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 Page 14 of 41 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 Page 15 of 41 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 Page 16 of 41 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 Page 17 of 41 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 Page 18 of 41 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 Page 19 of 41 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 Page 20 of 41 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 Page 21 of 41 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 Page 22 of 41 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 Page 23 of 41 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 Page 24 of 41 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 Page 25 of 41 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 Page 27 of 41 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 Page 28 of 41 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 Ɍɖ2É…. 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 24/11/06 10:27 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. 2 of 7 24/11/06 10:27 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 3 of 7 24/11/06 10:27 Chemical energetics: heat and work http://www.chem1.com/acad/webtext/energetics/CE01.html 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 4 of 7 24/11/06 10:27 Chemical energetics: heat and work http://www.chem1.com/acad/webtext/energetics/CE01.html 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: 5 of 7 24/11/06 10:27 Chemical energetics: heat and work http://www.chem1.com/acad/webtext/energetics/CE01.html 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 6 of 7 24/11/06 10:27 Chemical energetics: heat and work http://www.chem1.com/acad/webtext/energetics/CE01.html 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 For information about this Web site or to contact the author, please go to the home page. See here for other Chem1 Virtual Textbook topics and for PDF versions suitable for printing © 2005; Last update 02.01.2006 7 of 7 24/11/06 10:27 The First Law of Thermodynamics http://www.chem1.com/acad/webtext/energetics/CE02.html 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 1 of 12 24/11/06 10:28 The First Law of Thermodynamics http://www.chem1.com/acad/webtext/energetics/CE02.html 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 2 of 12 24/11/06 10:28 The First Law of Thermodynamics http://www.chem1.com/acad/webtext/energetics/CE02.html 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 3 of 12 24/11/06 10:28 The First Law of Thermodynamics http://www.chem1.com/acad/webtext/energetics/CE02.html 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: 4 of 12 24/11/06 10:28 The First Law of Thermodynamics http://www.chem1.com/acad/webtext/energetics/CE02.html 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 5 of 12 24/11/06 10:28 The First Law of Thermodynamics http://www.chem1.com/acad/webtext/energetics/CE02.html 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. 6 of 12 24/11/06 10:28 The First Law of Thermodynamics http://www.chem1.com/acad/webtext/energetics/CE02.html 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 7 of 12 24/11/06 10:28 The First Law of Thermodynamics http://www.chem1.com/acad/webtext/energetics/CE02.html 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. 8 of 12 24/11/06 10:28 The First Law of Thermodynamics http://www.chem1.com/acad/webtext/energetics/CE02.html 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 9 of 12 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 24/11/06 10:28 The First Law of Thermodynamics http://www.chem1.com/acad/webtext/energetics/CE02.html 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 10 of 12 24/11/06 10:28 The First Law of Thermodynamics 11 of 12 http://www.chem1.com/acad/webtext/energetics/CE02.html 24/11/06 10:28 The First Law of Thermodynamics http://www.chem1.com/acad/webtext/energetics/CE02.html 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 For information about this Web site or to contact the author, please go to the home page. See here for other Chem1 Virtual Textbook topics and for PDF versions suitable for printing © 2005; Last update 06.06.2006 12 of 12 24/11/06 10:28 Molecules as energy carriers and converters http://www.chem1.com/acad/webtext/energetics/CE03.html 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 1 of 11 24/11/06 10:34 Molecules as energy carriers and converters http://www.chem1.com/acad/webtext/energetics/CE03.html 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 2 of 11 24/11/06 10:34 Molecules as energy carriers and converters http://www.chem1.com/acad/webtext/energetics/CE03.html 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 3 of 11 24/11/06 10:34 Molecules as energy carriers and converters http://www.chem1.com/acad/webtext/energetics/CE03.html 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. 