An Introduction to Thermodynamics phần 7 pot

7/10/07 3- 79 f = p(c - 1) + 2 - c(p - 1) = c - p + 2 (36) Figure 5. Each area corresponds to a single phase. The lines are the common borders, and thus represent two phases in equilibrium (or three phases at a point) which is Gibbs’ phase rule, for two external degrees of freedom. You will recall from experience with drawing graphs that a point (or any finite number of points) corresponds to no degrees of freedom. A line (or a finite number of lines) represents one degree of freedom; setting one variable determines the other. Two degrees of freedom yield an area (or more than one area). ONE-COMPONENT PHASE DIAGRAMS. Water is a single component (c = 1), so the degrees of freedom may vary from zero to two. (c = 1) f = c - p + 2 = 3 - p (37) A single phase (vapor, liquid, or solid) is therefore represented by an area. Two phases (solid- liquid, solid-vapor, or vapor-liquid) are restricted to a line. Three phases yield a single point, called the triple point (solid + liquid + vapor). The phase diagram (with T on the vertical axis, to match following diagrams) is shown in Figure 5. You can calculate the slopes of the three lines from the Clapeyron and Clausius-Clapeyron equations, knowing the heats of fusion and of vaporization or sublimation and the densities of the phases. The upper end of the liquid-vapor line simply stops — above the critical point the distinction between liquid and vapor disappears. TWO-COMPONENT PHASE DIAGRAMS AT CONSTANT PRESSURE. We could not represent a single phase with two components on a two-dimensional page if we kept both temperature and pressure as external variables. We consider here the restricted example where pressure is maintained constant (e.g., at atmospheric pressure). Then Gibbs’ phase rule tells us the degrees of freedom are, again, with c = 2, (c = 2; const P) f = c - p + 1 = 3 - p (38) It follows that wherever there is a single phase present, that phase must be represented by an area (two degrees of freedom). If there are two phases present, the compositions are restricted to a line (or sometimes more than one line, but still one degree of freedom). Three phases can coexist 7/10/07 3- 80 only at a point, or a set of points, with no remaining degrees of freedom. Accordingly, an area can represent only a region in which there is a single phase, of variable composition. There are two apparent exceptions. First, the solubility of one solid in another is typically so Figure 3.6. Benzene-toluene equilibrium (liquid-vapor, at constant pressure). a) Calculated equilibrium states. b) Phase diagram. The area between the curved lines is not part of the equilibrium phase diagram. There can be no phase of such temperature and composition. small that the area of a “pure” solid will look like a single, vertical line. Second, when two phases, solid or liquid, are in equilibrium, those two phases will have different compositions and therefore different areas, bordered by lines that are separated from each other, as in the benzene- toluene diagram of Figure 6. Each line is simply the border of its respective area. If the overall composition falls between those lines, or areas, the material will split into variable amounts of the left and right areas (phases). The region between phases has been given the unfortunate label of a “two phase” region; a better description is a forbidden area. There can be no phase present with a composition falling in such a forbidden area. The relative amounts of the left phase and right phase are given by the lever-arm rule. If C and C are the compositions (as percentages of either L R phase — say the left-hand phase) and C is the overall composition of the mix at some point, the relative amounts (masses) of the left-hand and right-hand phases are given by m (C - C ) = m (C - C) (39) 1 L 2 R Similar behavior is shown by two immiscible solids or two immiscible liquids. Carbon tetrachloride and water liquids split into two layers, with very little solubility of either liquid in the other. At higher temperatures, however, the vapors mix freely. Thus the basic phase diagram is as shown in Figure 7a. There are three possible phases: the vapor, at higher temperatures, liquid CCl at room temperature and below, and liquid water. The two liquid phases can co-exist, but 4 there is no liquid phase with composition falling in the general range of 95% CCl /5% water to 4 95% water/5% CCl (where the numbers are illustrative only, and depend somewhat on 4 temperature). Figure 7a is somewhat difficult to interpret, so it is customary to add a horizontal line as a marker, to dramatize the lowest vapor temperature of the mixture, as in Figure 7b. (There can be no equilibrium across a horizontal line, which