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Exercícios resolvidos da 10ª edição do livro Físico-Química - Atkins

Tipologia: Exercícios

2020

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Baixe Físico-Química - Atkins 10ed - Exercícios resolvidos e outras Exercícios em PDF para Físico-Química, somente na Docsity! 1 The properties of gases 1A The perfect gas Answers to discussion questions 1A.2 The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it occupied alone the same container as the mixture at the same temperature. Dalton’s law is a limiting law because it holds exactly only under conditions where the gases have no effect upon each other. This can only be true in the limit of zero pressure where the molecules of the gas are very far apart. Hence, Dalton’s law holds exactly only for a mixture of perfect gases; for real gases, the law is only an approximation. Solutions to exercises 1A.1(b) The perfect gas law [1A.5] is pV = nRT, implying that the pressure would be p = nRT V All quantities on the right are given to us except n, which can be computed from the given mass of Ar. n = 25 g 39.95 −1g mol = 0.626 mol so p = (0.626 mol) × (8.31× 10−2 dm3 bar K−1 mol−1) × (30 + 273) K 1.5 dm3 = 10.5bar So no, the sample would not exert a pressure of 2.0 bar. 1A.2(b) Boyle’s law [1A.4a] applies. pV = constant so pfVf = piVi Solve for the initial pressure: (i) p i = p f V f V i = (1.97 bar) × (2.14dm3) (2.14 + 1.80)dm3 = 1.07 bar (ii) The original pressure in Torr is p i = (1.07 bar) × 1 atm 1.013 bar × 760 Torr 1 atm = 803 Torr 1A.3(b) The relation between pressure and temperature at constant volume can be derived from the perfect gas law, pV = nRT [1A.5] so p ∝ T and p i T i = p f T f The final pressure, then, ought to be p f = p i T f T i = (125 kPa) × (11+ 273)K (23 + 273)K = 120 kPa 1A.4(b) According to the perfect gas law [1.8], one can compute the amount of gas from pressure, temperature, and volume. pV = nRT so n = pV RT = (1.00 atm) × (1.013× 105 Pa atm−1) × (4.00 × 103m3) (8.3145 J K−1mol−1) × (20 + 273)K = 1.66 × 105 mol Once this is done, the mass of the gas can be computed from the amount and the molar mass: m = (1.66 × 105 mol) × (16.04 −1g mol ) = 2.67 × 106 g = 2.67 × 103 kg 1A.5(b) The total pressure is the external pressure plus the hydrostatic pressure [1A.1], making the total pressure 1 p = pex + ρgh . Let pex be the pressure at the top of the straw and p the pressure on the surface of the liquid (atmospheric pressure). Thus the pressure difference is p − p ex = ρgh = (1.0 −3g cm ) × 1 kg 103 g × 1 cm 10−2 m 3 × (9.81 m s−2 ) × (0.15m) = 1.5 × 103 Pa = 1.5 × 10−2 atm 1A.6(b) The pressure in the apparatus is given by p = pex + ρgh [1A.1] where pex = 760 Torr = 1 atm = 1.013×10 5 Pa, and ρgh = 13.55 g cm−3 × 1 kg 103 g × 1 cm 10−2 m 3 × 0.100 m × 9.806 m s−2 = 1.33 × 104 Pa p = 1.013× 10 5 Pa + 1.33 × 104 Pa = 1.146 × 105 Pa = 115 kPa 1A.7(b) Rearrange the perfect gas equation [1A.5] to give R = pV nT = pV m T All gases are perfect in the limit of zero pressure. Therefore the value of pVm/T extrapolated to zero pressure will give the best value of R. The molar mass can be introduced through pV = nRT = m M RT which upon rearrangement gives M = m V RT p = ρ RT p The best value of M is obtained from an extrapolation of ρ/p versus p to zero pressure; the intercept is M/RT. Draw up the following table: From Figure 1A.1(a), R = lim p→0 pV m T = 0.082 062 dm3 atm K−1 mol−1 Figure 1A.1 (a) p/atm (pVm/T)/(dm 3 atm K–1 mol–1) (ρ/p)/(g dm–3 atm–1) 0.750 000 0.082 0014 1.428 59 0.500 000 0.082 0227 1.428 22 0.250 000 0.082 0414 1.427 90 2 so p tot = n tot RT V = (4.0 mol) × (0.08206 dm3 atm mol−1 K−1) × (273.15 K) 22.4 dm3 = 4.0 atm (ii) The partial pressures are p N = x N p tot = (0.63) × (4.0 atm) = 2.5 atm and p H = (0.37) × (4.0 atm) = 1.5 atm (iii) p = p H + p N [1A.10] = (2.5 + 1.5) atm = 4.0 atm Solutions to problems 1A.2 Solving for n from the perfect gas equation [1A.5] yields pV n RT = . From the definition of molar mass mn M = , hence ρ = m V = Mp RT . Rearrangement yields the desired relation, namely RTp M ρ= . Therefore, for ideal gases p ρ = RT M and M = RT p / ρ . For real gases, find the zero-pressure limit of p ρ by plotting it against p. Draw up the following table. p/(kPa) 12.223 25.20 36.97 60.37 85.23 101.3 ρ/(kg m–3) 0.225 0.456 0.664 1.062 1.468 1.734 p / ρ 103 m2 s−2 54.3 55.3 55.7 56.8 58.1 58.4 Bear in mind that 1 kPa = 103 kg m–1 s–2. p ρ is plotted in Figure 1A.2. A straight line fits the data rather well. The extrapolation to p = 0 yields an intercept of 54.0×103 m2 s–2 . Then M = RT 5.40 × 104 m2 s−2 = (8.3145 J K−1 mol−1) × (298.15K) 5.40 × 104 m2 s−2 = 0.0459 kg mol−1 = 45.9 −1g mol Figure 1A.2 5 Comment. This method of the determination of the molar masses of gaseous compounds is due to Cannizarro who presented it at the Karlsruhe Congress of 1860. That conference had been called to resolve the problem of the determination of the molar masses of atoms and molecules and the molecular formulas of compounds. 