Chapter 2 The First Law of Thermodynamics

1. Chapter 2 The First Law of Thermodynamics ~Chapter 1 was mainly concerned with macroscopic properties such as pressure, volume, and temperature. We have seen how some of the relationships between these properties of ideal and real gases can be interpreted in terms of the behavior of molecules, that is, of the microscopic properties. ~This kinetic-molecular theory is of great value, but it is possible to interpret many of the relationships between macroscopic properties without any reference to the behavior of molecules, or even without assuming that molecules exist. This is what is done in the most formal treatments of the science of thermodynamics, which is concerned with the general relationships between the various forms of energy, including heat. ~At first it might appear that the formal study of thermodynamics, with no regard to molecular behavior, would not lead us very far. The contrary, however, is true. It has proved possible to develop some very far-reaching conclusions on the basis of purely thermodynamic arguments, and these conclusions are all the more convincing because they do not depend on the truth or falsity of any theories of atomic and molecular behavior. ~Pure thermodynamics starts with a small number of assumptions that are based on very well-established experimental results and makes logical deductions from them, finally giving a set of relationships that are bound to be true provided that the original premises are true.

2. 2.1 Origin of the First Law The Nature of Heat ~heat is a form of energy There are three laws of thermodynamics (aside from the zeroth law, which was mentioned in Section 1.4). The first law is essentially the principle of conservation of energy. seventeenth century heat is a manifestation of the motions of the particles of which matter is composed, but the evidence was not then compelling. nineteenth century ~work performed by humans is derived from the food they eat. ~precise experiments on the interconversion of work and heat, under a variety of conditions, were carried out . eighteenth century ~heat is a substance, and even listed it as one of the chemical elements. ~expenditure of work causes the production of heat. ~a numerical value for the amount of heat generated by a given amount of work was obtained. ~The energy of the universe remains constant, which is a compact statement of the first law of thermodynamics. ~Both work and heat are quantities that describe the transfer of energy from one system to another. If two systems are at different temperatures, heat can pass from one to the other directly, and there can also be transfer of matter from one to the other. Energy can also be transferred from one place to another in the form of work. No matter how these transfers occur, the total energy of the universe remains the same.

3. 2.2 States and State Functions The distinction between a system and its surroundings is particularly important in thermodynamics, since we are constantly concerned with transfer of heat between a system and its surroundings. We are also concerned with the work done by the system on its surroundings or by the surroundings on the system. In all cases the system must be carefully defined. Certain of the macroscopic properties have fixed values for a particular state of the system, whereas others do not. Suppose, for example, that we maintain 1 g of water in a vessel at 25 C and a pressure of 105 Pa (1 bar); it will have a volume of close to 1 cm3. These quantities, 1 g of H2O, 25 C, 105 Pa , and 1 cm3, all specify the state of the system. Whenever we satisfy these four conditions, we have the water in precisely the same state, and this means that the total amount of energy in the molecules is the same. As long as the system is in this state, it will have these particular specifications. These macroscopic properties that we have mentioned (mass, pressure, temperature, and volume) are known as state functions or state variables.

4. Characteristic of a State Function ~Once we have specified the state of a system by giving the values of some of the state functions, the values of all other state functions are fixed. Thus, in the example just given, once we have specified the mass, temperature, and pressure of the water, the volume is fixed. So, too, is the total energy in the molecules that make up the system, and energy is therefore another state function. The pressure and temperature, in fact, depend on the molecular motion in the system. ~When the state of a system is changed, the change in any state function depends only on the initial and final states of the system, and not on the path followed in making the change. For example, if we heat the water from 25 C to 26 C, the change in temperature is equal to the difference between the initial and final temperatures: The way in which the temperature change is brought about has no effect on this result. ~This example may seem trivial, but it is to be emphasized that not all functions have this characteristic. For example, raising the temperature of water from 25 C to 26 C can be done in various ways; the simplest is to add heat. Alternatively, we could stir the water vigorously with a paddle until the desired temperature rise had been achieved; this means that we are doing work on the system. We could also add some heat and do some work in addition. This shows that heat and work are not state functions.

