentration C2 is accompanied by a decrease in the free energy, and thus will occur spontaneously.
Fig. 12: The equilibrium
state for mixing
If identical numbers of
molecules of two gases are
allowed to mix as in c-d of
Fig. 10 above, the
equilibrium mole fraction of
each gas will be 0.5.
By the same token, the spontaneous “un-dilution” of a solution will not occur (we do not expect the tea to
diffuse back into the teabag!) However, un-dilution can be forced to occur if some means can be found to
supply to the system an amount of energy (in the form of work) equal to ΔGdil. An important practical
example of this is the metabolic work performed by the kidney in concentrating substances from the
blood for excretion in the urine.
To find the free energy of a solute at some arbitrary concentration, we proceed in very much the same
way as we did for a gas: we take the sum of the standard free energy, and any change in the free energy
that would accompany a change in concentration from the standard state to the actual state of the solution.
Using Eq. 24 it is easy to derive an expression analogous to Eq. 22
G = G° + RT ln C (25)
which gives the free energy of a solute at some arbitrary concentration C in terms of its value G° in its
standard state.
Although this expression has the same simple form as Eq. 22, its practical application is
fraught with difficulties, the major one being that it doesn’t usually give values of G that are
consistent with experiment, especially for solutes that are ionic or are slightly soluble. In such
solutions, intermolecular interactions between solute molecules and between solute and
solvent bring back the enthalpy term that we left out in deriving Eq. 22 (and thus Eq. 25). In
addition, the structural organization of the solution becomes concentration dependent, so that
the entropy depends on concentraiton in a more complicated way than is implied by Eq. 12 in
Part 2.
Activity and standard state of the solute
Instead of complicating G° by trying to correct for all of these effects, chemists have chosen to retain its
simple form by making a single small change in the form of 25:
G = G° + RT ln a (26)
This equation is guaranteed to work, because a, the activity of the solute, is its thermodynamically
effective concentration. The relation between the activity and the concentration is given by
a = gC (27)
where g (gamma) is the activity coefficient. As the solution becomes more dilute, the activity coefficient
approaches unity:
(28)
The price we pay for the simplicity of 26 is that the relation between the concentration and the activity at
higher concentrations can be quite complicated, and must be determined experimentally for every
different solution.
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The question of what standard state we choose for the solute (that is, at what concentration is
G° defined, and in what units is it expressed?) is one that you will wish you had never asked!
We might be tempted to use a concentration of 1 molar, but a solution this concentrated
would be subject to all kinds of intermolecular interaction effects, and would not make a very
practical standard state. These effects could be eliminated by going to the opposite extreme of
an “infinitely dilute” solution, but by Eq. 25 this would imply a free energy of minus infinity
for the solute, which would be awkward. Chemists have therefore agreed to define the
standard state of a solute as one in which the concentration is 1 molar, but all
solute-solute interactions are magically switched off, so that g is effectively unity. Since this is
physically impossible to achieve, no solution corresponding to this standard state can actually
exist, but this turns out to be only a small drawback, and seems to be the best compromise
between convenience, utility, and reality.
Summary: the key concepts developed on this page
(You are expected to be able to define and explain the significance of terms identified in green sans-serif type.)
1. The Gibbs free energy is a state function defined as G = H – TS .
2. The Gibbs free energy is the maximum useful work that a system can do on the surroundings when the process
occurs reversibly at constant temperature and pressure.
3. The practical utility of the Gibbs function is that ΔG for any process is negative if it leads to an increase in
the entropy of the world. Thus spontaneous change at a given temperature and pressure can only occur when
it would lead to a decrease in G.
4. The sign of the standard free energy change ΔG° of a chemical reaction determines whether the reaction will
tend to proceed in the forward or reverse direction.
5. Similarly, the relative signs of ΔG° and ΔS° determine whether the spontaniety of a chemical reaction will be
affected by the temperature, and if so, in what way.
6. The existence of sharp melting and boiling points reflects the differing temperature dependancies of the free
energies of the solid, liquid, and vapor phases of a pure substance, which are in turn reflect their differing
entropies.