1A.4 The mass of displaced gas is ρV, where V is the volume of the bulb and ρ is the density of the displaced gas. The balance condition for the two gases is m(bulb) = ρV(bulb) and m(bulb) = ρ′V(bulb) which implies that ρ = ρ′. Because [Problem 1.2] ρ = pM RT the balance condition is pM = p′M′ , which implies that ′M = p ′p × M This relation is valid in the limit of zero pressure (for a gas behaving perfectly). In experiment 1, p = 423.22 Torr, p′ = 327.10 Torr; hence ′M = 423.22 Torr 327.10 Torr × 70.014 g mol−1 = 90.59 g mol−1 In experiment 2, p = 427.22 Torr, p′ = 293.22 Torr; hence ′M = 427.22 Torr 293.22 Torr × 70.014 g mol−1 = 102.0 g mol−1 In a proper series of experiments one should reduce the pressure (e.g. by adjusting the balanced weight). Experiment 2 is closer to zero pressure than experiment 1, so it is more likely to be close to the true value: ′M ≈ 102 g mol−1 The molecules CH2FCF3 and CHF2CHF2 have molar mass of 102 g mol –1. Comment. The substantial difference in molar mass between the two experiments ought to make us wary of confidently accepting the result of Experiment 2, even if it is the more likely estimate. 1A.6 We assume that no H2 remains after the reaction has gone to completion. The balanced equation is N2 + 3 H2 → 2 NH3 . We can draw up the following table N2 H2 NH3 Total Initial amount n n′ 0 n + n′ Final amount n − 13 ′n 0 2 3 ′n n + 1 3 ′n Specifically 0.33 mol 0 1.33 mol 1.66 mol Mole fraction 0.20 0 0.80 1.00 p = nRT V = (1.66 mol) × (0.08206 dm3 atm mol−1 K−1) × (273.15K) 22.4dm3 = 1.66 atm p(H2) = x(H2)p = 0 p(N2) = x(N2)p = 0.20 × 1.66 atm = 0.33 atm p(NH3) = x(NH3)p = 0.80 × 1.66 atm = 1.33 atm 1A.8 The perfect gas law is pV = nRT so n = pV RT At mid-latitudes n = (1.00atm) ×{(1.00dm2 ) × (250 × 10−3 cm) / 10 cm dm−1} (0.08206dm3 atm K−1mol−1) × (273K) = 1.12 × 10−3 mol In the ozone hole n = (1.00atm) ×{(1.00dm2 ) × (100 × 10−3 cm) / 10cm dm−1} (0.08206dm3 atm K−1mol−1) × (273 K) = 4.46 × 10−4 mol The corresponding concentrations are 6 n V = 1.12 × 10−3 mol (1.00dm2 ) × (40 × 103 m) × (10dm m−1) = 2.8 × 10−9 moldm−3 and n V = 4.46 × 10−4 mol (1.00dm2 ) × (40 × 103 m) × (10dm m−1) = 1.1× 10−9 moldm−3 respectively. 1A.10 The perfect gas law [1A.5] can be rearranged to n = pV RT The volume of the balloon is V = 4π 3 r3 = 4π 3 × (3.0 m)3 = 113 m3 (a) n = (1.0atm) × (113 × 103 dm3) (0.08206 dm3 atm mol−1 K−1) × (298 K) = 4.62 × 103 mol (b) The mass that the balloon can lift is the difference between the mass of displaced air and the mass of the balloon. We assume that the mass of the balloon is essentially that of the gas it encloses: m = m(H 2 ) = nM (H 2 ) = (4.62 × 103 mol) × (2.02 −1g mol ) = 9.33 × 103 g Mass of displaced air = (113 m 3) × (1.22 −3kg m ) = 1.38 × 102 kg Therefore, the mass of the maximum payload is 138 kg − 9.33 kg = 1.3× 102 kg (c) For helium, m = nM (He) = (4.62 × 10 3 mol) × (4.00 g mol−1) = 18kg The maximum payload is now 138 kg − 18kg = 1.2 × 102 kg 1A.12 Avogadro’s principle states that equal volumes of gases contain equal amounts (moles) of the gases, so the volume mixing ratio is equal to the mole fraction. The definition of partial pressures is pJ = xJp . The perfect gas law is pV = nRT so n J V = p J RT = x J p RT (a) n(CCl 3 F) V = (261× 10−12 ) × (1.0atm) (0.08206dm3 atm K−1mol−1) × (10 + 273) K = 1.1× 10−11 moldm-3 and n(CCl 2 F 2 ) V = (509 × 10−12 ) × (1.0atm) (0.08206dm3 atm K−1mol−1) × (10 + 273) K = 2.2 × 10−11 moldm-3 (b) n(CCl 3 F) V = (261× 10−12 ) × (0.050atm) (0.08206dm3 atm K−1mol−1) × (200 K) = 8.0 × 10−13 moldm-3 and n(CCl 2 F 2 ) V = (509 × 10−12 ) × (0.050atm) (0.08206dm3 atm K−1mol−1) × (200 K) = 1.6 × 10−12 moldm-3 1B The kinetic model Answers to discussion questions 1B.2 The formula for the mean free path [eqn 1B.13] is λ = kT σ p In a container of constant volume, the mean free path is directly proportional to temperature and inversely proportional to pressure. The former dependence can be rationalized by noting that the faster the molecules travel, the farther on average they go between collisions. The latter also makes sense in that the lower the pressure, the less frequent are collisions, 7 p = kT σλ = (1.381× 10−23 J K−1)(293 K) 0.36 × (10−9 m)2 (10 × 0.34 × 10−9 m) = 3.3 × 106 J m−3 = 3.3 MPa Comment. This pressure works out to 33 bar (about 33 atm), conditions under which the assumption of perfect gas behavior and kinetic model applicability at least begins to come into question. 1B.7(b) The mean free path [1B.13] is λ = kT σ p = (1.381× 10−23 J K−1)(217 K) 0.43 × (10−9 m)2 (12.1× 103 Pa atm−1) = 5.8 × 10−7 m Solutions to problems 1B.2 The number of molecules that escape in unit time is the number per unit time that would have collided with a wall section of area A equal to the area of the small hole. This quantity is readily expressed in terms of ZW, the collision flux (collisions per unit time with a unit area), given in eqn 19A.6. That is, dN dt = −Z W A = − Ap (2π mkT )1/2 where p is the (constant) vapour pressure of the solid. The change in the number of molecules inside the cell in an interval t∆ is therefore WN Z A t∆ = − ∆ , and so the mass loss is ∆w = m∆N = − Ap m 2π kT 1/2 ∆t = − Ap M 2π RT 1/2 ∆t Therefore, the vapour pressure of the substance in the cell is 1/ 2 2 RTw p A t M π−∆ = × ∆ For the vapour pressure of germanium p = 43 × 10−9 kg π (0.50 × 10−3 m)(7200 s) × 2π (8.3145 J K−1 mol−1)(1273 K) 72.64 × 10−3 kg mol−1 1/2 = 7.3× 10−3 Pa = 7.3 mPa 1B.4 We proceed as in Justification 1B.2 except that, instead of