5. Suppose that there is a point A on the earth's surface that is 1000 m above sea level and another point B that is 4000 m above sea level. The difference, 3000 m, is the height of B with respect to A. In other words, the difference in height can be expressed as where hA and hB are the heights of A and B above sea level. Height above sea level is thus a state function, the difference h being in no way dependent on the path chosen. However, the distance we have to travel in order to go from A to B is dependent on the path; we can go by the shortest route or take a longer route. Distance traveled is therefore not a state function.

6. 2.3 Equilibrium States and Reversibility Thermodynamics is directly concerned only with equilibrium states, in which the state functions have constant values throughout the system. It provides us with information about the circumstances under which nonequilibrium states will move toward equilibrium, but by itself it tells us nothing about the nonequilibrium states. Figure 2.1 ~Suppose, for example, that we have a gas confined in a cylinder having a frictionless movable piston (Figure 2.1). ~If the piston is motionless, the state of the gas can be specified by giving the values of pressure, volume, and temperature. ~However, if the gas is compressed very rapidly, it passes through states for which pressure and temperature cannot be specified, there being a variation of these properties throughout the gas; the gas near to the piston is at first more compressed and heated than the gas at the far end of the cylinder. The gas then would be said to be in a nonequilibrium state. Pure thermodynamics could not deal with such a state, although it could tell us what kind of a change would spontaneously occur in order for equilibrium to be attained. The criteria for equilibrium are very important. The mechanical properties, the chemical properties, and the temperature must be uniform throughout the system and constant in time. The force acting on the system must be exactly balanced by the force exerted by the system, as otherwise the volume would be changing.

7. ~If we consider the system illustrated in Figure 2.1, we see that for the system to be at equilibrium the force F exerted on the piston must exactly balance the pressure P of the gas; if A is the area of the piston, PA = F If we increase the force, for example, by adding mass to the piston, the gas will be compressed; if we decrease it, by removing mass, the gas will expand. ~Suppose that we increase the force F by an infinitesimal amount dF. The pressure that we are exerting on the gas will now be infinitesimally greater than thepressure of the gas (i.e., it will be P+dP). The gas will therefore be compressed. We can make dP as small as we like, and at all stages during the infinitely slow compression we are therefore maintaining the gas in a state of equilibrium. We refer to a process of this kind as a reversible process. If we reduce the pressure to P-dP, the gas will expand infinitely slowly, that is, reversibly. Figure 2.1 Reversible processes play very important roles in thermodynamic arguments. All processes that actually occur are, however, irreversible; since they do not occur infinitely slowly, there is necessarily some departure from equilibrium.

8. 2.4 Energy, Heat, and Work ~According to which the total amount of energy in the universe is conserved, suppose that we add heat q to a system such as a gas confined in a cylinder (Figure 2.1). If nothing else is done to the system, the internal energy U increases by an amount that is exactly equal to the heat supplied: This increase in internal energy is the increase in the energy of the molecules which comprise the system. ~Suppose instead that no heat is transferred to the system but that by the addition of mass to the piston an amount of work w is performed on it. The system then gains internal energy by an amount equal to the work done: ~In general, if heat q is supplied to the system, and an amount of work w is also performed on the system, the increase in internal energy is given by A statement of the first law of thermodynamics

9. ~We can understand the law by noting that a collection of molecules, on absorbing some heat, can store some of it internally, and can do some work on the surroundings. According to the IUPAC convention the work done by the system is -w, so that ~It is, of course, necessary to employ the same units for U, q, and w. The SI unit of energy is the joule (J = kg m2 s-2); it is the energy corresponding to a force of one newton (N = kg m s-2) operating over a distance of one metre. 1 cal=4.184 J q>0, heat is added to the system q<0, heat flow out of the system w>0, work is done on the system w<0, work is done by the system Heat and work sign convention

10. ~We should note that the above equation leaves the absolute value of the internal energy U indefinite, in that we are dealing with only the energy change U, and for most practical purposes this is adequate. Absolute values can in principle be calculated, although they must always be referred to some arbitrary zero of energy. Thermodynamics is concerned almost entirely with energy changes. ~The internal energy U is a state function of the system; that is, it depends only on the state of the system and not on how the system achieved its particular state. ~Earlier we saw that a change from one state to another, such as from 25 C to 26 C, can be achieved by adding heat, by doing work, or by a combination of the two. It is found experimentally that however we bring about the temperature rise, the sum q+w is always the same. In other words, for a particular change in state, the quantity U, equal to q+w, is independent of the way in which the change is brought about. This behavior is characteristic of a state function. ~This example demonstrates that heat q and work w are not state functions since the change can be brought about by various divisions of the energy between heat and work; only the sum q+w is fixed.

