An Introduction to Thermodynamics phần 4 ppsx

1 1 2 1 1 1 dV dV dS dV dV dS dS += (13) 0 1 2 2 1 1 ≥         −= dV T P T P dS (14) T H S ∆ =∆ KJ/mol 96.108 K 15.373 J/mol 657,40 ⋅==∆ vap S It is assumed that T = T and thus the total entropy is unchanged by a transfer of thermal 5 1 2 energy between the two parts. The constant-energy restriction of equation 12 is satisfied because any work done by one part on the other can be compensated by such a transfer of thermal energy. 7/10/07 2- 40 Again consider a system constrained to constant total volume, and to have a uniform temperature. Then, as before, we can write the total change in entropy that might result from a change in volume of either part in the form (T, V) and, because total volume is constant, dV = - dV . Substitution of equation 12 gives 1 2 5 This proves that two bodies in equilibrium must have not only the same temperature but also the same pressure. It also proves that, when two bodies in contact, at the same temperature, have different pressures, the body at the higher pressure will tend to expand and compress the body at the lower pressure. ISOTHERMAL ENTROPY CHANGES. Entropy changes at a single temperature follow directly from equation 5. (Calculations involving a temperature difference or temperature change will be treated later.) A change of phase, under equilibrium conditions, will be at constant temperature and constant pressure. The thermal energy transfer, Q = Q , will be ∆H. Therefore, rev For example, the entropy change for the melting of ice was found to be 1.22 J/g·K or 22.0 J/mol·K. The entropy of vaporization of water at 100 C is o More typical liquids, not involving such strong intermolecular forces, have ∆S values of about vap 92 J/mol·K at their normal boiling points. This is known as Trouton’s rule. The work done in a reversible, isothermal expansion of an ideal gas was found to be (equation 10, Chapter 1) W = - nRT ln V /V . For an ideal gas at constant temperature ∆E = 0 rev 2 1 and therefore Q = -W . The entropy change is thus rev rev ( ) (15) /ln rev. I.G., , 12 VVnR T Q ST rev ==∆ Diabatic, like its better-known twin diabetic, implies that something “passes through” 6 (thermal energy or sugar). Thus adiabatic tells us there is no thermal energy transfer; Q = 0. In practice we achieve adiabatic conditions by insulating the system, by maintaining system and surroundings at the same temperature, or by carrying out the process very rapidly. Most gas expansions are fast enough to be nearly adiabatic. One sometimes sees the statement that ∆S = Q/T for a reversible process. This is not 7 really wrong, but it is doubly misleading. It de-emphasizes the need to find Q (which is often rev different from Q) and it provides no clue to finding ∆S for an irreversible process. Equation 5, ∆S = Q /T, covers both irreversible and reversible processes. rev 7/10/07 2- 41 If the gas expands through a pinhole or stopcock (an effusion-controlled leak) into an evacuated container (Figure 2), no work is done because the gas exerts forces only on the immovable walls of the container. By insulating the containers we can ensure that Q = 0. The energy is unchanged and therefore, if the gas is ideal, the temperature remains constant. This serves as an experimental demonstration (performed by Joule, in 1844) of the dependence of energy only on the temperature, although it is not as sensitive as a later method by Joule and Thomson. The entropy change of an ideal gas in such an irreversible, adiabatic expansion can be 6 calculated as follows. Entropy is a state function, so the entropy change depends only on the initial state and the final state. Entropy changes are independent of the path taken between the initial and final states. When an ideal gas expands at constant temperature from V to V , the final state is the same 1 2 whether the process is reversible or irreversible. Therefore, the entropy change for the adiabatic expansion of the ideal gas into a vacuum is (I.G., T) ∆S = nR ln V /V (16) 2 1 even though Q = -W = 0 in this process. The example just cited illustrates a general method for evaluating entropy changes in irreversible processes. Having determined what are the initial and final states, a path is found between those states that will be entirely reversible. Then equation 5 tells us ∆S = Q /T. rev 7 It is of interest to calculate the entropy change for the surroundings in the reversible and irreversible expansions above. In the reversible expansion, Q = Q = - Q = -(Q ) . rev surr rev