The Second Law
Free energy and equilibrium
Stephen Lower is a retired member of the
Dept of Chemistry, Simon Fraser University
Burnaby/Vancouver Canada
contact the author
Steve's personal Web page
© 2003 by Stephen Lower
all rights reserved
Last modifed 10.9.2003
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Previous page: Gibbs free energy
Page 5 of 6. - Index
Next page: Applications of free energy
Free energy and equilibrium
Approaching equilibrium: the free energy can only fall
The equilibrium constant
Equilibrium and temperature
Equilibrium without mixing: it's all or nothing
Coupled reactions
All natural processes tend to proceed in a direction that leads to the maximum possible spreading and
sharing of thermal energy. This principle, which we developed in Part 1, led to the definition of entropy
(in Part 2) as a measure of the degree of energy dispersal. We saw that every process and change that
happens in the world is accompanied by an increase in the entropy of the system-plus-surroundings. In
the previous part of this lesson we introduced the concept of the Gibbs free energy (the Gibbs function)
that expresses the tendency for change in terms of the properties of the system alone under conditions of
constant temperature and pressure: chemical change will tend to occur in whatever direction leads to a
decrease in the value of the Gibbs function.
Approaching equilibrium: the free energy can only fall
This means, of course, that if the total free energy G of a mixture of reactants and products goes through a
minimum value as the composition changes, then all net change will cease— the reaction system will be in
a state of chemical equilibrium. You will recall that the relative concentrations of reactants and products
in the equilibrium state is expressed by the equilibrium constant. In this lesson we will examine the
relation between the Gibbs free energy change for a reaction and the equilibrium constant.
To keep things as simple as possible, we will consider a homogeneous chemical reaction of the form
A+B
C+D
in which all components are gases at the temperature of interest. If the standard free energies of the
products is less than that of the reactants, ΔG° for the reaction will be negative and the reaction will
proceed to the right. But how far? If the reactants are completely transformed into products, the
equilibrium constant would be infinity. The equilibrium constants we actually observe all have finite
values, implying that even if the products have a lower free energy than the reactants, some of the latter
will always remain when the process comes to equilibrium.
In order to understand how equilibrium constants relate to ΔG° values, assume that all of the reactants are
gases, so that the free energy of gas A, for example, is given at all times by
GA = GA° + RT ln PA (29)
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(This relation was developed here in Part 4.)The free energy change for the reaction is sum of the free
energies of the products, minus that of the reactants:
ΔG = GC + GD – GA – GB (30)
Using Eq. 29 to expand each term on the right, we have
ΔG = (G°C + RT ln PC) + (G°D + RT ln PD) – (G°B + RT ln PB) – (G°A + RT ln PA) (31)
We can now express the G° terms collectively as ΔG°, and combine the logarithmic terms into a single
fraction
(32)
which is more conveniently expressed in terms of the reaction quotient QP
ΔG = ΔG° + RT ln Q (33) (important equation!)
What this all means: when G is as small as it can get, the reaction is in equilibrium
The free energy G is a quantity that becomes more negative during the course of any natural process.
Thus as a chemical reaction takes place, G only falls and will never become more positive. Eventually a
point is reached where any further transformation of reactants into products would cause G to increase. At
this point G is at a minimum (see the plot below), and no further change can take place; the reaction is at
equilibrium.
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in the dissociation of N2 O4
The gas-phase reaction N2 O4
2 NO2
furnishes a simple example of the free
energy relationships in a homogeneous
reaction. The free energy of 1 mole of
N2O4 (1) is smaller than that of 2 moles
of NO2 (2) by 5.3 kJ; thus ΔG° = +5.3 kJ
for the complete transformation of
reactants into products. The staight
diagonal line shows the free energy of
all possible compositions if the two
gases were prevented from mixing. The
red curved line show the free energy of
the actual reaction mixture. This passes
through a minimum at (3) where 0.814
mol of N2O4 are in equilibrium with
0.372 mol of NO2. The difference (4)
corresponds to the free energy of mixing
of reactants and products which always
results in an equilibrium mixture whose
free energy is lower than that of either
pure reactants or pure products. Thus
some amount of reaction will occur even
if ΔG° for the process is positive.
What’s the difference between ΔG and
ΔG°?