would connect points at different temperatures.) Also, if the diagram represents solid-liquid equilibrium, as for silver and copper in Figure 7c, it was recognized long before Gibbs’ work of the late 19 century that when the liquid th 7/10/07 3- 81 solution is cooled, either silver or copper will precipitate out first (depending on whether the composition lies to the left or right of the low point, called the eutectic). The composition of the remaining liquid accordingly moves toward the eutectic composition. In the last stage of precipitation, or freezing, the mixture that comes down is a fine mix of two phases, silver and copper, with the overall composition of the eutectic. Accordingly, a vertical line was drawn below the eutectic point to represent this eutectic mixture — two distinct phases, but with a distinctive appearance. Over decades, the vertical line was changed to a dotted line (to a b c Figure 3.7. a) Bare-bones phase diagram for two immiscible phases with miscible high- temperature phase (e.g., CCl and H O, liquid-vapor, or Silver-Copper, solid-liquid). b) The 4 2 horizontal line guides the eye. c) Early diagrams showed the eutectic mixture as if it were a separate phase, or (later) as a dotted line. indicate there is no phase with composition falling in this broad forbidden region) and more recently the vertical line is largely omitted. Examining a large number of such phase diagrams, the two patterns shown here will be recognized as appearing in combination. Often a solid compound forms, so the diagram can conceptually be divided in two, element one plus compound on the left and compound plus element two on the right. There may be regions of solubility, resembling the benzene-toluene diagram, within a larger phase diagram. Some compounds may spontaneously decompose as the temperature rises, yielding separate phases. The variety is seemingly endless. But almost all can be recognized as relatively simple combinations of these two basic patterns. Problems 1. For the transformation of graphite to diamond at 25 C, ∆G = 2.9 kJ/mole. The densities are o 2.25 and 3.51 g/cm . What is the minimum pressure required to make diamond 3 thermodynamically stable at this temperature? 2. At 1114 C the vapor pressure of Ni was observed to be 7.50 x 10 torr and at 1142 C it was o -8 o 14.33 x 10 torr. What is the heat of vaporization of nickel in this range? -8 3. Is the fugacity of ice increased or decreased by exerting pressure on the ice? Which will show the greatest change in fugacity for a pressure increase of 10 atm: ice, or liquid water? Explain. 4. The vapor pressure of a liquid compound, Y, obeys the equation ln P = a + bT , with P the -1 pressure in atm and T in kelvin. .90.0 )H/CC(HgBr O)/HC(HgBr 66 2 2 2 = 7/10/07 3- 82 a. What is the boiling point of Y? b. What is the heat of vaporization at the normal boiling point? 5. Find the mole fractions of the several components if 24 g of methane, CH , is mixed with 7 g 4 of CO, 14 g of N , and 8 g of He. 2 6. A saturated solution of benzoic acid, C H COOH, contains 0.32 g in 100 g of water. The 6 5 density of the solution is 1.3 g/cm . What is the molarity of the solution? 3 7. Find the mole fraction, molarity, and molality of iodine when 0.10 g of iodine is dissolved in 100 g of CCl (density 1.59 g/cm ). 4 3 8. a. What is the molarity of water in pure water? b. What is the molality of water in pure water? c. What is the molarity of CCl in pure CCl (density 1.59 g/cm )? 4 4 3 d. What is the molality of CCl in pure CCl ? 4 4 e. What is the molarity of an ideal gas at 1 atm and 25 C? o 9. A solution contains, by weight, 40% water, 35% ethanol (C H OH), and 25% acetone 2 5 (CH COCH ). 3 3 a. Find the mole fraction of each component. b. Find the molality of the ethanol and acetone if water is considered to be the solvent. c. Find the molality of the water and acetone if the ethanol is considered to be the solvent. 10. The solubility of CCl in water at room temperature is about 0.9 g/L and the vapor pressure 4 of CCl is about 100 torr. What would be the equilibrium vapor pressure of CCl above a beaker 4 4 containing CCl covered with a layer of water? 4 11. Water in equilibrium with air (20.9% O , 79.1% N by pressure, volume, or mole fraction) at 2 2 0 C contains 1.28 x 10 mole of air per liter, of which 34.91% is O . Calculate the Henry’s law o -3 2 constant a. for O 2 b. for N 2 12. Is the Henry’s law constant for benzene in water large or small? 13. The distribution coefficient for mercuric bromide, HgBr , between water and benzene is 0.90. 2 That is, with concentrations expressed as molarities, An aqueous solution contains 0.010 M HgBr . 