4 of 11 24/11/06 10:34 Molecules as energy carriers and converters http://www.chem1.com/acad/webtext/energetics/CE03.html 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 5 of 11 24/11/06 10:34 Molecules as energy carriers and converters http://www.chem1.com/acad/webtext/energetics/CE03.html 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 6 of 11 24/11/06 10:34 Molecules as energy carriers and converters http://www.chem1.com/acad/webtext/energetics/CE03.html 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 É¢U and É¢V 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 7 of 11 24/11/06 10:34 Molecules as energy carriers and converters http://www.chem1.com/acad/webtext/energetics/CE03.html 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 8 of 11 24/11/06 10:34 Molecules as energy carriers and converters http://www.chem1.com/acad/webtext/energetics/CE03.html 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 9 of 11 24/11/06 10:34 Molecules as energy carriers and converters http://www.chem1.com/acad/webtext/energetics/CE03.html 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 10 of 11 24/11/06 10:34 Molecules as energy carriers and converters http://www.chem1.com/acad/webtext/energetics/CE03.html 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 For information about this Web site or to contact the author, please go to the home page. See here for other Chem1 Virtual Textbook topics and for PDF versions suitable for printing © 2005 - Last update 2-01-2006 11 of 11 24/11/06 10:34 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html 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: 1 of 13 24/11/06 10:40 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html • 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 2 of 13 24/11/06 10:40 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html 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 3 of 13 24/11/06 10:40 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html 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. 4 of 13 ΔH° values is just the 24/11/06 10:40 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html • 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 5 of 13 24/11/06 10:40 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html 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 6 of 13 24/11/06 10:40 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html 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 7 of 13 24/11/06 10:40 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html 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, 8 of 13 24/11/06 10:40 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html 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 9 of 13 24/11/06 10:40 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html 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... 10 of 13 ... or even entire rooms! 24/11/06 10:40 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html 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 11 of 13 24/11/06 10:40 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html 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 12 of 13 24/11/06 10:40 Thermochemistry and calorimetry http://www.chem1.com/acad/webtext/energetics/CE04.html 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 For information about this Web site or to contact the author, please go to the home page. See here for other Chem1 Virtual Textbook topics and for PDF versions suitable for printing © 2005; Last revised 02.01.2006 13 of 13 24/11/06 10:40 Some applications of enthalpy and the First Law http://www.chem1.com/acad/webtext/energetics/CE05.html 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. 1 of 12 24/11/06 10:41 Some applications of enthalpy and the First Law http://www.chem1.com/acad/webtext/energetics/CE05.html 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 2 of 12 24/11/06 10:41 Some applications of enthalpy and the First Law http://www.chem1.com/acad/webtext/energetics/CE05.html 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 3 of 12 24/11/06 10:41 Some applications of enthalpy and the First Law http://www.chem1.com/acad/webtext/energetics/CE05.html 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 4 of 12 24/11/06 10:41 Some applications of enthalpy and the First Law http://www.chem1.com/acad/webtext/energetics/CE05.html 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 5 of 12 24/11/06 10:41 Some applications of enthalpy and the First Law http://www.chem1.com/acad/webtext/energetics/CE05.html 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 6 of 12 24/11/06 10:41 Some applications of enthalpy and the First Law http://www.chem1.com/acad/webtext/energetics/CE05.html 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. 