It’s very important to be aware of this
distinction; that little ° symbol makes a
world of difference! First, the standard free
energy change ΔG° has a single value for a
particular reaction at a given temperature
and pressure; this is the difference
(ΣG°f, products – ΣG°f, reactants)
that you get from tables. It corresponds to
the free energy change for a process that
never really happens: the complete
transformation of pure N2O4 into pure NO2
at a constant pressure of 1 atm.
The other quantity ΔG, defined by Eq. 33,
represents the total free energies of all
substances in the reaction mixture at any
particular system composition.
How ΔG varies with composition (see Figure above)
ΔG is the “distance” (in free energy) from the equilibrium state of a given reaction.
Thus for a sample of pure N2O4 or NO2 (as far from the equilibrium state as the system
can be!),
Q = [NO2]2/[N2O4] is +
or 0, respectively, making the logarithm in Eq 33, and thus
the value of ΔG, +
(1) or – (2). As the reaction proceeds in the appropriate
direction ΔG approaches zero; once there (3), the system is at its equilibrium
composition and no further net change will occur.
In contrast to ΔG° which is a constant for a given reaction, ΔG varies continuously as the composition
changes, finally reaching zero at equilibrium.
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The importance of mixing of reactants and products.
We are now in a position to answer the question posed earlier: if ΔG° for a reaction is negative, meaning
that the free energies of the products are more negative than those of the reactants, why will some of the
latter remain after equilibrium is reached? The answer is that no matter how low the free energy of the
products, the free energy of the system can be reduced even more by allowing some of the products to be
contaminated (i.e., diluted) by some reactants. Owing to the entropy associated with mixing of reactants
and products, no homogeneous reaction will be 100% complete. An interesting corollary of this is that
any reaction for which a balanced chemical equation can be written can in principle take place to some
extent.
3.4 The equilibrium constant
Now let us return to Eq. 33 which we reproduce here:
ΔG = ΔG° + RT ln Qp
(33)
As the reaction approaches equilibrium, ΔG becomes less negative and finally reaches zero. At
equilibrium ΔG = 0 and Q = K, so we can write
ΔG° = RT ln Kp (34) (Must know this!)
in which Kp ,the equilibrium constant expressed in pressure units, is the special value of Q that corresponds
to the equlibrium composition.
This equation is one of the most important in chemistry because it relates the equilibrium composition of a
chemical reaction system to measurable physical properties of the reactants and products. If you know the
entropies and the enthalpies of formation of a set of substances, you can predict the equilibrium constant
of any reaction involving these substances without the need to know anything about the mechanism of the
reaction.
Instead of writing Eq. 34 in terms of Kp , we can use any of the other forms of the equilibrium
constant such as Kc (concentrations), Kx (mole fractions), Kn (numbers of moles), etc.
Remember, however, that for ionic solutions especially, only the Ka, in which activities are
used, will be strictly valid.
It is often useful to solve Eq. 34 for the equilibrium constant, yielding
(35) (Must be able to derive this !)
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Fig. 14: The equilibrium
constant and ΔG°
These two plots show this
relation linearly (left) and
logarithmically (right). Notice
that an equilibrium constant
of unity implies a standard
free energy change of zero,
and that positive values of
ΔG° lead to values of K less
than unity.
3.5 Equilibrium and temperature
We have already discussed how changing the temperature will increase or decrease the tendency for a
process to take place, depending on the sign of ΔS°. This relation can be developed formally by
differentiating the relation
ΔG °= ΔH °– TΔS° (36)
with respect to the temperature:
(37)
... so the sign of the entropy change determines whether the reaction becomes more or less allowed as the
temperature increases.
We often want to know how a change in the temperature will affect the value of an equilibrium constant
whose value is known at some fixed temperature. Suppose that the equilibrium constant has the value K1
at temperature T1 and we wish to estimate K2 at temperature T2 . Expanding Eq. 34 in terms of ΔH° and
ΔS°, we obtain
–RT1 ln K1 = ΔH ° – T1 ΔS° and –RT2 ln K2 = ΔH ° – T2 ΔS°
Dividing both sides by RT and subtracting, we obtain
(38)
Which is most conveniently expressed as the ratio
(39) Important equation!