taking a product of three one- dimensional distributions in order to get the three-dimensional distribution, we make a product of two one-dimensional distributions. f (v x ,v y )dv x dv y = f (v x 2 ) f (v y 2 )dv x dv y = m 2π kT e− mv 2 /2kT dv x dv y where v2 = v x 2 + v y 2 . The probability f(v)dv that the molecules have a two-dimensional speed, v, in the range v to v + dv is the sum of the probabilities that it is in any of the area elements dvxdvy in the circular shell of radius v. The sum of the area elements is the area of the circular shell of radius v and thickness dv which is π(ν+dν)2 – πν2 = 2πνdν . Therefore, f (v) = m kT ve− mv 2 /2kT = M RT ve− Mv 2 /2 RT M R = m k The mean speed is determined as v mean = vf (v)dv 0 ∞ ∫ = m kT v2e− mv 2 /2kT dv 0 ∞ ∫ Using integral G.3 from the Resource Section yields v mean = m kT × π 1/2 4 × 2kT m 3/2 = π kT 2m 1/2 = π RT 2M 1/2 1B.6 The distribution [1B.4] is 10 f (v) = 4π 3/2 M 2π RT v2e− Mv 2 /2 RT . The proportion of molecules with speeds less than vrms is P = f (v)dv 0 vrms ∫ = 4π 3/2 M 2π RT v2e− Mv 2 /2 RT dv 0 vrms ∫ Defining a ≡ R / 2RT , P = 4π 3/2 a π v2e− av 2 dv 0 vrms ∫ = −4π 3/2 a π d da e− av 2 dv 0 vrms ∫ Defining 2 2 1/ 2. Then, d d andav v aχ χ−≡ = P = −4π 3/2 a π d da 1 a1/2 e− χ 2 dχ 0 vrmsa 1/2 ∫{ } = −4π 3/2 a π − 1 2 3/2 1 a() e− χ2 dχ0vrmsa 1/2 ∫ + 1/2 1 a() dda e− χ2 dχ0vrmsa 1/2 ∫ Then we use the error function [Integral G.6]: e− χ 2 dχ 0 vrmsa 1/2 ∫ = π 1/2 / 2( )erf (vrmsa1/2 ) . d da e− χ 2 dχ 0 vrmsa 1/2 ∫ = dv rms a1/2 da × (e − avrms 2 ) = 1 2 c a1/2 e− avrms 2 where we have used d dz f ( y)d y0 z ∫ = f (z) Substituting and cancelling we obtain P = erf (v rms a1/2 ) − 2v rms a1/2 / π 1/2( )e− avrms2 Now v rms = 3RT M 1/2 so v rms a1/2 = 3RT M 1/2 × M 2RT 1/2 = 3 2 1/2 and P = erf 3 2 − 1/2 6 π e−3/2 = 0.92 − 0.31 = 0.61 Therefore, (a) 1 – P = 39% have a speed greater than the root mean square speed. (b) P = 61% of the molecules have a speed less than the root mean square speed. (c) For the proportions in terms of the mean speed vmean, replace vrms by v mean = 8kT / π m( )1/2 = 8 / 3π( )1/2 vrms so vmeana1/2 = 2/π1/2 . Then P = erf (v mean a1/2 ) − 2v mean a1/2 / π 1/2( )× (e− av2mean ) = erf 2 / π 1/2( )− 4 / π( )e−4/π = 0.889 − 0.356 = 0.533 That is, 53% of the molecules have a speed less than the mean, and 47% have a speed greater than the mean. 1B.8 The average is obtained by substituting the distribution (eqn 1B.4) into eqn 1B.7: vn = vn f (v)dv 0 ∞ ∫ = 4π M 2π RT 3/2 vn+2e− Mv 2 /2 RT dv 0 ∞ ∫ For even values of n, use Integral G.8: vn = 4π M 2π RT 3/2 (n + 1)!! 2 n+4 2 2RT M n+2 2 2π RT M 1/2 = (n + 1)!! RT M n 2 where (n+1)!! = 1 × 3 × 5 ... × (n+1) Thus vn 1/n = (n + 1)!! RT M 1/2 even n 11 For odd values of n, use Integral G.7: vn = 4π M 2π RT 3/2 n + 1 2 ! 2 2RT M n+3 2 = 2 π 1/2 2RT M n/2 Thus vn 1/n = 2 π 1/2 2RT M n/2 1/n = 21/n π 1/2n 2RT M 1/2 odd n Question. Show that these expressions reduce to vmean and vrms for n = 1 and 2 respectively. 1B.10 Dry atmospheric air is 78.08% N2, 20.95% O2, 0.93% Ar, 0.04% CO2, plus traces of other gases. Nitrogen, oxygen, and carbon dioxide contribute 99.06% of the molecules in a volume with each molecule contributing an average rotational energy equal to kT. (Linear molecules can rotate in two dimensions, contributing two “quadratic terms” of rotational energy, or kT by the equipartition theorem [Topic B.3(b)]. The rotational energy density is given by ρ R = E R V = 0.9906N ε R V = 0.9906NkT V = 0.9906 p = 0.9906(1.013× 105 Pa) = 1.004 × 105 J m−3 = 0.1004 J cm−3 The total energy density is translational plus rotational (vibrational energy contributing negligibly): ρ tot = ρ T + ρ R = 0.15 J cm−3 + 0.10 J cm−3 = 0.25 J cm−3 1B.12 The fraction of molecules (call it F) between speeds a and b is given by F(a,b) = f (v)dv a b ∫ where f(v) is given by eqn 1B.4. This integral can be approximated by a sum over a discrete set of velocity values. For convenience, let the velocities vi be evenly spaced within the interval such that vi+1 = vi + ∆v: ( , ) ( )ΔiF a b f v v≈ ∑ On a spreadsheet or other mathematical software, make a column of velocity values and then a column for f(v) [1B.4] at 300 K and at 1000 K. Figure 1B.1 shows f(v) plotted against v for these two temperatures. Each curve is labeled with the numerical value of T/K, and each is shaded under the curve between the speeds of 100 and 200 m s–1. F(a,b) is simply the area under the curve between v = a and v = b. One should take some care to avoid double counting at the edges of the interval, that is, not including both endpoints of the interval with full weight. example, beginning the sum with the area under the curve at those speeds. Using a spreadsheet that evaluates f(v) at 5-m s–1 intervals, and including points at both 100 and 200 m s–1 with half weight, F(100 m s–1, 200 m s–1) ≈ 0.281 at 300 K and 0.066 at 1000 K. Figure 1B.1 12 and a = 3 × (0.148 dm 3 mol−1)2 × (48.20atm) = 3.17 dm6 −2atm mol But this problem is overdetermined. We have another piece of information T c = 8a 27Rb If we use Tc along with Vc as above, we would arrive at the same value of b along with a = 27RbT c 8 = 9RV c T c 8 = 9(0.08206 dm3 atm mol−1 K−1)(0.148 dm3 mol−1)(305.4 K) 8 = 4.17 dm6 atm mol−2 Or we could use Tc along with