50 ml of this solution is to be extracted with 150 2 ml of benzene. a. What fraction of the HgBr is left in the aqueous phase if the extraction is made with one 2 150 ml portion of benzene? b. What fraction is left if the extraction is made with three successive 50 ml portions of benzene? 14. Over a certain range of concentrations, the volume of a solution containing m moles of NaCl in 1 kg water has been found to be, in ml, V = 1002.935 + 16.670 m + 1.636 m + 0.170 m 3/2 2 Find the partial molal volume of a. NaCl in a 2 m solution b. water in a 2 m NaCl solution c. pure NaCl (density 2.165 g/cm ) 3 2 1 1 RT TH N ∆∆ =∆ )( 111 BA HHH −=∆ 7/10/07 3- 83 15. The vapor pressures of benzene and toluene, at 27 C, are 120 and 40 torr. Assuming an ideal o solution, what is the composition of the vapor above a solution containing 100 g of benzene (C H ) and 150 g of toluene (C H CH )? 6 6 6 5 3 16. In your work for Adulterated Chemicals, Inc., you have isolated a new antibiotic through a lengthy series of extractions, biological tests, and so forth. A few milligrams are available, and by the ultra-centrifuge method it has been found that the molar mass is 10,000. It is desired to check this by another method. For a 1% by weight solution of the substance in water, calculate a. the freezing-point depression b. the boiling-point elevation c. the change in vapor pressure at 25 C o d. the osmotic pressure, at 25 C, in cm H O. (The density of mercury is 13.6 g/cm .) o 3 2 Which method would you recommend? 17. Boiling occurs in a solution when the sum of the vapor pressures of the components is equal to the atmospheric pressure (or the applied pressure), but these vapor pressures are lower than for the pure materials (cf. Raoult’s law, equation 29c). If two liquids are immiscible, each exhibits its own vapor pressure (in each phase), and “steam distillation” occurs when the sum of the vapor pressures is equal to the atmospheric pressure. a. Calculate the vapor pressure of pure water at 99 C. o b. What vapor pressure must a compound, immiscible with water, have to steam distill at 99 C when the atmospheric pressure is 745 torr? o c. What would be the composition of the vapor in such a steam distillation? 18. A solution of 3.795 g of sulfur in 100 g of CS boils at 46.66 C. The boiling point of pure 2 o CS is 46.30 C and the heat of vaporization of CS is 26.8 kJ/mol. From this experimental result, 2 2 o what is the probable formula of the sulfur molecule in the solution? 19. Show that if two phases (A and B) in equilibrium are both slightly impure, the resultant change in temperature (change in melting point or boiling point, relative to pure phases), ∆T, is given by where ∆N is the difference in mole fraction of the major component (∆N = N - N ) and ∆H is 1 1 1 1 1 A B the enthalpy of transition . (1) reactantsproducts BADC jjii GbGaGdGc GnGnG −−+= −=∆ ∑ ∑ (2) o B o A o D o C o GbGaGdGcG −−+=∆ 7/10/07 4- 84 4 Chemical Equilibrium Chemical equilibrium is one of the most important applications of thermodynamics. The theory follows very simply from the equations already derived. Setting up the necessary equations in practice requires only an understanding of the meaning of a chemical equation. Most mistakes arise in the elementary process of counting the molecules or moles of reactants and products. For some systems the solution of the mathematical equations requires some skill and, especially, some understanding of the physical implications so that reasonable assumptions and approximations can be made. For the types of problems considered here, the primary requirement is to follow a simple, methodical procedure in finding the appropriate equations, and then to insert numerical values into these equations. Solution of the equations is then quite easy. Free-Energy Changes in Chemical Reactions A completely general chemical reaction may be represented in the form aA + bB 6 cC + dD This should be interpreted as follows: Under certain conditions of overall temperature, pressure, and concentrations of A, B, C, and D (not yet specified here), a moles of substance A react with b moles of substance B to form c moles of substance C and d moles of substance D. Although some of the reactants (A and B and perhaps others) are consumed and some products (C and D and perhaps others) are formed, it is assumed that there is no change in temperature, in pressure, or in concentrations of any of the substances as a consequence of the chemical reaction. The free-energy change, if the reaction proceeds as written, will be The values of the partial molal, or effective, free energies on the right-hand side depend on the states of these compounds, including the temperature, the pressure, and the individual concentrations. Thus ∆G may have any value, depending on the states of reactants and products. Remember