11. ~The definite integral of a state function such as U, is an exact differential because it has a value of U2-U1 = U, which is independent of the path by which the process occurs. ~A state function such as the internal energy U has a significance in relation to a particular state. ~heat and work are the integrals of inexact differentials. These integrals are not fixed but depend on the process by which the change from state 1 to state 2 occurs. ~It would therefore be wrong to write q2-q1 = q or w2-w1 = w; the quantities q and w have no meaning. ~Heat and work make themselves evident only during a change from one state to another and have no significance when a system remains in a particular state; they are properties of the path and not of the state. slashes indicating that the integrals are inexact

12. ~If U were not a state function, we could have violations of the principle of conservation of energy, something that has never been observed. ~To see how a violation could occur, consider two states A and B, and suppose that there are two alternative paths between them. Suppose that for one of these paths U is 10 J; and for the other, 30 J: ~We could go from A to B by the first path and expend 10 J of heat. If we then returned from B to A by the second path, we would gain 30 J. We would then have the system in its original state, and would have a net gain of 20 J. ~Energy would therefore have been created from nothing. The process could be continued indefinitely, with a gain of energy at each completion of the cycle. ~Many attempts have been made to create energy in this way, by the construction of perpetual motion machines of the first kind, but all have ended in failure-patent offices are constantly rejecting devices that would only work if the first law of thermodynamics were violated! The inability to make perpetual motion machines provides convincing evidence that energy cannot be created or destroyed.

13. Nature of Internal Energy In purely thermodynamic studies it is not necessary to consider what internal energy really consists of; however, most of us like to have some answer to this question, in terms of molecular energies. There are contributions to the internal energy of a substance from 1. the kinetic energy of motion of the individual molecules, 2. the potential energy that arises from interactions between molecules, 3. the kinetic and potential energy of the nuclei and electrons within the individual molecules. The precise treatment of these factors is somewhat complicated and it is a great strength of thermodynamics that we can make use of the concept of internal energy without having to deal with it on a detailed molecular basis.

14. Nature of Work There are various ways in which a system may do work, or by which work may be done on a system. ~For example, if we pass a current through a solution and electrolyze it, we are performing one form of work ~ electrical work. ~Chemical work is usually, but not quite always, involved when larger molecules are synthesized from smaller ones, as in living organisms. ~Osmotic work is the work required to transport and concentrate chemical substances. It is involved, for example, when seawater is purified by reverse osmosis and in the formation of the gastric juice, where the acid concentration is much higher than that of the surroundings. ~Mechanical work is performed, for example, when a weight is lifted.

15. Figure 2.2 One simple way in which work is done is when an external force brings about a compression of a system. Suppose, for example, that we have an arrangement in which a gas or liquid is maintained at constant pressure P, which it exerts against a movable piston (Figure 2.2). ~Suppose that the force is increased by an infinitesimal amount dF, so that the piston moves infinitely slowly, the process being reversible. If the piston moves to the left a distance l, the reversible work wrev on the system is wrev = Fl = PAl ~However, Al is the volume swept out by the movement of the piston, that is, the decrease in volume of the gas, which is -V. The work done on the system is thus wrev = -PV In order for the system to be at equilibrium we must apply a force P to the piston, the force being related to the pressure by the relationship F = PA where A is the area of the position. Note: V<0, the gas compresses  wrev>0 V>0, the gas expands  wrev<0

16. Example 2.1 Suppose that a chemical reaction is caused to occur in a bulb to which is attached a capillary tube having a cross-sectional area of 2.50 mm2. The tube is open to the atmosphere (pressure = 101.325 kPa), and during the course of the reaction the rise in the capillary is 2.40 cm. Calculate the work done by the reaction system. Solution The volume increase is 2.50 10-6 m22.4010-2 m = 6.0010-8 m3 The work done by the system, which following the IUPAC convention must be written as -w, is PV: -w= PV= 1.01325l05 Pa6.0010-8 m3 =6.0810-3 J [Pa=N m-2;N m=J]