pc. In that case, we can solve the pair of equations for a and b by first setting the two expressions for a equal to each other: a = 27b2 p c = 27RbT c 8 Solving the resulting equation for b yields b = RT c 8p c = (0.08206 dm3 atm mol−1 K−1)(305.4 K) 8(48.20 atm) = 0.06499 dm3 mol−1 and then a = 27(0.06499 dm3 mol–1)2(48.20 atm) = 5.497 dm6 atm mol–2 These results are summarized in the following table Using a/dm6 atm mol–2 b/dm3 mol–1 Vc & pc 3.17 0.0493 Vc & Tc 4.17 0.0493 pc & Tc 5.497 0.06499 One way of selecting best values for these parameters would be to take the mean of the three determinations, namely a = 4.28 dm6 atm mol–2 and b = 0.0546 dm3 mol–1 . By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e., twice their radius); that volume times the Avogadro constant is the molar excluded volume b b = N A 4π (2r)3 3 so r = 1 2 1/3 3b 4π N A r = 1 2 1/3 3(0.0546 dm3 mol−1) 4π (6.022 × 1023 mol−1) = 1.39 × 10−9 dm = 0.139 nm 1C.7(b) The Boyle temperature, TB, is the temperature at which the virial coefficient B = 0. In order to express TB in terms of a and b, the van der Waals equation [1C.5b] must be recast into the form of the virial equation. p = RT V m − b − a V m 2 Factoring out RT V m yields p = RT V m 1 1− b /V m − a RTV m So long as b/Vm < 1, the first term inside the brackets can be expanded using (1–x)–1 = 1 + x + x2 + ... , which gives p = RT V m 1+ b − a RT × 1 V m +L We can now identify the second virial coefficient as B = b − a RT 15 At the Boyle temperature B = 0 = b − a RT B so T B = a bR = 27T c 8 (i) From Table 1C.3, a = 4.484 dm6 atm mol–2 and b = 0.0434 dm3 mol–1. Therefore, T B = (4.484 dm6 atm mol−2 ) (0.08206 L atm mol−1 K−1) × (0.0434 dm3 mol−1) = 1259 K (ii) As in Exercise 1C.6(b), b = N A 4π (2r)3 3 so r = 1 2 1/3 3b 4π N A r = 1 2 1/3 3(0.0434 dm3 mol−1) 4π(6.022 × 1023 mol−1) = 1.29 × 10−9 dm = 1.29 × 10−10 m = 0.129 nm 1C.8(b) States that have the same reduced pressure, temperature, and volume [1C.8] are said to correspond. The reduced pressure and temperature for N2 at 1.0 atm and 25°C are [Table 1C.2] p r = p p c = 1.0atm 33.54atm = 0.030 and T r = T T c = (25 + 273) K 126.3K = 2.36 The corresponding states are (i) For H2S (critical constants obtained from NIST Chemistry WebBook) T = 2.36(373.3 K) = 881 K p = 0.030(89.7 atm) = 2.67 atm (ii) For CO2 T = 2.36(304.2 K) = 718 K p = 0.030(72.9 atm) = 2.2 atm (iii) For Ar T = 2.36(150.7 K) = 356 K p = 0.030(48.0 atm) = 1.4 atm 1C.9(b) The van der Waals equation [1C.5b] is p = RT V m − b − a V m 2 which can be solved for b b = V m − RT p + a V m 2 = 4.00 × 10−4 m3 mol−1 − (8.3145 J K−1 mol−1) × (288 K) 4.0 × 106 Pa + 0.76 m 6 Pa mol−2 (4.00 × 10−4 m3 mol−1)2 = 1.3× 10−4 m3 mol−1 The compression factor is Z = pV m RT [1C.2] = (4.0 × 106 Pa) × (4.00 × 10−4 m3 mol−1) (8.3145 J K−1 mol−1) × (288 K) = 0.67 Solutions to problems 1C.2 From the definition of Z [1C.1] and the virial equation [1C.3b], Z may be expressed in virial form as Z = 1+ B 1 V m + C 2 1 V m +L 16 Since V m = RT p (by assumption of approximate perfect gas behavior), 1 V m = p RT ; hence upon substitution, and dropping terms beyond the second power of 1 V m Z = 1+ B p RT + C p RT 2 = 1+ (−21.7 × 10−3 dm3 mol−1) × 100atm (0.08206 dm3 atm mol−1 K−1) × (273K) +(1.200 × 10−3 dm6 mol−2 ) × 2 100atm (0.08206 dm3 atm mol−1 K−1) × (273K) = 1− 0.0968 + 0.0239 = 0.927 V m = (0.927) RT p = (0.927) (0.08206 dm3 atm mol−1 K−1)(273 K) 100 atm = 0.208 dm3 Question. What is the value of Z obtained from the next approximation using the value of Vm just calculated? Which value of Z is likely to be more accurate? 1C.4 Since B′(TB) = 0 at the Boyle temperature [Topic 1.3b]: ′B (TB ) = a + be −c/TB 2 = 0 Solving for TB: ( ) 1/2 1/2 2 2 B 1 1 (1131K ) 5.0 10 K ( 0 1993bar )ln ln (0 2002bar ) c T a b − − − − = = = × − − − . . 1C.6 From Table 1C.4 T c = 2 3 × 2a 3bR 1/2 , p c = 1 12 × 2aR 3b3 1/2 ( )1 223 abR / may be solved for from the expression for pc and yields 12bp c R . Thus T c = 2 3 × 12 p c b R = 8 3 × p c V c R = 8 3 × (40 atm) × (160 × 10−3 dm3 mol−1) 0.08206 dm3 atm mol−1 K−1 = 210K By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e., twice their radius); that volume times the Avogadro constant is the molar excluded volume b b = N A 4π(2r)3 3 so r = 1 2 1/3 3b 4πN A [Exercise 1C.6(b)] = 1 2 1/3 V c 4πN A r = 1 2 1/3 160 cm3 mol−1 4π(6.022 × 1023 mol−1) = 1.38 × 10−8 cm = 0.138 nm 1C.8 Substitute the van der Waals equation [1C.5b] into the definition of the compression factor [1C.2] 17 or m 2 m m 1 pV B C RT V V = + + + (a) If we assume that the series may be truncated after the B term, then a plot of pV m RT vs 1 V m will have B as its slope and 1 as its y-intercept. Transforming the data gives p/MPa Vm/(dm 3 mol–1) (1/Vm)/(mol dm –3) pVm/RT 0.4000 6.2208 0.1608 0.9976 0.5000 4.9736 0.2011 0.9970 0.6000 4.1423 0.2414 0.9964 0.8000 3.1031 0.3223 0.9952 1.000 2.4795 0.4033 0.9941 1.500 1.6483 0.6067 0.9912 2.000 1.2328 0.8112 0.9885 2.500 0.98357 1.017 0.9858 3.000 0.81746 1.223 0.9832 4.000 0.60998 1.639 0.9782 Figure 1C.1(a) The data are plotted in Figure 1C.1(a). The data fit a straight line reasonably well, and the y- intercept is very close to 1. The regression yields B = –1.324×10–2 dm3 mol–1. (b) A quadratic function fits the data somewhat better (Figure 1C.1(b)) with a slightly better correlation coefficient and a y-intercept closer to 1. This fit implies that truncation of the virial series after the term with C is more accurate than after just the B term. The regression then yields 20 Figure 1C.1(b) B = –1.503×10–2 dm3 mol–1 and C = –1.06×10–3 dm6 mol–2 1C.20 The perfect gas equation [1A.5] gives V m = RT p = (8.3145 J K−1 mol−1)(250 K) 150 × 103 Pa = 0.0139 m3 = 13.9 dm3 The van der Waals equation [1C.5b] is a cubic equation in Vm. Cubic equations can be solved analytically. However, this approach is cumbersome, so we proceed as in Example 1C.1. The van der Waals equation is rearranged to the cubic form V m 3 − b + RT p V m 2 + a p V m − ab p = 0 or x3 − b + RT p x2 + a p x − ab p = 0 with x = Vm/(dm3 mol–1) . It will be convenient to have the pressure in atm: 150 kPa × 1 atm 101.3 kPa = 1.481 atm The coefficients in the equation are b + RT p = (5.42 × 10−2 dm3 mol−1) + (0.08206 dm3 atm mol−1 K−1) × (250 K) 1.481 atm = (5.42 × 10−2 + 13.85) dm3 mol−1 = 13.91 dm3 mol−1 a p = 6.260 dm6 atm mol−2 1.481 atm = 4.23 dm6 mol−2 ab p = (6.260 dm6 atm mol−2 ) × (5.42 × 10−2 dm3 mol−1) 1.481 atm = 2.291 × 10−2 dm9 mol−3 Thus, the equation to be solved is x 3 − 13.91x2 + 4.23x − (2.291 × 10−2 ) = 0 . Calculators and computer software for the solution of polynomials are readily available. In this case we find x = 13.6 and Vm = 13.6 dm 3 mol–1 . Taking the van der Waals result to be more accurate, the error in the perfect-gas value is 13.9 − 13.6 13.6 × 100% = 2% 21 Assume all gases are perfect unless stated otherwise. Unless otherwise stated, thermochemical data are for 298.15 K. 2 The First Law 2A Internal energy Answers to discussion questions 2A.2 Work is a precisely defined mechanical concept. It is produced from the application of a force through a distance. The technical definition is based on the realization that both force and displacement are vector quantities and it is the component of the force acting in the direction of the displacement that is used in the calculation of the amount of work, that is, work is the scalar product of the two vectors. In vector notation cosw fd θ= − ⋅ = −dF , where θ is the angle between the force and the displacement. The negative sign is inserted to conform to the standard thermodynamic convention. Heat is associated with a non-adiabatic process and is defined as the difference between the adiabatic work and the non-adiabatic work associated with the same change in state of the system. This is the formal (and best) definition of heat and is based on the definition of work. A less precise definition of heat is the statement that heat is the form of energy that is transferred between bodies in thermal contact with each other by virtue of a difference in temperature. The interpretations of heat and work in terms of energy levels and populations is based upon the change in the total energy of a system that arises from a change in the molecular energy levels of a system and from a change in the populations of those levels as explained more fully in Chapter 15 of this text. The statistical thermodynamics of Chapter 15 allows us to express the change in total energy of a system in the following form: d d di i i i i i N N Nε ε ε〈 〉 = +∑ ∑ The work done by the system in a reversible, isothermal expansion can be identified with the second term on the right of this expression, since there is no change in the populations of the levels which depend only on temperature; hence, the first term on the right is zero. Because the influx of energy as heat does not change the energy levels of a system, but does result in a change in temperature, the second term on the right of the above equation is zero and the heat associated with the process (a constant volume process, with no additional work) can be identified with the first term. The change in populations is due to the change in temperature, which redistributes the molecules over the fixed energy levels. Solutions to exercises 2A.1(b) See the solution to Exercise 2A.1(a) where we introduced the following equation based on the material of Chapter 15. C V ,m = 12 (3 + vR * + 2v V * )R with a mode active if T > θ M (where M is T, R, or V). (i) O 3 : C V ,m = 12 (3+ 3+ 0)R = 3R [experimental = 3.7R] 1 1 13 3 8.314 J K mol 298.15 K= 7.436 kJ molE RT − − −= = × × (ii) C 2 H 6 : C V ,m = 12 (3+ 3+ 2 × 1)R = 4R [experimental = 6.3R] 1 1 14 4 8.314 J K mol 298.15 K= 9.915 kJ molRTE − − −= × ×= 2:1 r r c c c c 8 3 [Table 1C.4] 27 T V a T V T V nb T V Rb = , = , = , = w = − 8na 27b × (T r ) × ln V r,2 − 1 3 V r,1 − 1 3 − na 3b × 1 V r,2 − 1 V r,1 The van der Waals constants can be eliminated by defining w r = 3bw a , then w = aw r 3b and w r = − 8 9 nT r ln V r,2 − 1 / 3 V r,1 − 1 / 3 − n 1 V r,2 − 1 V r,1 Along the critical isotherm, Tr = 1, Vr,1 = 1, and Vr,2 = x. Hence w r n = − 8 9 ln 3x − 1 2 − 1 x + 1 2A.6 One obvious limitation is that the model treats only displacements along the chain, not displacements that take an end away from the chain. (See Fig. 2A.2 in the Student’s Solutions Manual) (a) The displacement is twice the persistence length, so x = 2l, n = 2, ν = n/N = 2/200 = 1/100 and F = kT 2l ln 1+ ν 1− ν = (1.381× 10−23 J K−1)(298 K) 2 × 45 × 10−9 m ln 1.01 0.99 = 9.1× 10−16 N Figure 2A.1 (b) Fig. 2A.1 displays a plot of force vs. displacement for Hooke’s law and for the one-dimensional freely jointed chain. For small displacements the plots very nearly coincide. However, for large displacements, the magnitude of the force in the one-dimensional model grows much faster. In fact, in the one-dimensional model, the magnitude of the force approaches infinity for a finite displacement, -5 -4 -3 -2 -1 0 1 2 3 4 5 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 fo rc e displacement Hooke 1-D model 2:4 namely a displacement the size of the chain itself (|ν| = 1). (For Hooke’s law, the force approaches infinity only for infinitely large displacements.) (c) Work is dw = − F dx = kT 2l ln 1+ ν 1− ν dx = kNT 2 ln 1+ ν 1− ν dν This integrates to w = kNT 2 ln 1+ ν 1− ν dν 0 νf ∫ = kNT 2 [ln(1+ ν ) − ln(1− ν )]dν 0 νf ∫ = kNT 2 [(1+ ν ) ln(1+ ν ) − ν + (1− ν ) ln(1− ν ) + ν] 0 νf = kNT 2 [(1+ ν f ) ln(1+ ν f ) + (1− ν f ) ln(1− ν f )] (d) The expression for work is well behaved for displacements less than the length of the chain; however, for νf = ±1, we must be a bit more careful, for the expression above is indeterminate at these points. In particular, for expansion to the full length of the chain w = lim ν→1 kNT 2 [(1+ ν ) ln(1+ ν ) + (1− ν ) ln(1− ν )] = kNT 2 (1+ 1) ln(1+ 1) + lim ν→1 (1− ν ) ln(1− ν ) = kNT 2 2 ln 2 + lim ν→1 ln(1− ν ) (1− ν )−1 where we have written the indeterminate term in the form of a ratio in order to apply l’Hospital’s rule. Focusing on the problematic limit and taking the required derivatives of numerator and denominator yields: lim ν→1 ln(1− ν ) (1− ν )−1 = lim ν→1 −(1− ν )−1 (1− ν )−2 = lim ν→1 [−(1− ν )] = 0 Therefore; w = kNT 2 (2 ln 2) = kNT ln 2 2B Enthalpy Answers to discussion questions 2B.2 See figure 2B.3 of the text. There are two related reasons that can be given as to why Cp is greater than CV. For ideal gases Cp − CV = nR. For other gases that can be considered roughly ideal the difference is still approximately nR. Upon examination of figure 2B.3, we see that the slope of the curve of enthalpy against temperature is in most cases greater that the slope of the curve of energy against temperature; hence Cp is in most cases greater than CV. Solutions to exercises 2B.1(b) q p = nC p,m∆T [2B.7] C p,m = q p n∆T = 178J 1.9 mol × 1.78 K = 53J K−1 mol−1 C V ,m = Cp,m − R = (53− 8.3) J K −1 mol−1 = 45J K−1 mol−1 2:5 2B.2(b) (i) At constant pressure, q = ∆H. ( ) 100 273K 1 p 25 273K 373K 2 1 298K 2 2 3 d [20 17 (0 4001) K]d J K 1 20 17 (0 4001) J K 2 K 1 (20 17) (373 298) (0 4001) (373 298 ) J 11 6 10 J 2 q C T T T T T H + − + − = = . + . / = . + . × = . × − + . × − = . × = ∆ ∫ ∫ ( ) ( ) ( )1 11.00 mol 8.3145 J K mol 75 K 623 Jw p V nR T − −= − ∆ = − ∆ = − × × = − ( )11.6 0.623 kJ 11.0 kJU q w∆ = + = − = (ii) The energy and enthalpy of a perfect gas depend on temperature alone. Thus, 11.6kJH∆ = and 11.0 kJU∆ = , as above. At constant volume, w = 0 and ∆U = q , so 11.0 kJq = + . 2B.3(b) ,m[2B.2, 2B.7]p p pH q C T nC T∆ = = ∆ = ∆ ∆H = q p = (2.0 mol) × (37.11J K−1 mol−1) × (277 − 250) K = 2.0 × 103 J mol−1 ∆H = ∆U + ∆( pV ) = ∆U + nR∆T so ∆U = ∆H − nR∆T ∆U = 2.0 × 103 J mol−1 − (2.0 mol) × (8.3145J K−1 mol−1) × (277 − 250) K = 1.6 × 103 J mol−1 Solutions to problems 2B.2 In order to explore which of the two proposed equations best fit the data we have used PSI- PLOT®. The parameters obtained with the fitting process to eqn. 2B.8 along with their standard deviations are given in the following table. parameters a b/10-3 K-1 c/105 K2 values 28.796 27.89 -1.490 std dev of parameter 0.820 0.91 0.6480 The correlation coefficient is 0.99947. The parameters and their standard deviations obtained with the fitting process to the suggested alternate equation are as follows: parameters α β/10-3 K-1 γ/10-6 K-2 values 24.636 38.18 -6.495 std dev of parameter 0.437 1.45 1.106 2:6 1 4.375 kJ 0.663 K 6.60 kJ K q T C − ∆ = = = + 2C.6(b) (a) reaction(3) = (–2) × reaction(1) + reaction(2) and ∆n g = −1 The enthalpies of reactions are combined in the same manner as the equations (Hess’s law). ∆ r H O (3) = (−2) × ∆ r H O (1) + ∆ r H O (2) = [(−2) × (52.96) + (−483.64)]kJ mol−1 = −589.56 kJ mol−1 ∆ r U O = ∆ r H O − ∆n g RT = −589.56 kJ mol−1 − (−3) × (8.314 J K−1 mol−1) × (298 K) = −589.56 kJ mol−1 + 7.43 kJ mol−1 = −582.13 kJ mol−1 (b) ∆ f H O refers to the formation of one mole of the compound, so ( )O 1 1f 1(HI) 52.96 kJ mol 26.48 kJ mol2H − −∆ = = ( )O 1 1f 2 1(H O) 483.64 kJ mol 241.82 kJ mol2H − −∆ = − = − 2C.7(b) O Or r [2B.4]gH U RT n∆ = ∆ + ∆ = −772.7 kJ mol−1 + (5) × (8.3145 × 10−3 kJ K−1 mol−1) × (298 K) = −760.3 kJ mol−1 2C.8(b)The hydrogenation reaction is (1)C 2 H 2 (g) + H 2 (g) → C 2 H 4 (g) ∆ r H O (T ) = ? The reactions and accompanying data which are to be combined in order to yield reaction (1) and ∆ r H O (T ) are (2) H 2 (g) + 1 2 O 2 (g) → H 2 O(l) ∆ c H O (2) = −285.83kJ mol−1 (3) C 2 H 4 (g) + 3O 2 (g) → 2H 2 O(l) + 2CO 2 (g) ∆ c H O (3) = −1411kJ mol−1 (4) C 2 H 2 (g) + 5 2 O 2 (g) → H 2 O(l) + 2CO 2 (g) ∆ c H O (4) = −1300 kJ mol−1 reaction (1) = reaction (2) − reaction (3) + reaction (4) Hence, at 298 K: (i) ∆ r H O = ∆ c H O (2) − ∆ c H O (3) + ∆ c H O (4) = [(−285.83) − (−1411) + (−1300)]kJ mol−1 = −175kJ mol−1 