that, for purposes of the calculation, the states of reactants and products are unchanged by the reaction. The temperature, pressure, and concentration of each substance is the same before and after the reaction. The reaction may involve a negligible fraction of the substances present, or reactants may be added and products removed as the reaction proceeds. Tables give ∆G = ∆G for the reaction occurring with all reactants and products in their o standard states. ( ) ( ) ( ) ( ) (3) o BB o AA o DD o CC o GGbGGaGGdGGcGG −−−−−+−=∆−∆ ( ) ( ) TBfRTG TBfRTG i o i o i iii += += ln and ln (4) ln o i i o ii f f RTGG =−         −−+=∆−∆ o B B o A A o D D o C C o f f b f f a f f d f f cRTGG lnlnlnln (5) o i i i f f a = ( ) b B a A d D c C o aaaaRTGG lnlnlnln −−+=∆−∆ (6) ln b B a A d D c C o aa aa RTGG +∆=∆ ( ) (7) ln aRTGG o Q +∆=∆ ( ) (8) b B a A d D c C aa aa a =Q 7/10/07 4- 85 Subtraction of equation 2 from equation 1 gives At this point it is convenient to substitute the fugacities (equation 7a, Chapter 3) for the free- energy terms. Therefore, Substitution of equation 4 into equation 3, for each of the substances, gives The ratio of the fugacity of any substance, f , to the fugacity of the same substance in its standard i state, f , is called the “activity”, a . i i o It follows from the definition that activity is always dimensionless and without units. Substitution of activity for the ratio of fugacities and replacement of n ln x by ln x gives n which can be further simplified to the form It will be convenient to write this in the form defining Q (a) to be ( ) ( ) (9) atm 1 atm atm P P f f a o === More generally, the standard state of any gas is taken as the ideal gas at 1 atm pressure; 1 that is, it is a fictitious state with properties obtained from the properties of the actual gas, at sufficiently low pressures that the gas is ideal, extrapolated to 1 atm pressure using the equations for behavior of an ideal gas. The numerical value of the activity coefficient is the ratio of the fugacity of the real gas to that of the idealized gas. The dimensions are reciprocal pressure. 7/10/07 4- 86 or the corresponding form appropriate to the particular reaction. Equation 6 is a very important result with obvious physical interpretation. If all substances (reactants and products) are in their standard states, all activities are, by definition, equal to 1. Then Q (a) = 1, ln Q (a) = 0, and ∆G = ∆G . The correction term depends on the value of Q (a), o which depends on the individual activities and therefore on the fugacities, or on the pressure and concentration of each substance at the given temperature. Standard States The choice of standard states affects the value of ∆G as it appears in tables, and determines o the relationship between the state of a compound and its activity. There are certain conventions that are customarily followed, so that it is generally considered unnecessary to explain in detail the choice of standard states that has been made. Unless the conventions are understood, therefore, it may not be possible to apply thermodynamic values found in the literature. Standard states are chosen to give a convenient form for the activity. GASES. For an ideal gas, fugacity is equal to pressure. We choose the standard state to be 1 atm pressure. Then f = P, f = P = 1 atm: o o That is, the activity, which is a dimensionless ratio, is equal to the numerical value of the pressure, with the pressure expressed in atmospheres. This simple relationship can be extended to cover real gases by the introduction of an “activity coefficient”, γ (Greek gamma), that will equal 1 if the gas is ideal, and will differ from 1 as the gas deviates from ideality. a = γP (10) The activity coefficient defined in this way has units of atm . Equation 10 is applicable to any -1 gas, real or ideal. If the gas is non-ideal, it is necessary to know the value of γ. 1 LIQUIDS AND SOLUTIONS. The major component of an ideal solution has a fugacity given by Raoult’s law — (11) 1 1 1 1 N f f a o == (12) 111 Na γ = (13) 2 2 2 2 2 c k f f f a o === 7/10/07 4- 87 f = f N 1 1 1 o — in which f is the fugacity of the pure solvent. We choose the pure solvent as the standard 1 o state so that f is the fugacity in the standard state. Then 1 o Thus the activity of the solvent is equal to its mole fraction, in the ideal solution, and the activity of any pure liquid is 1. Not all solutions are ideal, so it is convenient to allow for deviations from Raoult’s law by introducing an activity coefficient, as for gases, defined by the equation This activity coefficient is dimensionless. It will be 1 for a pure liquid, for a solvent in an infinitely dilute solution, or for a component of an ideal solution. It may differ appreciably from 1 