17. Figure 2.3 If the pressure P varies during a volume change, we must obtain the work done by a process of integration. The work done on the system while an external pressure P moves the piston so that the volume of the gas changes by an infinitesimal volume dV is If, as illustrated in Figure 2.3, the volume changes from a value V1 to a value V2, the reversible work done on the system is P is constant P is not constant expressing pressure as a function of V before performing the integration Note: V1>V2 (i.e., we have compressed the gas) and the reversible work is positive.

18. Figure 2.4 ~We have already noted that work done is not a state function, and this may be further stressed with reference to the mechanical work of expansion. ~The previous derivation has shown that the work is related to the process carried out rather than to the initial and final states. ~We can consider the reversible expansion of a gas from volume V1 to volume V2, and can also consider an irreversible process, as illustrated in Figure 2.4. ~Figure 2.4a shows the expansion of a gas, in which the pressure is falling as the volume increases. ~The reversible work done by the system is given by the integral ~Figure 2.4b presents that the process is performed irreversibly by instantaneously dropping the external pressure to the final pressure P2. ~The work done by the system is now against this pressure P2 throughout the whole expansion, and is given by • The irreversible work is less than the reversible work. • In both processes the state of the system has changed from A to B, the work done is different. • The work done by the system in a reversible expansion from A to B represents the maximum work that the system can perform in changing from A to B.

19. Example 2.2 Suppose that water at its boiling point is maintained in a cylinder that has a frictionless piston. For equilibrium to be established, the pressure that must be applied to the piston is 1 atm (101.325 kPa). Suppose that we now reduce the external pressure by an infinitesimal amount in order to have a reversible expansion. If the piston sweeps out a volume of 2.00 dm3, what is the work done by the system? Solution The external pressure remains constant at 101.325 kPa, and, therefore, the reversible work done by the system is -wrev=PV = 101 325 Pa2.00 dm3=202.65 Pa m3 Since Pa = kg m-1 s-2 (see Appendix A), the units are kg m2 s-2 = J; thus the work done by the system is 202.65 J.

20. Problem 2.8 A balloon is 0.50 m in diameter and contains air at 25 C and 1 bar pressure. It is then filled with air isothermally and reversibly until the pressure reaches 5 bar. Assume that the pressure is proportional to the diameter of the balloon and calculate (a) the final diameter of the balloon and (b) the work done in the process. Solution (a) (b)

21. ~For many purposes it is convenient to express the first law of thermodynamics with respect to an infinitesimal change. We have where again the symbol denotes an inexact differential. ~However, if only PV work is involved and P is a constant, may be written as -P dV, where dV is the infinitesimal increase in volume; thus, Processes at Constant Volume ~If an infinitesimal process occurs at constant volume, and only PV work is involved, where the subscript V indicates that the heat is supplied at constant volume. (Note that under these circumstances dqv is an exact differential, so that the d has lost its slash.) ~This equation integrates to ~The increase of internal energy of a system in a rigid container (i.e., at constant volume) is thus equal to the heat qv that is supplied to it.

22. Processes at Constant Pressure: Enthalpy ~In most chemical systems we are concerned with processes occurring in open vessels, which means that they occur at constant pressure rather than at constant volume. ~For an infinitesimal process at constant pressure the heat absorbed dqP is given by ~If the process involves a change from state 1 to state 2, this equation integrates as follows: P is constant Enthalpy is defined as H=U+PV This equation is valid only if the work is all PV work. ~Under these circumstances the increase in enthalpy H of a system is equal to the heat qP that is supplied to it at constant pressure. ~Since U, P, and V are all state functions, that enthalpy is also a state function. ~H>0a positive amount of heat is absorbed by the system, endothermic processes. H<0a positive amount of heat is evolved by the system, exothermic processes.