surr 7/10/07 2- 42 Therefore (rev) ∆S = - nR ln V /V surr 2 1 The adiabatic expansion into a vacuum produces no change at all in the surroundings, and therefore ∆S = 0 surr Adding together the entropy changes for system and surroundings, we obtain, for the reversible expansion, (∆S) = 0 system + surroundings and for the irreversible expansion, into a vacuum, (∆S) = nR ln V /V > 0 system + surroundings 2 1 This result agrees with the requirements of the second law. The particular irreversible path chosen is an extreme case. If the gas had been allowed to expand against a constant external pressure (as in Figure 1, Chapter 1) the entropy change for system plus surroundings would have been positive, but less than that for the expansion into a vacuum. The magnitude of the total entropy change (system plus surroundings) can be taken as a measure of the degree of thermodynamic irreversibility of any process. INTERPRETATION OF ENTROPY. We have seen that we can measure the increase in entropy by measuring the amount of energy transferred to the system, as thermal energy transfer, Q, divided by the (absolute) temperature, provided we measure Q along a path between initial and final states that is completely reversible. To know whether a process will actually occur, we must make a similar measurement for changes in the surroundings, as we will emphasize below. Without getting involved in the detailed calculations of statistical mechanics, it is important to look more carefully at the meaning of entropy as a spread function. In particular, there are two ways of measuring the spreading, or the randomness, of a state. We have seen that entropy is appropriately called the spread function because it is a measure of the amount of spreading of energy among the molecules of a substance. However, in addition to considering where the packets of energy are located we should look also at where the molecules are (like the yellow and white rice molecules we mixed earlier). Remarkably, the same measurement of Q /T gives rev information on this second type of spreading. Consider a sample of an ideal gas, occupying an initial volume V (e.g., 1 m ). Assume, quite i 3 arbitrarily for the moment, that each molecule requires some very small volume, V (e.g., 2 Å , o 3 most of which volume elements are unoccupied at any given instant). Then we can say there are spaces for V /V molecules (5 x 10 in our example). If we let the gas expand to a larger volume, i o 29 V , then by the same reasoning there are now spaces for V /V molecules. The important thing is f f o 7/10/07 2- 43 that we have increased the number of available spaces by the ratio V /V . But we already know f i that if we expand an ideal gas, the entropy increases by ∆S = nR ln V /V . In other words, when f i we increased the number of available, or accessible, spaces by the ratio V /V , the entropy f i increased by the logarithm of this same ratio. We could equally well make the same kind of argument for molecules in a crystal, where the number of available spaces is more readily counted. If some number of “stray” molecules are introduced and given an opportunity to diffuse throughout the crystal, we would obtain a substantial increase in entropy per molecule. Or we may simply drop a crystal of salt into water. When we return, very likely the salt molecules will have diffused throughout the water. The salt molecules, or their constituent ions, are attracted to each other and roughly equally attracted to water molecules, so there is negligible energy difference, but there is a very high probability of the salt diffusing into the water, never to return to the crystal unless external conditions are changed. It is like a drop of water on the pavement. Even though energetically the water drop would prefer to remain with its neighbors, the lure of wide open spaces is enough to make the drop evaporate. The process is driven by an increase in entropy, or in probability. The crystal is highly ordered; the solution is highly disordered, or random. There are more states accessible to the salt when it is dissolved in water, or accessible to gas molecules when a larger volume is available to the molecules. More available, or accessible, states can be equated to a greater probability. Thus entropy is variously described as measuring disorder, or randomness, or probability. We can always organize molecules, forming crystals or condensing a gas to a liquid or arranging conditions such that molecules will form living structures. To do so always