Do you remember the Le Châtelier Principle? Here is its theoretical foundation with respect
to the effect of the temperature on equilibrium: if the reaction is exothermic (ΔH° < 0), then
increasing temperature will make the second exponential term smaller and K will decrease—
that is, the equilibrium will “shift to the left”. If ΔH° > 0 then increasing T will make the
exponent less negative and K will increase.
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This is an extremely important relationship, but not just because of its use in calculating the temperature
dependence of an equilibrium constant. Even more important is its application in the “reverse” direction
to experimentally determine ΔH° from two values of the equilibrium constant measured at different
temperatures. Direct calorimetric determinations of heats of reaction are not easy to make; relatively few
chemists have the equipment and experience required for this rather exacting task. Measurement of an
equilibrium constant is generally much easier, and often well within the capabilities of anyone who has
had an introductory Chemistry course. Once the value of ΔH° is determined it can be combined with the
Gibbs free energy change (from a single observation of K, through Eq. 34) to allow ΔS° to be calculated
through Eq. 36.
Problem Example 3
The vapor pressure of solid ammonium perchlorate was found to be 0.026 torr at 520 and 2.32
torr at 620 K. Addition of NH3 gas was observed to repress the vaporization, suggesting that
the equilibrium under study is
NH4ClO4(s)
NH3(g) + HClO4(g)
Use this information to calculate ΔH° and ΔS° for this process.
Solution. In problems of this kind the first step is usually to express the equilibrium constant
in terms of the observed pressures. From the stoichiometry of this reaction it is apparent that
Kp = P(NH3) × P(HClO4)
and that each of these partial pressures is just half the vapor pressure of the solid P, so that
Kp = (.026)/(760 × 2). This gives K1 = 2.92 × 10–10 and K2 = 2.33 × 10–6. Substituting these
into Eq. 39 and solving for the enthalpy change gives ΔH° = 241 kJ mol–1. Next, we
substitute the two equilibrium constants into Eq. 34 and obtain ΔG1° = 94.9 kJ mol–1 and
ΔG2° = 56.1 kJ mol–1. The entropy change is estimated from Eq. 37, using the non-calculus
form Δ(ΔG°)/ΔT since the functional relation between ΔS° and the temperature is not known.
This gives ΔS° = +338 J mol–1K –1. Notice that the sign of this entropy change could have
been anticipated by inspection of the reaction equation, since the volume increase due to the
gaseous products will dominate other entropy differences.
Equilibrium without mixing: it's all or nothing
You should now understand that for homogeneous reactions (those that take place entirely in the gas
phase or in solution) the equilibrium composition will never be 100% products, no matter how much
lower their free energy relative to the reactants. As was summarized in Fig.13, this is due to "dilution" of
the products by the reactants. In heterogeneous reactions (those which involve more than one phase)
this dilution, and the effects that flow from it, may not be possible.
A particularly simple but important type of a heterogeneous process is phase change.
Consider, for example, an equilibrium mixture of ice and liquid water. The concentration of
H2 O in each phase is dependent only on the density of the phase; there is no way that ice can
be “diluted” with water, or vice versa. This means that at all temperatures other than the
freezing point, the lowest free energy state will be that corresponding to pure ice or pure
liquid. Only at the freezing point, where the free energies of water and ice are identical, can
both phases coexist, and they may do so in any proportion.
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Fig. 15: Free energy of ice-water system
Only at 0°C can ice and liquid water coexist in any proportion. Note that in contrast to
Fig. 13 (above), there is no free energy minimum at intermediate compositions.
3.6 Coupled reactions
Two reactions are said to be coupled when the product of one of them is the reactant in the other:
A
B
B
C
If the standard free energy of the first reaction is positive but that of the second reaction is sufficiently
negative, then for the overall process will be negative and we say that the first reaction is “driven” by the
second one. This, of course, is just another way of describing an effect that you already know as the Le
Châtelier principle: the removal of substance B by the second reaction causes the equilibrium of the first to
“shift to the right”. Similarly, the equilibrium constant of the overall reaction is the product of the
equilibrium constants of the two steps.