23. Problem 2.2 The densities of ice and water at 0 C are 0.9168 and 0.9998 g cm-3, respectively. If H for the fusion process at atmospheric pressure is 6.025 kJ mol-1, what is U? How much work is done on the system? Solution 1 mol of ice has a volume of 18.01 g/0.9168 g cm-3=19.64 cm3 1 mol of water has a volume of 18.01 g/0.9998 g cm-3=18.01 cm3

24. Heat Capacity ~The amount of heat required to raise the temperature of any substance by 1 K (which of course is the same as 1 C) is known as its heat capacity, and is given the symbol C; its SI unit is J K-1. ~The word specific before the name of any extensive physical quantity refers to the quantity per unit mass. The term specific heat capacity is thus the amount of heat required to raise the temperature of unit mass of a material by 1 K; if the unit mass is 1 kg, the unit is J K-1 kg-1, which is the SI unit for specific heat capacity. ~The word molar before the name of a quantity refers to the quantity divided by the amount of substance. The SI unit for the molar heat capacity is J K-1 mol-1. ~Since heat is not a state function, neither is the heat capacity. It is therefore always necessary, when stating a heat capacity, to specify the process by which the temperature is raised by 1 K.

25. Isochoric Process The heat capacity related to a process occurring at constant volume (an isochoric process); this is denoted by CV and is defined by qV is the heat supplied at constant volume. Note: CV,m represents the molar heat capacity at constant volume. Isobaric Process The heat capacity related to a process occurring at constant pressure (an isobaric process); this is denoted by CP and is defined by qP is the heat supplied at constant volume. Note: CP,m represents the molar heat capacity at constant pressure.

26. The heat required to raise the temperature of 1 mol of material from T1 to T2 at constant volume is CV,m is independent of temperature The heat required to raise the temperature of 1 mol of material from T1 to T2 at constant pressure is CP,m is independent of temperature ~For liquids and solids, Um and Hm are very close to one another. Consequently, CV,m and CP,m are essentially the same for solids and liquids. ~For gases, however, the (PV) term is appreciable, and there is a significant difference between CV,m and CP,m. For 1 mol of an ideal gas,

27. Problem 2.4 The latent heat of fusion of water at 0 C is 6.025 kJ mol-1 and the molar heat capacities (CP,m) of water and ice are 75.3 and 37.7 J K-1 mol-1, respectively. The CP values can be taken to be independent of temperature. Calculate H for the freezing of 1 mol of supercooled water at -10.0 C. Solution Heat the water from -10 C to 0 C: Freeze the water at 0 C: Cool the ice from 0 C to -10 C:

28. 2.5 Thermochemistry The study of enthalpy changes in chemical processes is known as thermochemistry. Extent of Reaction In dealing with enthalpy changes in chemical processes it is very convenient to make use of a quantity known as the extent of reaction; it is given the symbol . The extent of reaction must be related to a specified stoichiometric equation for a reaction. A chemical reaction can be written in general as It can also be written as where A, B, Y, and Zare known as stoichiometric coefficients. By definition the stoichiometric coefficient is positive for a product and negative for a reactant.

29. The extent of reaction is defined by where ni,0 is the initial amount of the substance i and ni is the amount at any time. What makes the extent of reaction so useful is that it is the same for every reactant and product. Thus the extent of reaction is the amount of any product formed divided by its stoichiometric coefficient: It is also the change in the amount of any reactant (a negative quantity), divided by its stoichiometric coefficient (also a negative quantity): These quantities are all equal.

30. Example 2.3 When 10 mol of nitrogen and 20 mol of hydrogen are passed through a catalytic converter, after a certain time 5 mol of ammonia are produced. Calculate the amounts of nitrogen and hydrogen that remain unreacted. Calculate also the extent of reaction: a. on the basis of the stoichiometric equation b. on the basis of the stoichiometric equation Solution The amounts are N2 H2 NH3 Initially 10 20 0 mol Finally 7.5 12.5 5 mol a. The extent of reaction is the amount of ammonia formed, 5 mol, divided by the stoichiometric coefficient for NH3: The same answer is obtained if we divide the amounts of N2 and H2 consumed, 2.5 mol and 7.5 mol, by the respective stoichiometric coefficients: b. The extent of reaction is now doubled, since the stoichiometric coefficients are halved:

31. ~The SI unit for the extent of reaction is the mole, and the mole referred to relates to the stoichiometric equation. For example, if the equation is specified to be the mole relates to N2, 3H2, or 2NH3. If, therefore, the H for this reaction is stated to be -46.0 kJ mol-1, it is to be understood that this value refers to the removal of 1 mol of N2 and 1 mol of 3H2, which is the same as 3 mol of H2. It also refers to the formation of 1 mol of 2NH3, which is the same as 2 mol of NH3. In other words, the H value relates to the reaction as written in the stoichiometric equation. ~This recommended IUPAC procedure, which we shall use throughout this book, avoids the necessity of saying, for example, -46.0 kJ per mol of nitrogen, or -23.0 kJ per mol of ammonia. It cannot be emphasized too strongly that when this IUPAC procedure is used, the stoichiometric equation must be specified.

32. Standard States Enthalpy is a state function, and the enthalpy change that occurs in a chemical process depends on the states of the reactants and products. Consider, for example, the complete combustion of ethanol, in which 1 mol is oxidized to carbon dioxide and water: ~The enthalpy change in this reaction depends on whether we start with liquid ethanol or with ethanol in the vapor phase. ~It also depends on whether liquid or gaseous water is produced in the reaction. ~Another factor is the pressure of the reactants and products. ~Also, the enthalpy change in a reaction varies with the temperature at which the process occurs. In giving a value for an enthalpy change it is therefore necessary to specify (1) the state of matter of the reactants and products (gaseous, liquid, or solid; if the last, the allotropic form), (2) the pressure, and (3) the temperature. If the reaction occurs in solution, the concentrations must also be specified.

33. By general agreement the standard state of a substance is the form in which it is most stable at 25.00 C (298.15 K) and 1 bar (105 Pa) pressure. ~For example, the standard state of oxygen is the gas, and we specify this by writing O2(g). ~Since mercury, water, and ethanol are liquids at 25 C, their standard states are Hg(l), H2O(l), and C2H5OH(l). ~The standard state of carbon isgraphite. These standard states should be specified if there is any ambiguity; for example, not involving standard states If a reaction involves species in solution, their standard state is 1 mol kg-1(1 m); for example,

34. Enthalpy changes depend somewhat on the temperature at which the process occurs. Standard thermodynamic data are commonly quoted for a temperature of 25.00 C (298.15 K), and this can be given as a subscript or in parentheses; thus ~The superscript  on the H specifies that we are dealing with standard states, so that a pressure of 1 bar is assumed and need not be stated. ~The subscript c on the  is commonly used to indicate complete combustion, and the modern practice is to attach such subscripts to the  and not to the H. ~As emphasized in our discussion of extent of reaction, the value 1357.7 kJ mol-1 relates to the combustion of 1 mol of ethanol, since that is what appears in the equation. ~Standard thermodynamic values can be given for a temperature other than 25 C; for example, we could give a value for H (100 C), and it would be understood that the pressure was again 1 bar and that reactants and products were in their standard states but at 100 C.

35. Measurement of Enthalpy Changes The enthalpy changes occurring in chemical processes may be measured by three main methods: Direct Calorimetry Some reactions occur to completion and without side reactions, and it is therefore possible to measure their H values by causing the reactions to occur in a calorimeter(熱卡計). The neutralization of an aqueous solution of a strong acid by a solution of a strong base is an example of such a process, the reaction that occurs being Combustion processes also frequently occur to completion with simple stoichiometry. When an organic compound is burnt in excess of oxygen, the carbon is practically all converted into CO2 and the hydrogen into H2O, while the nitrogen is usually present as N2 in the final products. Often such combustions of organic compounds occur cleanly, and much thermochemical information has been obtained by burning organic compounds in calorimeters.

36. Indirect Calorimetry Few reactions occur in a simple manner, following a simple chemical equation, with the result that the enthalpy changes corresponding to a simple chemical equation often cannot be measured directly. For many of these, the enthalpy changes can be calculated from the values for other reactions by making use of Hess's law. According to this law, it is permissible to write stoichiometric equations, together with the enthalpy changes, and to treat them as mathematical equations, thereby obtaining a thermochemically valid result. For example, suppose that a substance A reacts with B according to the equation 1. A+BX H1 = -10 kJ mol-1 Suppose that X reacts with an additional molecule of A to give another product Y: 2. A+XY H2= -20 kJ mol-1 According to Hess's law, it is permissible to add these two equations and obtain: 3. 2A+BY H3 =H1 + H2 = -30 kJ mol-1 The law follows at once from the principle of conservation of energy and from the fact that enthalpy is a state function. Thus, if reactions 1 and 2 occur, there is a net evolution of 30 kJ when 1 mol of Y is produced. In principle we could reconvert Y into 2A + B by the reverse of reaction 3. If the heat required to do this were different from 30 kJ, we should have obtained the starting materials with a net gain or loss of heat, and this would violate the principle of conservation of energy.