has a cost involved, specifically the cost of increased randomness in the surroundings. The second law of thermodynamics does not tell us we cannot decrease entropy in any given sample. It tells us only that, for the universe as a whole (system plus surroundings), entropy will always increase. A study of how signals may be communicated from one place to another gave rise to a new understanding of probability under the heading of information theory. Information theory turned out to be of major importance in understanding the meaning of entropy as well as in discovering better ways to organize the storage and distribution of information over telephone lines, as radio or television signals, or in computers. Gibbs Free Energy When investigating equilibrium or reversibility, it was necessary to find the change for the system but also for the surroundings. Usually we prefer to ignore the surroundings most of the time. The free energy functions provide a means of doing just that. We began with the variables P, V, and T, to which we added energy, E, and subsequently have added enthalpy, H, entropy, S, and now the two free energy functions, F and G, as shown in equations 17-19 and in Figure 3. H = E + PV (17) F = E - TS (18) G = H - TS (19) (22) V P G T =       ∂ ∂ (23) S T G P −=       ∂ ∂ For many years the Gibbs free energy and the Helmholtz free energy were not clearly 8 distinguished, and thus both were designated by the symbol F. In recent decades the symbol F has been selected for Helmholtz free energy and the symbol G for Gibbs free energy. Unless specifically indicated otherwise, the term “free energy” will always mean the Gibbs free energy in this book. [Fig. 3: A becomes F.] 7/10/07 2- 44 G = F + PV (19a) G = E + PV - TS (19b) We introduced the Helmholtz free energy from the relationship (eqn. 4) ∆A = ∆F = W and 8 rev the observation that a change in F represents the minimum amount of work that must be done on the system, at constant temperature, to get from the initial to the final state. For example, you cannot put a box on a shelf without doing some work on the box (and gravitational field), nor can you compress a gas without doing work on the gas. There is good reason for picking the correct function. Recall that we found enthalpy, H, to be much more convenient than energy, E, when measuring thermal energy transfers under conditions of constant pressure. For much the same reasons, we will find G to be generally more convenient than F. The Helmholtz free energy is more appropriate at constant volume; Gibbs free energy is better for constant pressure. FREE ENERGY AND EQUILIBRIUM. Assume for the moment that the only work done is work of expansion or compression, and that the composition of the system does not change. Then, from equation 19b, dG = dE + d(PV) - d(TS) As we have seen, because E is a state function, dE is independent of path, so dE = q + w = q + w rev rev regardless of path. We may therefore substitute q + w for dE, to obtain rev rev dG = q + w + P dV + V dP - T dS - S dT (20) rev rev But q = T dS and w = - P dV. Therefore, quite generally, rev rev dG = V dP - S dT (21) We effectively split this equation into two parts, by holding constant first the temperature, then the pressure: We introduced the assumption that w # w by considering an expansion or compression 9 rev of an ideal gas, and from that and the independence of dE on path, we concluded that q = dE - rev w $ q. Now we may work backward from the second law, which is one of our basic postulates, rev (dS) = q /T + (q ) /T $ 0. This inequality must be valid for any process occurring in syst + surr rev rev surr surr the system, including if q = (q ) , which may therefore be assumed without loss of generality. surr rev surr But we know that q = - q (by combining the first law with the first-law equation), and because surr we are assuming temperature is constant, we may set T = T. Therefore (dS) = q /T - surr syst + surr rev q/T $ 0 and thus q $ q. It follows from this that w # w . rev rev 7/10/07 2- 45 Note, of course, that P and V naturally go together, as do T and S. An equally valid alternative to equation 20 is obtained by substituting dE = q + w, giving dG = q + w + P dV + V dP - T dS - S dT Then, for constant temperature and pressure, (T,P) dG = q + w + P dV - T dS (24) But at constant pressure we may safely assume the pressure of the system is the same as the pressure of the surroundings (meaning the part of the surroundings that is in actual contact with, and therefore may