when the fugacity of the liquid in solution deviates from Raoult’s law. If a solute obeys Henry’s law, the fugacity is proportional to concentration and the standard state of the solute may be chosen such that f = k, the Henry’s law constant (even though such a 2 o standard state may be physically unattainable). Then with |c | representing the numerical value of c . For any particular concentration, the value of the 2 2 activity depends on the units chosen for expressing concentration. Common choices are molarity, molality, or sometimes mole fraction (especially when the choice between solute and solvent is ambiguous). Activity is dimensionless but the activity coefficient has dimensions of inverse concentration. For example, if c is in mole per liter, γ has units of liter per mole. In any case, 2 2 a = γ c 2 2 2 Molality has some advantages in accommodating changing temperature. Nevertheless, in subsequent discussions we will assume the concentration of solute is expressed as molarity. For dilute water solutions the difference between molarity and molality is negligible. Concentrated solutions may show large deviations from Henry’s law, as a consequence of large intermolecular forces. This may also occur with uncharged solutes, leading, for example, to a maximum-boiling or minimum-boiling solution, known as an azeotrope. It is also expected for multiply charged ions (which typically form complex ions of smaller charge). In dilute solutions, activity coefficients of ions are less than 1, approaching 1 at infinite dilution. Activity coefficients of ions in concentrated solutions (on the order of 1 M) may increase and become greater than 1. Strong electrolytes fully dissociate in aqueous solution, so the concentration of undissociated electrolyte may be assumed to be vanishingly small. We can eliminate consideration of the undissociated electrolyte by defining its activity to be equal to the product of the activities of the ions into which it dissociates. Then the equilibrium constant for the dissociation is 1, ∆G for the o ( ) ( ) dTCTHTH T P ∫ += 2 1 T 12 S T G P −=       ∂ ∂ It is possible to assign an arbitrary absolute value for entropy, but such methods always 2 omit some parts of S, and thus should be regarded as indeterminate. Note, however, that this limitation does not apply to finding the change in ∆G (as, for example, in finding the change in ∆G ) with change in temperature. This change depends on ∆S, which can be measured. The reaction temperature dependence of fugacity can also be found. 7/10/07 4- 88 dissociation is zero, and the irrelevant quantities disappear from the calculations. SOLIDS. The fugacity of a solid is very nearly constant, at a given temperature, independent of the applied pressure (except for extreme pressures). We choose this normal value of the fugacity as f , so the solid has an activity of 1. Appreciable deviations from unit activity for a solid may o be expected only under very high pressures, when impurities are present (as in mercury-metal amalgams), when the solid is severely strained, or when the solid exists as extremely small particles for which surface effects cannot be neglected. VARIATIONS AND LIMITATION. Other choices of standard states are possible and may be convenient for special circumstances. For example, the standard state of a gas may be taken as a concentration of 1 mol/L rather than at 1 atm. Such a choice seldom, if ever, appears in tables of standard free energies, but may occasionally be found, especially in introductory discussions, for applications to chemical equilibria. There is an important distinction between the choices of standard states for enthalpy and for free energy. Both are commonly taken for 1 atm pressure and for the most stable phase under the standard conditions. However, enthalpies are given for a fixed temperature, usually 25 C (but o sometimes 18 C). Corrections for other temperatures are made by calculations involving the heat o capacity. On the other hand, the temperature dependence of the free energy is given by equation 23, Chapter 2, We have defined entropy only through ∆S, and therefore subject to an arbitrary additive constant. The change in free energy with temperature is thus also subject to an arbitrary additive constant, and therefore is unknown. It is therefore necessary to choose the standard state for the 2 specification of free energy to be at the temperature of interest. CALCULATIONS. We are now prepared to calculate free-energy changes for specific chemical reactions for particular choices of conditions. Consider first the reaction for the synthesis of ammonia at 25 C when each of the three gases is present at a partial pressure of 5 atm. The o standard free-energy change, ∆G , is - 16.64 J/mol (NH ). Assume the gases are ideal. o 3 [...]