37. Example 2.4 The enthalpy changes in the complete combustion of crystalline -D-glucose and maltose at 298 K, with the formation of gaseous CO2 and liquid H2O, are: cH/kJ mol-1 -D-Glucose, C6H12O6(c) -2809.1 Maltose, C12H22O11(c) -5645.5 Calculate the enthalpy change accompanying the conversion of 1 mol of crystalline glucose into crystalline maltose. Solution

38. Variation of Equilibrium Constant with Temperature A third general method of measuring H will be mentioned here only very briefly, since it is based on the second law of thermodynamics and is considered in Section 4.8. This method is based on the equation for the variation of the equilibrium constant K with the temperature: ~If, therefore, we measure K at a series of temperatures and plot ln K against 1/T, the slope of the line at any temperature will be -H/8.3145 J mol-1, and hence Hcan be calculated. ~Whenever an equilibrium constant for a reaction can be measured satisfactorily at various temperatures, this method thus provides a very useful way of obtaining H. ~The method cannot be used for reactions that go essentially to completion, in which case a reliable K cannot be obtained, or for reactions that are complicated by side reactions.

39. Calorimetry The heats evolved in combustion processes are determined in bomb calorimeters, two types of which are shown in Figure 2.5. A weighed sample of the material to be burnt is placed in the cup supported in the reaction vessel, or bomb, which is designed to withstand high pressures. The heavy- walled steel bomb of about 400 ml volume is filled with oxygen at a pressure of perhaps 25 atm, this being more than enough to cause complete combustion. The reaction is initiated by passing an electric current through the ignition wire. Heat is evolved rapidly in the combustion process and is determined in two different ways in the two types of calorimeter. Figure 2.5

40. Figure 2.5a ~In the type of calorimeter shown in Figure 2.5a, the bomb is surrounded by a water jacket that is insulated as much as possible from the surroundings. ~The water in the jacket is stirred, and the rise in temperature brought about by the combustion is measured. ~From the thermal characteristics of the apparatus the heat evolved can be calculated. ~A correction is made for the heat produced in the ignition wire, and it is customary to calibrate the apparatus by burning a sample having a known heat of combustion.

41. Figure 2.5b ~The type of calorimeter illustrated in Figure 2.5b is known as an adiabatic calorimeter. An adiabatic process is thus one in which there is no flow of heat. ~In the adiabatic calorimeter this is achieved by surrounding the inner water jacket by an outer water jacket which by means of a heating coilis maintained at the same temperature as the inner jacket. ~When this is done the amount of heat supplied to the outer jacket just cancels the heat loss to the surroundings. ~This allows a simpler determination of the temperature rise due to the combustion; the measured (Tfinal –Tinitial) is directly related to the amount of heat evolved in the combustion.

42. ~By the use of calorimeters of these types, heats of combustion can be measured with an accuracy of better than 0.01%. Such high precision is necessary, since heats evolved in combustion are large, and sometimes we are more interested in the differences between the values for two compounds than in the absolute values. ~Sometimes the heat changes occurring in chemical reactions are exceedingly small, and it is then necessary to employ very sensitive calorimeters. Such instruments are known as microcalorimeters. ~Another type of microcalorimeter is the continuous flow calorimeter; which permits two reactant solutions to be thermally equilibrated during passage through separate platinum tubes and then brought together in a mixing chamber; the heat change in the reaction is measured.