1 Cu2 S(s)
2 S(s) + O2 (g)
3 Cu2 S(s)
2 Cu(s) + S(s)
ΔG° = + 86.2 kJ
ΔH° = + 76.3 kJ
SO2 (g)
ΔG° = –300.1 kJ ΔH° = + 296.8 kJ
2 Cu(s) + SO2 (g)
ΔG° = –213.9 kJ ΔH° = – 217.3 kJ
In the above example, reaction 1 is the first step in obtaining metallic copper
is obtained from one of its principal ores. This reaction is both endothermic
and has a positive free energy change, so it will not proceed spontaneously at
any temperature. If Cu2S is heated in the air, however, the sulfur is removed
as rapidly as it is formed by oxidation in the highly spontaneous reaction 2,
which supplies the free energy required to drive 1. The combined process,
known as roasting, is of considerable industrial importance and is one of a
large class of processes employed for winning metals from their ores which are
described in more detail in Part 6 of this lesson.
Summary: the key concepts developed on this page
(You are expected to be able to define and explain the significance of terms identified in green sans-serif type.)
1. As a homogeneous chemical reaction proceeds, the free energies of the reactants become more negative and
those of the products more positive as the composition of the system changes.
2. The total free energy of the system (reactants + products) always becomes more negative as the reaction
proceeds. Eventually it reaches a minimum value at a system composition that defines the equilibrium
composition of the system, after which time no further net change will occur ( Fig. 13.)
3. The equilibrium constant for the reaction is determined the standard free energy change:
ΔG° = RT ln Kp
4. The sign of the temperature dependence of the equilibrium constant is governed by the sign of ΔH°. This is
the basis of the Le Châtelier Principle.
5. The free energies of solid and liquid components are constants that do not change with composition. Thus in
heterogeneous reactions such as phase changes, the total free energy does not pass through a minimum and
when the system is not at equilibrium only all-products or all-reactants will be stable.
6. Two reactions are coupled when the product of one reaction is consumed in the other. If ΔG° for the first
reaction is positive, the overall process can still be spontaneous if ΔG° for the second reaction is sufficiently
negative— in which case the second reaction is said to "drive" the first reaction.
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What is free energy?
Stephen Lower is a retired member of the
Dept of Chemistry, Simon Fraser University
Burnaby/Vancouver Canada
contact the author
Steve's personal Web page
Some applications of free energy
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Previous page: Free energy and
equilibrium
Page 6 of 6. - Index
Some applications of entropy and free energy
Colligative properties of solutions
Extraction of metals from their oxides
Bioenergetics
Oxidation-reduction: the fall of the electron
Acid-base reactions: the fall of the proton
4.1 Colligative properties of solutions
Vapor pressure lowering, boiling point elevation, freezing point depression and osmosis are well-known
phenomena that occur when a non-volatile solute such as sugar or a salt is dissolved in a volatile solvent
such as water. All these effects result from “dilution” of the solvent by the added solute, and because of
this commonality they are referred to as colligative properties (Lat. co ligare, connected to.) The key
role of the solvent concentration is obscured by the expressions used to calculate the magnitude of these
effects, in which only the solute concentration appears. The details of how to carry out these calculations
and the many important applications of colligative properties are covered in the unit on solutions. Our
purpose here is to offer a more complete explanation of why these phenomena occur.
Basically, these all result from the effect of dilution (of the solvent) on its entropy, and thus in the increase
in the density of quantum states of the system in the solution compared to that in the pure liquid.
Equilibrium between two phases (liquid-gas for boiling and solid-liquid for freezing) occurs when equal
numbers of microstates are occupied in each phase. The temperatures at which this occurs are depicted by
the shading in the Figure below. (See here in Part 1 for a more complete explanation of this type of
diagram.)
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Diution of the solvent adds new
Dilution of the solvent adds new
energy states to the liquid, but
quantum states to the liquid, but
does not affect the vapor phase.
does not affect the solid phase.
This raises the temperature
This reduces the temperature
required to make equal numbers of
required to make equal numbers of
microstates accessible in the two
states accessible in the two
phases.
phases.