... true equilibrium constant (or “thermodynamic equilibrium constant”) is rigorously constant! 7/ 10/ 07 4- 90 ∆Go = - RT ln K ( 17) This equation was derived without benefit of arbitrary assumptions, so it must be satisfied for all chemical equilibria There is a strong tendency to misinterpret equation 17 The left-hand side refers to standard state conditions, the right-hand side to equilibrium conditions,... equilibrium point, and the equilibrium constant is a measure of how much the equilibrium concentrations (or activities) differ from their standardstate values (activities equal to 1) The minus sign shows that the left-hand (standard state) side deviation from equilibrium is equal and opposite to the right-hand (equilibrium) side deviation from standard states Application of the equilibrium constant is straightforward... confusion) is that many textbooks state that the equilibrium constant depends on pressure or on the presence of other substances, but this is because they are not talking about the equilibrium constant but rather the value of Q (c) at equilibrium, which is not necessarily a constant The “changes in the equilibrium constant” referred to are really changes in the activity coefficients, or changes in the deviations... ⋅ 298 ln 10 −4 10 −3 ) 2 = 55900 + 2 478 ln10 −10 = - 1.150 kJ/mol The dissolution of HCl gas, at a pressure of 0.10 atm, is somewhat less spontaneous if the water contains 0.50 M HCl The standard free energies of HCl(g) and HCl(aq) are - 95. 27 and - 131.2 kJ/mol, so ∆Go = - 131.2 - (- 95. 27) = - 35.9 kJ/mol, and (by chosen convention) a HCl(aq) = aH+· aCl- 7/ 10/ 07 4- 89 ∆G = ∆G o + RT ln a H + aCl −... Constants At equilibrium, under constant temperature and pressure, the free-energy change is zero and equation 6 becomes  ac ad 0 = ∆ G o + RT ln  C D  aaab  A B   = ∆ G o + RT ln Q (a )eq   eq or  ac ad ∆ G o = − RT ln  C D  aaab  A B     eq (15) For any given reaction, the left-hand side, ∆G o, is a constant, obtainable from handbook tables It depends only on (1) the choice of standard... choice of standard states, and (2) the temperature Therefore, at any given temperature the right-hand side of equation 15 — and therefore Q (a) eq — must also be a constant, which is called the equilibrium constant, K (or K eq) Q (a )eq c d  aC a D  a b =  a AaB   = K   eq (16) This equation is really quite a remarkable result, and one that students are often unwilling to accept in its full implications... helpful to think of the equation in an expanded form: (∆G)std st - (∆G)equil = RT ln Q std st - RT ln Q equil Now we recognize that (∆G)std st = ∆Go, (∆G)equil = 0, Q std st = 1 and ln 1 = 0, and Q equil = K Therefore the equation may be written as ∆Go - 0 = 0 - RT ln K which may be interpreted as saying that ∆G o is a measure of how far the standard-state concentrations (or activities) differ from an equilibrium... says that, despite changes of pressure, concentrations of reactants or products, or the presence of other substances, the particular function of activities represented by Q (a) can assume only one value at equilibrium, for a given temperature and choice of standard states by which the activities are defined 3 The value of Q (a) at equilibrium is determined by equation 15, which can be rewritten as 3... N2 3/ 2 = − 16 ,640 + 2, 478 ln H2 5 25 = − 20 62 kJ/mol (NH 3 ) The standard free-energy change for the decomposition of Ag 2O at 25 oC is 10.84 kJ/mol The free-energy change for Ag2O in air (20% O2) will be slightly different: Ag2O 6 2 Ag + ½ O2 (a ) (a ) 2 ∆G = ∆G + RT ln o 1/ 2 Ag O2 aAg2O Activities of the solids are 1 and the activity of the oxygen is 0.20 (equal to the pressure in atmospheres)... ln 5 = 8.846 kJ/mol The free-energy change for the dissolution of CaF 2 in water at 25 oC, containing 0.001 M NaF and 0.0001 M CaCl2 may be calculated as follows The standard free energies of solid CaF2 and of CaF2 as a solute are -1,162 and -1,106 kJ/mol Thus ∆Go for the dissolution process is 55.90 kJ/mol CaF2 -6 Ca++ + 2 FSetting aCa++ = |cCa++| = 0.0001 and aF- = |cF-| = 0.001 gives ( )( ∆G . equilibrium constant”) is rigorously constant! 7/ 10/ 07 4- 90 Equilibrium Constants At equilibrium, under constant temperature and pressure, the free-energy change is zero and equation 6 becomes For any given. g/cm ) 3 2 1 1 RT TH N ∆∆ =∆ )( 111 BA HHH −=∆ 7/ 10/ 07 3- 83 15. The vapor pressures of benzene and toluene, at 27 C, are 120 and 40 torr. Assuming an ideal o solution, what is the composition. pressure, and concentrations of A, B, C, and D (not yet specified here), a moles of substance A react with b moles of substance B to form c moles of substance C and d moles of substance D. Although some