7 of 12 24/11/06 10:41 Some applications of enthalpy and the First Law http://www.chem1.com/acad/webtext/energetics/CE05.html 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 8 of 12 The above images are from the NDBC site. For a more detailed 24/11/06 10:41 Some applications of enthalpy and the First Law http://www.chem1.com/acad/webtext/energetics/CE05.html 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 9 of 12 24/11/06 10:41 Some applications of enthalpy and the First Law http://www.chem1.com/acad/webtext/energetics/CE05.html 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 10 of 12 24/11/06 10:41 Some applications of enthalpy and the First Law http://www.chem1.com/acad/webtext/energetics/CE05.html (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 11 of 12 24/11/06 10:41 Some applications of enthalpy and the First Law http://www.chem1.com/acad/webtext/energetics/CE05.html 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 For information about this Web site or to contact the author, please go to the home page. See here for other Chem1 Virtual Textbook topics and for PDF versions suitable for printing © 2005 - Last updated 02.01.2006 12 of 12 24/11/06 10:41 Reading(s) #3 Energy spreading and spontaneous change http://www.chem1.com/acad/webtext/thermeq/TE1.html 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.) 1 of 9 24/11/06 10:51 Energy spreading and spontaneous change http://www.chem1.com/acad/webtext/thermeq/TE1.html 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 2 of 9 24/11/06 10:51 Energy spreading and spontaneous change http://www.chem1.com/acad/webtext/thermeq/TE1.html 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 3 of 9 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 24/11/06 10:51 Energy spreading and spontaneous change http://www.chem1.com/acad/webtext/thermeq/TE1.html 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 4 of 9 24/11/06 10:51 Energy spreading and spontaneous change http://www.chem1.com/acad/webtext/thermeq/TE1.html 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: 5 of 9 24/11/06 10:51 Energy spreading and spontaneous change http://www.chem1.com/acad/webtext/thermeq/TE1.html 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? 6 of 9 24/11/06 10:51 Energy spreading and spontaneous change http://www.chem1.com/acad/webtext/thermeq/TE1.html 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 7 of 9 24/11/06 10:51 Energy spreading and spontaneous change http://www.chem1.com/acad/webtext/thermeq/TE1.html 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. 8 of 9 24/11/06 10:51 Energy spreading and spontaneous change http://www.chem1.com/acad/webtext/thermeq/TE1.html 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 9 of 9 24/11/06 10:51 What is entropy? http://www.chem1.com/acad/webtext/thermeq/TE2.html 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. 1 of 8 24/11/06 10:52 What is entropy? http://www.chem1.com/acad/webtext/thermeq/TE2.html 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. 2 of 8 24/11/06 10:52 What is entropy? http://www.chem1.com/acad/webtext/thermeq/TE2.html 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. 3 of 8 24/11/06 10:52 What is entropy? http://www.chem1.com/acad/webtext/thermeq/TE2.html 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 4 of 8 24/11/06 10:52 What is entropy? http://www.chem1.com/acad/webtext/thermeq/TE2.html 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 5 of 8 24/11/06 10:52 What is entropy? http://www.chem1.com/acad/webtext/thermeq/TE2.html 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 6 of 8 24/11/06 10:52 What is entropy? http://www.chem1.com/acad/webtext/thermeq/TE2.html (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 7 of 8 The Second Law of Thermodynamics 24/11/06 10:52 What is entropy? http://www.chem1.com/acad/webtext/thermeq/TE2.html 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 30.03.2004-->-->-->--> See here for other Chem1 Virtual Textbook topics A PDF version of this document is available for printing 8 of 8 24/11/06 10:52 The Second Law of Thermodynamics http://www.chem1.com/acad/webtext/thermeq/TE3.html 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 1 of 8 24/11/06 10:52 The Second Law of Thermodynamics http://www.chem1.com/acad/webtext/thermeq/TE3.html 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. 2 of 8 24/11/06 10:52 The Second Law of Thermodynamics http://www.chem1.com/acad/webtext/thermeq/TE3.html 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 3 of 8 24/11/06 10:52 The Second Law of Thermodynamics http://www.chem1.com/acad/webtext/thermeq/TE3.html 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 4 of 8 24/11/06 10:52 The Second Law of Thermodynamics http://www.chem1.com/acad/webtext/thermeq/TE3.html 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 5 of 8 24/11/06 10:52 The Second Law of Thermodynamics http://www.chem1.com/acad/webtext/thermeq/TE3.html 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 6 of 8 24/11/06 10:52 The Second Law of Thermodynamics http://www.chem1.com/acad/webtext/thermeq/TE3.html 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. 