affect, the system), so w = - P dV, where, as usual, P is the pressure of the system. Also, we know that q $ q. Therefore, 9 rev (T,P) dG = q - T dS = q - q # 0 rev Thus (T,P) dG # 0 (25) in general, and for the limiting case of the reversible, or equilibrium, process, (T,P) dG = 0 (25a) This shows that ∆G must be negative, or, in the limit of a reversible process, zero. That is, G is a minimum for an equilibrium state. This is the more convenient criterion for equilibrium that was sought. Looking for the change in free energy, G, under our normal operating conditions of ∫ ∫ ∫∫ ===       ∂ ∂ ==∆ 12 /ln / PPnRTPdPnRTdPVdP P G dGG T 7/10/07 2- 46 constant temperature and constant pressure, we need only look at changes of the system. It is not necessary to make separate calculations for the surroundings. We began this section with the assumption that the only work done was P dV work, or work of expansion or compression. Now we revisit that question. Retain the conditions of constant temperature and constant pressure, but from equation 20 we obtain (T, P) dG = P dV + w rev The total work done on the system may be broken into expansion/compression work and other forms (e.g., electrical work) which we label w’. For reversible processes, w = - P dV + w’ rev rev It follows that (T,P) dG = w’ (26) rev Thus, apart from providing a convenient measure of whether the system is or is not at equilibrium (with only P dV work), the free energy tells us how much other work can be expected from a process. In an electrochemical cell, w’ is the electrical energy generated by the chemical reaction. Work against a gravitational field or against a magnetic field would also be included in w’. IDEAL-GAS EXPANSIONS. The change in free energy of an ideal gas in an isothermal expansion or compression is particularly important because it has been found possible to express free energy changes of all systems, whether gases or not, in this same form. The broader theory will be developed in the discussion of physical equilibria. Start with the definition of G, equation 19. At constant temperature, (T) ∆G = ∆H - T ∆S (27) In an isothermal, reversible expansion of an ideal gas, ∆E = 0 and ∆H = 0. q = - w = InRT dV/V = nRT ln V /V rev rev 2 1 ∆S = q /T = nR ln V /V rev 2 1 ∆G = - T ∆S = - nRT ln V /V 2 1 or (T, I.G.) ∆G = nRT ln P /P (28) 2 1 The same result could be obtained from equation 22. FREE-ENERGY CHANGES IN CHEMICAL REACTIONS. A very important application of the free- energy function is to the description of chemical reactions. Because G is a state function, free energies can be added and subtracted in the same manner as enthalpies. It is not possible to 7/10/07 2- 47 assign meaningful absolute values to free energies, but handbook tables give values of G relative to the elements from which compounds are formed. Thus the free energy of formation of any element is, by definition, zero. From Table 1 we can see that the standard free energy of formation of SO in its standard state (gas, at 1 atm pressure, at 25 C) is - 370.4 kJ/mol, whereas 3 o ∆G (SO ) = - 300.2 kJ/mol. Subtracting (and setting ∆G (O ) = 0) gives ∆G = - 70.2 o o f 2 f 2 reaction kJ/mole. The negative value indicates that the reaction is spontaneous, showing that oxygen gas will combine with sulfur dioxide to form sulfur trioxide at this temperature and pressure, at least in the presence of a suitable catalyst. Table 1 STANDARD ENTHALPIES AND FREE ENERGIES OF FORMATION* Compound State ∆H ∆G Compound State ∆H ∆G f f f f o o o o AgBr c -100.4 -96.9 H O g -241.8 -228.6 2 AgCl c -127.0 -109.8 lq -285.8 -237.1 AgI c -61.8 -66.2 H O g -187.8 -120.4 2 2 Ag O c -31.1 -11.2 lq -136.3 -105.6 2 Al O c -1675.7 -1582.3 H S g -20.6 -33.4 2 3 2 Br g 30.9 3.1 I c(rh) 0 0 2 2 C diamond 1.9 2.9 NH g -45.9 -16.4 3 C graphite 0 0 aq -362.5 -236.5 CF g -933.6(639.9) 66.2 NH Cl c -314.4 -202.9 4 4 CCl lq -139.3 -68.6 N O g 81.6 103.2 4 2 CO g -110.5 -137.2 NO g 91.3 87.6 CO g -393.5 -394.4 NO g 33.2 51.3 2 2 CaCO calcite -1207.6 -1129.1 N O g 11.1 99.8 3 2 4 aragonite -1207.8 -1128.2 NaCl c -411.2 -384.1 CaCl c -795.4 -748.8 aq -407.3 -393.1 2 aq -877.2 -816. NaOH c -425.8 -379.5 CaF c -1228.0 -1175.6 aq -470.1 -419.1 2 aq -1208 -1031.2 O g 142.7 163.2 3 CaO c -634.9 -603.3 SO g -296.8 -300.1 2 Ca(OH) c -985.2 -897.5 SO g -395.7 -371.1 2 3 aq -1002.8 -858. CH g -74.6 -50.5 4 FeO c -272.0 -244.3 C H g 227.4 209.9 2 2 Fe O c -824. -742.2 C H g 52.4 68.4 2 3 2 4 Fe O c -1118.4 -1015.4 C H g -84.0 -32.0 3 4 2 6 HBr g -36.3 -53.4 CH OH g -201.0 -162.3 3 aq -121.6 -104.0 lq -239.2 -166.6 HF g -273.3 -275.4 C H OH g -234.8 -167.9 2 5 HI g 26.5 1.7 lq -277.6 -174.8 C H lq 49.1 124.5 6 6 g 82.9 129.7 *Values are in kJ/mol, for 25 C, 1 atm. The standard states for liquids (lq) and crystalline solids (c) are the pure o materials at 1 atm pressure and the standard states for gases (g) are the pure gases at 1 atm pressure in the “ideal gas state” (that is, extrapolated from low pressures assuming ideal gas behavior). “aq” refers to the (hypothetical ideal) 1 molal solution in water. ( ) T reactantsproducts T P GG P G         ∂ −∂ =       ∂ ∆∂ T reactants T products T P G P G P G       ∂ ∂ −         ∂ ∂ =       ∂ ∆∂ reactantsproducts T VV P G −=       ∂ ∆∂ (29) V P G T ∆=       ∂ ∆∂ 7/10/07 2- 48 Free energy values do not tell us anything about rates of reaction. There may be a significant activation energy required to get a reaction to proceed, even when it is thermodynamically allowed. (That is what catalysts are good for. By offering a different path, they may involve a substantially lower activation energy and therefore a faster rate.) On the other hand, if ∆G > 0, the reaction cannot proceed, under the specified conditions, regardless of catalysts or time. LECHATELIER’S PRINCIPLE AND EQUILIBRIUM. For every chemical reaction there exists an equilibrium point at which there is no further tendency for the reaction to proceed forward or backward. Although the equilibrium point of most reactions is far to one side or the other, a true equilibrium point can be calculated, and for nearly all systems the equilibrium point can be demonstrated if the experimental techniques are sufficiently sensitive. LeChatelier’s principle states that when any stress is applied to a system at equilibrium, the system will respond in a manner that will reduce the stress. For example, the reaction 2 NO + O !6 2 NO 2 2 releases thermal energy (∆H = - 56.53 kJ/mol(NO )) but causes a decrease of volume, or 2 pressure. Therefore an increase of temperature will tend to drive the reaction to the left, so that the reaction will absorb thermal energy, but an increase of pressure, by compression, will tend to drive the reaction to the right to decrease the number of moles of gas and hence the pressure. These relationships can be derived, quantitatively, from the second law. The change in ∆G = G - G with change of pressure is reaction products reactants The derivative of the sum is the sum of the derivatives: From equation 22, these derivatives are the respective volumes: or (30) S T G P ∆−=       ∂ ∆∂ (30a) T HG T G P ∆−∆ =       ∂ ∆∂ (31) T H T G P ∆ −=       ∂ ∆∂ It is assumed that there are no other changes that might cause a reverse effect. For 10 example, changes in the relative concentrations of the reactants, or addition of an inert gas, are not within the scope of equations 29 and 30 or 31. 7/10/07 2- 49 Similarly, from equation 23, This equation tells how ∆G (measured at some constant temperature and pressure) depends upon the temperature at which the process occurs. The ∆S on the right-hand side is similarly to be measured at a constant pressure and at a constant temperature (the temperature at which the derivative, or the slope of ∆G vs. T, is determined). We may therefore substitute for - ∆S the quantity (∆G - ∆H)/T. To find the shift of equilibrium point we set ∆G = 0 and obtain The first result, equation 29, says that if there is a volume increase during the process, a pressure increase will cause an increase in ∆G, making the process less spontaneous. The second equation (31) says that if the process is endothermic (∆H positive), a temperature increase will make ∆G more negative, thus making the process more spontaneous. The equations apply to phase changes or other equilibrium processes as well as to chemical reactions. The importance of the 10 equations is that they give a quantitative description of how ∆G changes, and therefore of how the equilibrium point changes. Free Energy and the Energy-Entropy Battle For the formation of ammonia, we may write the equation ½ N +3/2 H &6 NH 2 2 3 ∆G = - 16.4 kJ/mol, ∆H = - 45.9 kJ/mol, and ∆S = -99.2 J/mol·K. The reaction clearly is spontaneous at room temperature and pressure, but it doesn’t go. The activation barrier is too great. From LeChatelier’s rule, we can see that an increase in temperature, to overcome the [...]