Fig. 16: Explanation of bp elevation and fp depression in terms of energy
spreading (entropy.)
The shifts of the bp and fp in opposite directions reflects the higher energies of
vapor microstates and lower energies of solid microstates in relation to those of
the liquid. Note that these diagrams are purely schematic and nowhere near to
scale!
The more conventional explanation of bp elevation and fp depression utilizes plots of –TΔS° for each
phase in terms of the temperature.
Melting- and boiling points of a pure substance
The stable phase at any temperature will be the
one having the lowest free energy, and from
which molecules have the smallest escaping
tendency.
Melting- and boiling points of a solvent
containing a non-volatile solute.
The solute "dilutes" the solvent, reducing its
free energy and shifting the transition
temperatures in opposite directions.
Fig. 17: Conventional explanation of bp elevation and fp depression
Osmotic pressure: effects of pressure on the entropy
When a liquid is subjected to hydrostatic pressure— for example, by an inert, non-dissolving gas that
occupies the vapor space above the surface, the vapor pressure of the liquid is raised. The pressure acts to
compress the liquid very slightly, effectively narrowing the potential energy well in which the individual
molecules reside and thus increasing their tendency to escape from the liquid phase. (Because liquids are
not very compressible, the effect is quite small; a 100-atm applied pressure will raise the vapor pressure of
water at 25°C by only about 2 torr.) In terms of the entropy, we can say that the applied pressure reduces
the dimensions of the "box" within which the principal translational motions of the molecules are confined
within the liquid, thus reducing the density of the quantized energy states in the liquid phase.
This phenomenon can explain osmotic pressure. Osmotic pressure, students must be reminded, is not
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what "drives" osmosis, but is rather the hydrostatic pressure that must be applied to the more concentrated
solution (more dilute solvent) in order to stop osmotic flow of solvent into the solution. The effect of this
pressure P is to slightly increase the spacing of solvent energy states on the high-pressure (dilute-solvent)
side of the membrane to match that of the pure solvent, restoring osmotic equilibrium.
4.2 Extraction of metals from their oxides
Since ancient times, the recovery of metals from their ores has been
one of the most important applications of chemistry to civilization and
culture. The oldest, and still the most common smelting process for
oxide ores involves heating them in the presence of carbon. Originally,
charcoal was used, but industrial-scale smelting uses coke, a crude
form of carbon prepared by pyrolysis (heating) of coal.
See an interesting History of Metals Web page
(this illustration found on a page on copper smelting
shows a 16th-century smelting operation)
The basic reactions are:
MO + C
M + CO
MO + ½ O2
M + ½ CO2
MO + CO
M + CO2
(i)
(ii)
(iii)
Each of these can be regarded as a pair of coupled reactions
in which the metal M and the carbon are effectively competing
for the oxygen atom. Using (i) as an example, we can write
MO
M + ½ O2
C + ½ O2
CO
(iv)
(v)
At ordinary environmental temperatures, reaction (iv) is always
spontaneous in the reverse direction (that is why ores form in the first place!), so ΔG°(iv) will be positive.
ΔG°(v) is always negative, but at low temperatures it will not be sufficiently negative to drive (iv).
The smelting process depends on the different ways in which the free energies of reactions like (iv) and
(v) vary with the temperature. This temperature dependence is almost entirely dominated by the TΔS°
term in the Gibbs function, and thus by the entropy change. The latter depends mainly on Δn g , the
change in the number of moles of gas in the reaction. Removal of oxygen from the ore is always
accompanied by a large increase in the system volume so ΔS for this step is always positive and the
reaction becomes more spontaneous at higher temperatures. The temperature dependences of the reactions
that take up oxygen vary, however:
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reaction
C + ½ O2
CO
C + O2
CO2
CO + ½ O2
CO2
Δn g
d(ΔG°)/dT
.5
<0
0
0
–.5
>0
A plot of the temperature dependences of the free energies of these reactions, superimposed on similar
plots for the oxygen removal reactions (iv) is called an Ellingham diagram. In order for a given oxide
MO to be smeltable, the temperature must be high enough that (iv) falls below that of at least one of the
oxygen-consuming reactions. The slopes of the lines on this diagram are given by
Ellingham diagram
An ore can be reduced by carbon
only if its Gibbs free energy of
formation falls below that of
one of the carbon reduction
reactions. Practical refining
temperatures are generally
limited to about 1500°K.