7 of 8 24/11/06 10:52 The Second Law of Thermodynamics http://www.chem1.com/acad/webtext/thermeq/TE3.html 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 Last modifed 22.11.05-->-->-->--> See here for other Chem1 Virtual Textbook topics A PDF version of this document is available for printing 8 of 8 24/11/06 10:52 Free energy: the Gibbs function http://www.chem1.com/acad/webtext/thermeq/TE4.html 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 1 of 9 24/11/06 10:52 Free energy: the Gibbs function http://www.chem1.com/acad/webtext/thermeq/TE4.html 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? 2 of 9 24/11/06 10:52 Free energy: the Gibbs function http://www.chem1.com/acad/webtext/thermeq/TE4.html 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 3 of 9 B 24/11/06 10:52 Free energy: the Gibbs function 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: http://www.chem1.com/acad/webtext/thermeq/TE4.html 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. 4 of 9 24/11/06 10:52 Free energy: the Gibbs function http://www.chem1.com/acad/webtext/thermeq/TE4.html 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. 5 of 9 24/11/06 10:52 Free energy: the Gibbs function http://www.chem1.com/acad/webtext/thermeq/TE4.html 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 6 of 9 24/11/06 10:52 Free energy: the Gibbs function http://www.chem1.com/acad/webtext/thermeq/TE4.html 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) 7 of 9 24/11/06 10:52 Free energy: the Gibbs function http://www.chem1.com/acad/webtext/thermeq/TE4.html 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. 8 of 9 24/11/06 10:52 Free energy: the Gibbs function http://www.chem1.com/acad/webtext/thermeq/TE4.html 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 See here for other Chem1 Virtual Textbook topics A PDF version of this document is available for printing 9 of 9 24/11/06 10:52 Free energy and equilibrium http://www.chem1.com/acad/webtext/thermeq/TE5.html 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) 1 of 8 24/11/06 10:53 Free energy and equilibrium http://www.chem1.com/acad/webtext/thermeq/TE5.html (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. 2 of 8 24/11/06 10:53 Free energy and equilibrium http://www.chem1.com/acad/webtext/thermeq/TE5.html 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. 3 of 8 24/11/06 10:53 Free energy and equilibrium http://www.chem1.com/acad/webtext/thermeq/TE5.html 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 !) 4 of 8 24/11/06 10:53 Free energy and equilibrium http://www.chem1.com/acad/webtext/thermeq/TE5.html 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. 5 of 8 24/11/06 10:53 Free energy and equilibrium http://www.chem1.com/acad/webtext/thermeq/TE5.html 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. 6 of 8 24/11/06 10:53 Free energy and equilibrium http://www.chem1.com/acad/webtext/thermeq/TE5.html 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. 7 of 8 24/11/06 10:53 Free energy and equilibrium http://www.chem1.com/acad/webtext/thermeq/TE5.html 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 © 2003 by Stephen Lower all rights reserved Last modifed 02.01.2006-->--> See here for other Chem1 Virtual Textbook topics A PDF version of this document is available for printing + 8 of 8 24/11/06 10:53 Applications of Entropy and Free Energy http://www.chem1.com/acad/webtext/thermeq/TE6.html 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.) 1 of 9 24/11/06 10:53 Applications of Entropy and Free Energy http://www.chem1.com/acad/webtext/thermeq/TE6.html 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 2 of 9 24/11/06 10:53 Applications of Entropy and Free Energy http://www.chem1.com/acad/webtext/thermeq/TE6.html 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: 3 of 9 24/11/06 10:53 Applications of Entropy and Free Energy http://www.chem1.com/acad/webtext/thermeq/TE6.html 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 4 of 9 24/11/06 10:53 Applications of Entropy and Free Energy http://www.chem1.com/acad/webtext/thermeq/TE6.html 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. 5 of 9 24/11/06 10:53 Applications of Entropy and Free Energy http://www.chem1.com/acad/webtext/thermeq/TE6.html 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. 6 of 9 24/11/06 10:53 Applications of Entropy and Free Energy http://www.chem1.com/acad/webtext/thermeq/TE6.html 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. 7 of 9 24/11/06 10:53 Applications of Entropy and Free Energy http://www.chem1.com/acad/webtext/thermeq/TE6.html 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 8 of 9 24/11/06 10:53 Applications of Entropy and Free Energy http://www.chem1.com/acad/webtext/thermeq/TE6.html © 2003 by Stephen Lower all rights reserved Last modifed 02.01.2006-->-->-->-->-->--> See here for other Chem1 Virtual Textbook topics A PDF version of this document is available for printing + 9 of 9 24/11/06 10:53