... such as constant pressure or constant volume have thus far been specified If pressure is constant, CP is inserted in these equations; if volume is constant, CV is required If neither pressure nor volume is constant, it may be necessary to find an alternative path consisting of constant pressure and constant temperature segments, or constant volume and constant temperature segments It may be, too, that... too, that it will be simpler to return to equation 32 For example, in a reversible, adiabatic expansion or compression, q = q rev = 0, and therefore it is clear that the integral, and the entropy change, must be identically zero To apply equation 34 one would have to define an “adiabatic heat capacity” that would always be zero 7/10/07 2- 51 The entropy change for the transfer of thermal energy between... distance, as shown in the lower part of Figure 4, but the temperature will be constant at each point in the system even though a small amount of thermal energy is flowing past each point The entropy change is the sum of three terms: the entropy change of the left-hand body at T1; the entropy change for the insulator; and the entropy change for the right-hand body at temperature T2 ∆S = ∆S1 + ∆Sinsulator... -86.36 +19.6 + 14. 1 +166.2 -163.1 +43 . 14 -198.2 +11.2 -333.3 Entropy Changes with Change of Temperature 7/10/07 2- 50 The fundamental equation for entropy change, equation 5, gives the infinitesimal entropy increase for the absorption of an infinitesimal amount of thermal energy at some temperature T If the temperature changes continuously during the process, the sum of the entropy changes can be written...activation barrier and reaction kinetics, will make the reaction less spontaneous, but an increase in pressure will drive the reaction back to the right The synthesis of ammonia is therefore carried out at high temperature (to get the reaction to go) and at high pressure (to recover yield) We can look at the reaction in a somewhat different way Write ∆G = ∆H - T∆S = -45 .9 kJ/mol - T x (-99.05... insulator added to the problem of Figure 4 plays no significant part in the calculation It can make little difference to the state, or the entropy, of a body that loses a given amount of thermal energy at the temperature T 1 what happens to this energy after it is gone 11 Actually, the entropy change of the insulator is zero, after the steady state has been reached, because the state of the insulator... written (25 oC and 1 atm of each gas) is spontaneous because the reactants give off thermal energy (∆H = Q < 0 at const P) without a major change in the thermal energy of the system (T is constant) The system, as a whole, is falling into a potential well, giving off energy that is conducted to the surroundings as thermal energy On the other hand, this drop in internal energy drives the system to a more... called black body radiation and requires the radiation field to be in equilibrium with the radiating body, which cannot be the case if radiation is being “leaked” to the low-temperature body.) We must therefore look for a reasonable substitute that will give us an answer suitably close to reality We achieve this approximation to reality by a thought experiment Add a thermal insulator connecting the two... larger volume and as speeds increase, the spread in speeds will increase as well), so the condensation of elements to compound(s) becomes less probable They must be driven together by increasing the pressure Of course, common sense tells us much the same thing, qualitatively At high temperatures, compounds tend to decompose At low temperatures, substances tend to condense, to liquids or to crystals Energy... integral is to be evaluated, but it can be handled quite easily as follows: ∆S = ∫ qrev q dT q dT dT = ∫ rev = ∫ rev = ∫C T T dT dT T T (33) The heat capacity, C, is often nearly independent of temperature (over temperature intervals of tens of degrees) so the equation can be simplified to the form ∆ S = C ln T2 T1 ( 34) Note that the heat capacity has not been given a subscript in equations 33 or 34; this . -858. CH g - 74. 6 -50.5 4 FeO c -272.0 - 244 .3 C H g 227 .4 209.9 2 2 Fe O c -8 24. - 742 .2 C H g 52 .4 68 .4 2 3 2 4 Fe O c -1118 .4 -1015 .4 C H g - 84. 0 -32.0 3 4 2 6 HBr g -36.3 -53 .4 CH OH g -201.0. constant, it may be necessary to find an alternative path consisting of constant pressure and constant temperature segments, or constant volume and constant temperature segments. It may be, too,. consider a system constrained to constant total volume, and to have a uniform temperature. Then, as before, we can write the total change in entropy that might result from a change in volume of either