Examination of the Ellingham diagram shown above illustrates why the metals known to the ancients were
mainly those such as copper and lead, which can be obtained by smelting at the relatively low
temperatures that were obtainable by the methods available at the time in which a charcoal fire supplied
both the heat and the carbon. Thus the bronze age preceded the iron age; the latter had to await the
development of technology capable of producing higher temperatures, such as the blast furnace. Smelting
of aluminum oxide by carbon requires too high temperatures to be practical; commercial production of
aluminum is accomplished by electrolysis of the molten ore.
4.3 Bioenergetics
Many of the reactions that take place in living organisms require a source of free energy to drive them.
The immediate source of this energy in heterotrophic organisms, which include animals, fungi, and most
bacteria, is the sugar glucose. Oxidation of glucose to carbon dioxide and water is accompanied by a large
negative free energy change
C6 H1 2O6 + O2
6 CO2 + 6 H2 O
ΔG° = – 2880 kJ mol– 1 (40)
Of course it would not do to simply “burn” the glucose in the normal way; the energy change would be
wasted as heat, and rather too quickly for the well-being of the organism. Effective utilization of this free
energy requires a means of capturing it from the glucose and then releasing it in small amounts when and
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where it is needed. This is accomplished by breaking down the glucose in a series of a dozen or more
steps in which the energy liberated in each stage is captured by an “energy carrier” molecule, of which the
most important is adenosine diphosphate, known as ADP. At each step in the breakdown of glucose, an
ADP molecule reacts with inorganic phosphate (denoted by Pi ) and changes into ATP:
ADP + Pi
ATP
ΔG° = +30 kJ mol– 1
Structures of glucose and ATP
The 30 kJ mol– 1 of free energy stored in each ATP molecule is released when the molecule travels to a
site where it is needed and loses one of its phosphate groups, yielding inorganic phosphate and ADP,
which eventually finds its way back the site of glucose metabolism for recycling back into ATP. The
complete breakdown of one molecule of glucose is coupled with the production of 38 molecules of ATP
according to the overall reaction
C6 H1 2O6 + 6 O2 + 38 Pi + 38 ADP
38 ATP + 6CO2 + 44 H2 O
For each mole of glucose metabolized, 38 × (30 kJ) = 1140 kJ of free energy is captured as ATP,
representing an energy efficiency of 1140/2880 = 0.4. That is, 40% of the free energy obtainable from
the oxidation of glucose is made available to drive other metabolic processes. The rest is liberated as heat.
Where does the glucose come from? Animals obtain their glucose from their food, especially cellulose and
starches that, like glucose, have the empirical formula {CH2 O}. Animals obtain this food by eating plants
or other animals. Ultimately, all food comes from plants, most of which are able to make their own
glucose from CO2 and H2 O through the process of photosynthesis. This is just the reverse of Eq 40 in
which the free energy is supplied by the quanta of light absorbed by chlorophyll and other photosynthetic
pigments.
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The photosynthesis-respiration free enrgy cycle
The zigzags represent the individual steps of energy capture by ADP and its
delivery to metabolic processes by ATP. "Thank a green plant every day."
This describes aerobic respiration, which evolved after the development of photosynthetic life on
Earth began to raise the concentration of atmospheric oxygen. Oxygen is a poison to most life processes at
the cellular level, and it is believed that aerobic respiration developed as a means to protect organisms
from this peril. Those that did not adapt have literally “gone underground” and constitute the more
primitive anaerobic bacteria.
The function of oxygen in respiration is to serve as an acceptor of the electrons that glucose loses when it
undergoes oxidation. Other electron acceptors can fulfill the same function when oxygen is not available,
but none yields nearly as much free energy. For example, if oxygen cannot be supplied to mammalian
muscle cells as rapidly as it is needed, they switch over to an anaerobic process yielding lactic acid instead
of CO2 :
C6 H1 2O6 + 2 ADP
2 CH3 CH(OH)COOH
ΔG° = –218 kJ mol– 1
In this process, only (2 × 30 kJ) = 60 kJ of free energy is captured, so the efficiency is only 28% on the
basis of this reaction, and it is even lower in relation to glucose. In “aerobic” exercising, one tries to
maintain sufficient lung capacity and cardiac output to supply oxygen to muscle cells at a rate that
promotes the aerobic pathway.