43. Relationship between U and H Bomb calorimeters and other calorimeters in which the volume is constant give the internal energy change U. Other calorimeters operate at constant pressure and therefore give H values. Whether U or H is determined, the other quantity is easily calculated from the stoichiometric equation for the reaction. If all reactants and products are solids or liquids, the change in volume if a reaction occurs at constant pressure is quite small. Usually 1 mol of a solid or liquid has a volume of less than 1 dm3, and the volume change in a reaction will always be less than 1% (i.e., less than 0.01 dm3). At 1 bar pressure, with V=0.01 dm3, This is quite negligible compared with most heats of reaction, which are of the order of kilojoules, and is much less than the experimental error of most determinations. Note: If gases are involved in the reaction, however, either as reactants or products, U and H may differ significantly, as illustrated in the following example.

44. Example 2.5 For the complete combustion of ethanol, the amount of heat produced, as measured in a bomb calorimeter, is 1364.47 kJ mo-1 at 25 C. Calculate cH for the reaction. Solution Since the bomb calorimeter operates at constant volume, cU =-1364.47 kJ mol-1. The product of reaction contains 2 mol of gas and the reactants, 3 mol; the change n is therefore -1 mol. Assuming the ideal gas law to apply, (PV) is equal to nRT, and therefore to The difference between U and H is now large enough to be experimentally significant.

45. Problem 2.15 A sample of liquid benzene weighing 0.633 g is burned in a bomb calorimeter at 25 C, and 26.54 kJ of heat are evolved. a. Calculate U per mole of benzene. b. Calculate H per mole of benzene. Solution Molar mass of benzene = 612.01+61.008=78.11 g mol-1 Heat evolved in the combustion of 1 mol = 26.54 kJ78.11 g mol-1/0.633 g=3274.9 kJ a. b.

46. Temperature Dependence of Enthalpies of Reaction Enthalpy changes are commonly tabulated at 25 C, and it is frequently necessary to have the values at other temperatures. These can be calculated from the heat capacities of the reactants and products. The enthalpy change in a reaction can be written as partial differentiation with respect to temperature at constant pressure Similarly, For small changes in temperature the heat capacities, and hence CP and CV, may be taken as constant. integrating between two temperatures T1 and T2

47. If there is a large difference between the temperatures T1 and T2, it is necessary to take into account the variation of CP with temperature. This is often done by expressing the molar value Cp,m as a power series: Alternatively, and somewhat more satisfactorily, we can use an equation of the form

48. Example 2.6 Consider the gas-phase reaction A bomb-calorimetric study of this reaction at 25 C leads to H= -565.98 kJ mol-1. Calculate H for this reaction at 2000 K. Solution Note that when numerical values are given, it is permissible to drop the subscript m from H, since the unit kJ mol-1 avoids ambiguity. Remember that the mole referred to always relates to the reaction as written.

49. Enthalpies of Formation ~The total number of known chemical reactions is enormous, and it would be very inconvenient if one had to tabulate enthalpies of reaction for all of them. ~We can avoid having to do this by tabulating molar enthalpies of formation of chemical compounds, which are the enthalpy changes associated with the formation of 1 mol of the substance from the elements in their standard states. ~From these enthalpies of formation it is possible to calculate enthalpy changes in chemical reactions. ~We have seen that the standard state of each element and compound is taken to be the most stable form in which it occurs at 1 bar pressure and at 25 C. Suppose that we form methane, at 1 bar and 25 C, from C(graphite) and H2(g), which are the standard states; the stoichiometric equation is ~It is found that H for this reaction is -74.81 kJ mol-1, and this quantity is known as the standard molar enthalpy of formation fHof methane at 25 C (298.15 K). ~The term standard enthalpy of formation refers to the enthalpy change when the compound in its standard state is formed from the elements in their standard states. The standard enthalpy of formation of any element in its standard state is zero.

50. ~Enthalpies of formation of organic compounds are commonly obtained from their enthalpies of combustion, by application of Hess's law. ~When, for example, 1 mol of methane is burned in an excess of oxygen, 802.37 kJ of heat is evolved, and we can therefore write In addition, we have the following data: If we add reactions 2 and 3 and subtract reaction 1, the result is Enthalpies of formation of many other compounds can be deduced in a similar way. Appendix D gives some enthalpies of formation. The values, of course, depend on the state in which the substance occurs, and this is indicated in the table; the value for liquid ethanol, for example, is a little different from that for ethanol in aqueous solution.