The fall of the electron
Oxidation-reduction reactions proceed in a direction that allows the electron to “fall” (in free energy)
from a “source” to a “sink”. Later on when you study electrochemistry you will see how this free energy
can manifest itself as an electrical voltage and be extracted from the system as electrical work. In
organisms, the free energy is captured in a series of coupled reactions as described above, and eventually
made available for the various free energy-consuming processes associated with life processes.
The figure below shows the various electron sources (foodstuffs) and sinks (oxidizing agents) on a
vertical scale that illustrates the relative free energy of an electron in each sink. Notice that carbohydrate is
at the top of this scale and dioxygen is at the bottom, indicating that O2 is the best possible electron
acceptor in terms of free energy yield. For a similar chart relating to the familiar electromotive series of
the elements, see here.
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Electron free energy levels of bioenergetic interest
The zero for G° is arbitrarily set at the electron activity at which the H2/H+
couple is at equilibrium; this corresponds to E° = 0 volts on the ordinary
electromotive scale. The pE(w) scale on the left represents the electron
activity in air-saturated water and is analogous to the pH scale for proton
activity. See the link below for more information.
The pdf document Redox equilibria in natural
waters includes a detailed treatment of pE and
of pE-pH diagrams. Download / View this PDF
file now
Organisms that live in environments where oxygen is lacking, such as marshes, muddy soils, and the
intestinal tracts of animals, must utilize other electron acceptors to extract free energy from carbohydrate.
A wide variety of inorganic ions such as sulfate and nitrate, as well as other carbon compounds can serve
as electron acceptors, yielding the gaseous products like H2 S, NH3 and CH4 which are commonly noticed
in such locations. From the location of these acceptors on the scale, it is apparent that the amount of
energy they can extract from a given quantity of carbohydrate is much less than for O2 . One reason that
aerobic organisms have dominated the earth is believed to be the much greater energy-efficiency of
oxygen as an electron acceptor.
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What did organisms use for food before there was a widespread supply of carbohydrate in the world? Any
of the electron sources near the upper left of the table can in theory serve this function, although at
reduced energy efficiency. As a matter of fact, there are still a number of these autotrophic bacteria
around whose “food” is CH4 , CH3 OH, FeCO3 , and even H2 !
The fall of the proton
According to the widely useful Brønsted-Lowry concept, an acid is a proton donor and a base is a
proton acceptor. In 1953, Gurney showed how this idea could be made even more useful by placing
acid-base conjugate pairs on a proton-free energy scale.
In this view, acids are proton sources
and bases are proton sinks. Protons fall
spontaneously from acids to fill sinks
in which the proton free energy levels
are lower. The pH is a measure of the
average proton free energy in the
solution; when this quantity is the
same as the proton free energy level of
a conjugate pair, the two species are
present in equal concentrations (this
corresponds, of course to the equality
of pH and pKa in the conventional
theory.)
The proton-free energy concept is commonly employed in aquatic environmental chemistry in which
multiple acid-base systems must be dealt with on a semi-quantitative bases. It is, however, admirably
adapted to any presentation of acid-base chemistry, even at the first-year college level, and it seems a
shame that it never seems to have made its way into the ordinary curriculum. A nice overview and a more
complete diagram can be seen on this Web page.
See also the following PDF documents for much more detailed
treatments of the proton-free energy acid-base concept:
Acid-base equilibria and calculations covers the quantitative
treatment of acid-base equilibria at somewhat greater breadth and
depth than is available in standard textbooks. Download/view this
pdf file now
Acid-base chemistry of natural aquatic systems is similar to the item
listed above, but is more focussed on aquatic chemistry.
Download / View this pdf file now
Free energy and
equilibrium?
back to ThermEq home page
Stephen Lower is a retired member of the
Dept of Chemistry, Simon Fraser University
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