O O r r g g 1 1 1 [2B.4]; 1 175kJ mol ( 1) (2 48kJ mol ) 173kJ mol U H n RT n − − − ∆ = ∆ − ∆ ∆ = − = − − − × . = − (ii) At 427 K: O O Or r r(427 K) (298K) (427 K 298K)pH H C∆ = ∆ + ∆ − [Example 2C.2] 2:9 O O O O r J ,m ,m 2 4 ,m 2 2 ,m 2 J 3 1 1 3 1 1 (J)[2C.7c] (C H g) (C H g) (H g) (43 56 43 93 28 82) 10 kJ K mol 29 19 10 kJ K mol p p p p pC C C C Cν − − − − − − ∆ = = , − , − , = . − . − . × = − . × ∑ O 1 3 1 1 r 1 (427 K) ( 175kJ mol ) (29 19 10 kJ K mol ) (129K) 171 kJ mol H − − − − − ∆ = − − . × × = − 2C.9(b) For the reaction 8 2 2 210C H (l) + 12O (g) 10CO (g) + 4H O(g)→ O O O Or f 2 f 2 f 81010 (CO , g) + 4 (H O,g) (C H , l)H H H H∆ = ∆ × ∆ − ∆× In order to calculate the enthalpy of reaction at 478 K we first calculate its value at 298 K using data in Tables 2C.1 and 2C.2. Note at 298 K naphthalene is a solid. It melts at 80.2 °C = 353.4 K. O 1 1 1 r 1 ) 4(298 K) 10 393.51 kJ mol ( 241.82 kJ mol ) (78.53 kJ mol ) 4980.91 kJ mol(H − − −− +∆ = − × − − = −× Then, using data on the heat capacities and transition enthalpies of all the reacting substances, we can calculate the change in enthalpy, ΔH, of each substance as the temperature increases from 298 K to 478 K. The enthalpy of reaction at 478 K can be obtained by adding all these enthalpy changes to the enthalpy of reaction at 298 K. This process is shown below: O O r r 2 2 8 210(478 K) (298 K) 10 (CO , g) 4 (H O, g) (C H ) 12 (O , g)H H H H H H∆ = ∆ + ∆ + × ∆ − ∆ − × ∆× For H2O(g), CO2(g), and O2(g) we have O478KO O f f ,m 298K (478 K) (298 K) dpH H C T∆ = ∆ + ∫ For naphthalene we have to take into account the change in state from solid to liquid at 80.2 °C = 353.4 K. Then O O353.4K 478KO O f f ,m trs ,m 298K 353.4K (478 K) (298 K) d + dp pH H C T H C T∆ = ∆ + ∆ +∫ ∫ We will express the temperature dependence of the heat capacities in the form of the equation given in Problem 2C.7 because data for the heat capacities of the substances involved in this reaction are only available in that form. They are not available for all the substances in the form of the equation of Table 2B.1. We use O ,m 2 pC T Tα β γ= + + For H2O(g), CO2(g), and O2(g), α, β, and γ values are given in Problem 2C.7. For naphthalene, solid and liquid, γ is zero and the two forms of the heat capacity equation are then identical and we take α = a and β = b from Table 2B.1. O 1fus 10 8(C H ) 19.01 kJ molH −∆ = Using the data given in Problem 2C.7 we calculate 1 1 1 (CO ,g) 5.299 kJ mol , (H O, g) 6.168 kJ mol , and (O ,g) 5.430 2 2 2 kJ molH H H− − −∆ = ∆ = ∆ = Using the data from Table 2C.1 we calculate for naphthalene 110 8(C H ) 55.36 kJ molH −∆ = Collecting all these enthalpy changes we have O O 1 1(478 K) (298 K) (10 5.299 4 6.168 55.36 12 5.430)kJ mol 5023.77 kJ mol r r H H − −∆ = ∆ + × + × − − × = − 2C.10(b) The cycle is shown in Fig. 2C.1. Figure 2C.1 2:10 O 2 O O O hyd soln 2 f 2 sub O O O vap 2 diss 2 ion O O O ion eg hyd (Ca ) (CaBr ) (CaBr s) (Ca) (Br ) (Br ) (Ca) (Ca ) 2 (Br) 2 (Br ) [ ( 103 1) ( 682 8) 178 2 30 91 192 9 589 7 1145 2( 331 0) 2( 289)]kJ mol H H H H H H H H H H + + − − −∆ = −∆ − ∆ , + ∆ + ∆ + ∆ + ∆ + ∆ + ∆ + ∆ = − − . − − . + . + . + . + . + + − . + − 1 11684kJ mol−= so O 2 1hyd (Ca ) 1684kJ molH + −∆ = − Solutions to problems 2C.2 Cr(C 6 H 6 ) 2 (s) → Cr(s) + 2C 6 H 6 (g) ∆n g = +2 mol O O r r 1 1 1 1 2 from[2B.4] (8 0kJ mol ) (2) (8 314 J K mol ) (583K) 17 7 kJ mol H U RT − − − − ∆ = ∆ + , = . + × . × = + . In terms of enthalpies of formation ∆ r H O = (2) × ∆ f H O (benzene,583K) − ∆ f H O (metallocene,583K) or ∆ r H O (metallocene,583K) = 2∆ f H O (benzene,583K) − 17.7 kJ mol−1 The enthalpy of formation of benzene gas at 583 K is related to its value at 298 K by ∆ f H O (benzene,583K) = ∆ f H O (benzene,298 K) +(T b − 298 K)C p,m (l) + ∆ vap H O + (583K − T b )C p,m (g) −6 × (583K − 298 K)C p,m (gr) − 3× (583K − 298 K)C p,m (H 2 ,g) where Tb is the boiling temperature of benzene (353 K). We shall assume that the heat capacities of graphite and hydrogen are approximately constant in the range of interest and use their values from Tables 2B.1 and 2B.2. 2:11 2D State functions and exact differentials Answers to discussion questions 2D.2 An inversion temperature is the temperature at which the Joule-Thomson coefficient, µ, changes sign from negative to positive or vice-versa. For a perfect gas µ is always zero, thus it cannot have an inversion temperature. As explained in detail in Section 2D.3, the existence of the Joule-Thomson effect depends upon intermolecular attractions and repulsions. A perfect gas has by definition no intermolecular attractions and repulsions, so it cannot exhibit the Joule-Thomson effect. Solutions to exercises 2D.1(b) Also see exercises E2D.1(a) and E2D.2(a) and their solutions. The internal pressure of a van der Waals gas is 2m/ .T a Vπ = The molar volume can be estimated from the perfect gas equation: 3 1 1 3 1 m 0.08206 dm atm K mol 298 K 24.76 dm mol 1.000 atm 1.00 bar 1.013 bar RT V p − − −×= = = × 6 2 2 2 3 1 2 m 6.775 atm dm mol 1.11 10 atm 11.2 mbar (24.76 dm mol )T a V π − − −= = = × = 2D.2(b) The internal energy is a function of temperature and volume, Um = Um(T,Vm), so dU m = ∂U m ∂T Vm dT + T ∂U m ∂V m dV m [π T = (∂U m / ∂V ) T ] For an isothermal expansion dT = 0; hence dU m = T ∂U m ∂V m dV m = π T dV m = a V m 2 dV m 3 1 3 1 m,2 m,2 3 1 3 1m,1 m,2 30.00dm mol m m m m2 2 1 00dm mol m m 3 1 1 1 30.00dm mol 1.00 dm molm 3 3 3 d d d 29.00 0 9667 dm mol 30.00dm mol 1 00dm mol 30.00dm mol V V V V Va a U U V VV V a a a a a − − − −. − − − − ∆ = − + = . . = = = = − = ∫ ∫ ∫ From Table 1C.3, a = 1.337 dm6 atm mol–1 3 6 2 m 3 1 5 1 3 3 3 1 1 3 (0 9667 moldm ) (1 337atm dm mol ) 1m (1 2924atm mol ) (1 01325 10 Pa atm ) 10 dm 131.0Pa m mol 131.0 J mol dm U − − − − − ∆ = . . = . × . × × = = × w = − p dV m∫ where p = RT V m − b − a V m 2 for a van der Waals gas. Hence, w = − RT V m − b dV m + a V m 2 dV m∫∫ = −q + ∆Um Thus 2:14 3 13 1 3 13 1 30.00dm mol30.00dm mol m m 1.00dm mol1.00dm mol m 2 1 1 1 2 d ln( ) 30.00 3 20 10 (8 314J K mol ) (298K) ln 8.505kJ mol 1 00 3 20 10 RT q V RT V b V b −− −− − − − − − = = − | − − . × = . × × = + . − . × ∫ and 1 1 1 1m (8505J mol ) (131J mol ) 8374 J mol 8.37 kJ molw q U − − − −= − + ∆ = − + = − = − 2D.3(b) The expansion coefficient is α = 1 V ∂V ∂T p = ′V (3.7 × 10−4 K−1 + 2 × 1.52 × 10−6 T K−2 ) V = ′V [3.7 × 10−4 + 2 × 1.52 × 10−6 (T / K)]K−1 ′V [0.77 + 3.7 × 10−4 (T / K) + 1.52 × 10−6 (T / K)2 ] = [3.7 × 10−4 + 2 × 1.52 × 10−6 (310)]K−1 0.77 + 3.7 × 10−4 (310) + 1.52 × 10−6 (310)2 = 1.27 × 10−3 K−1 2D.4(b) Isothermal compressibility is κ T = − 1 V ∂V ∂p T ≈ − ∆V V ∆p so ∆p = − ∆V Vκ T A density increase of 0.10 per cent means 0 0010V V∆ / = − . . So the additional pressure that must be applied is 2 6 1 0 0010 4.5 10 atm 2 21 10 atm p − − . ∆ = = × . × 2D.5(b) The isothermal Joule-Thomson coefficient is 1 1 1 1 1m ,m (1 11K atm ) (37 11J K mol ) 41.2 J atm molp T H C p µ − − − − − ∂ = − = − . × . = − ∂ If this coefficient is constant in an isothermal Joule-Thomson experiment, then the heat which must be supplied to maintain constant temperature is ∆H in the following relationship 1 1 1 1 1 1 3 41 2J atm mol so (41 2J atm mol ) (41 2J atm mol ) (10 0mol) ( 75atm) 30.9 10 J H n H n p p H − − − − − − ∆ / = − . ∆ = − . ∆ ∆ ∆ = − . × . × − = × Solutions to problems 2D.2 c s = γ RT M 1 2 , γ = Cp,m CV ,m , C p,m = CV ,m + R (a) C V ,m = 1 2 R(3 + νR ∗ + 2ν V ∗ ) = 12 R(3+ 2) = 5 2 R C p,m = 5 2 R + R = 7 2 R γ = 7 5 = 1.40; hence c s = 1.40RT M 1 2 (b) C V ,m = 1 2 R(3+ 2) = 5 2 R, γ = 1.40, c s = 1.40RT M 1 2 2:15 (c) C V ,m = 1 2 R(3+ 3) = 3R C p,m = 3R + R = 4R, γ = 4 3 , c s = 4RT 3M 1 2 For air, M ≈ 29 g mol −1, T ≈ 298 K, γ = 1.40 1 21 1 s 3 1 (1 40) (2 48kJ mol ) 350 ms 29 10 kg mol c − − − − . × . = = × 2D.4 (a) V = V(p,T); hence, dV = T ∂V ∂p dp + p ∂V ∂T dT Likewise p = p(V,T), so dp = ∂p ∂V T dV + ∂p ∂T V dT (b) We use α = 1 V p ∂V ∂T [2D.6] and κ T = − 1 V T ∂V ∂p [2D.7] and obtain d lnV = 1 V dV = 1 V T ∂V ∂p dp + 1 V p ∂V ∂T dT = −κ T dp + α dT . Likewise d ln p = dp p = 1 p ∂p ∂V T dV + 1 p ∂p ∂T V dT We express ∂p ∂V T in terms of κT: κ T = − 1 V ∂V ∂p T = − V ∂p ∂V T −1 so ∂p ∂V T = − 1 κ T V We express ∂p ∂T V in terms of κT and α ( ) 1 so ( ) T p V p V TT V Tp T V p T V p T V p α κ ∂ / ∂ ∂ ∂ ∂ ∂ = − = − = ∂ ∂ ∂ ∂ ∂ / ∂ so d d 1 d d ln d T T T V T V p T p V p p V α α κ κ κ = − + = − 2D.6 ( ) ( ) 1 1 p p V V T TV V α ∂= = ∂ ∂ ∂ [reciprocal identity, Mathematical Background 2] ( ) 3 2 3 2 11 [Problem 2D.5] 2 ( ) ( ) ( ) ( ) (2 ) ( ) V naT V nb V nb RV RV V nb RTV na V nb α = × − × − − × − = − × − 2:16 With vibrations ( ) ( )1 1,m 2 2 7.5 / 3 2 (3 4 5) 6.5 and 1.15 6.5V C R γ= + + × − = = = Without vibrations ( ) ( )1 1,m 2 2 3.5 / 3 2 2.5 and 1.40 2.5V C R γ= + = = = Experimental 1 1 1 1 37.11 J mol K 1.29 (37.11 8.3145) J mol K γ − − − −= =− The experimental result is closer to that obtained by neglecting vibrations, but not so close that vibrations can be neglected entirely. 34 10 1 1 R 23 1 (6 626 10 J s) (2 998 10 cms ) (0.39cm ) 0.56K 298 K 1 381 10 J K hcB k θ − − − − − . × × . × × = = = << . × and therefore rotational contributions cannot be neglected. 2E.2(b) For reversible adiabatic expansion ( )i f 1 f i [2E.2a] cV VT T / = where ( ) 1 1,mm 1 1 37.11 8.3145 J K mol 3.463 8.3145 J K mol pV C RCc R R − − , − − − − = = = = ; therefore, the final temperature is T f = (298.15K) × 1/3.463 500 × 10−3 dm3 2.00dm3 = 200 K 2E.3(b) In an adiabatic process, the initial and final pressures are related by (eqn. 2E.3) pfVf γ = piVi γ where γ = C p,m C V ,m = C p,m C p,m − R = 20.8 J K−1 mol−1 (20.8 − 8.31) J K−1 mol−1 = 1.67 Find Vi from the perfect gas law: -1 -1 3i i 3 i (2.5 mol) (8.31 J K mol ) (325 K) 0.0281 m 240 10 Pa nRT V p × × = = = × so 1/ 1/1.67 3 3i f i f 240 kPa (0.0281 m ) 0.0372 m 150 kPa p V V p γ = = × = Find the final temperature from the perfect gas law: 3 3 f f f -1 -1 (150 10 Pa) (0.0372 m ) 269 K (2.5 mol) (8.31 J K mol ) p V T nR × × = = = × Adiabatic work is (eqn. 2E.1) 1 1 3(20.8 8.31) J K mol 2.5 mol (269 325) K 1.7 10 JVw C T − −= ∆ = − × × − = − × 2E.4(b) Reversible adiabatic work is m f i [2E.1] ( ) ( )V pw C T n C R T T,= ∆ = − × − where the temperatures are related by T f = T i 1/c V i V f [2E.2a] where c = C V ,m R = C p,m − R R = 2.503 2:19 So T f = 23.0 + 273.15( )K × 400 × 10−3dm3 2.00 dm3 1 2.503 = 156 K and w = 3.12 g 28.0 g mol−1 × 29.125 − 8.3145( )J K−1 mol−1 × 156 − 296( )K= −325 J 2E.5(b) For reversible adiabatic expansion p f V f γ = p i V i γ [2E.3] so ( ) 1.33 3 i f i 3 f 400 10 dm 97.3 Torr 3.6 Torr 5.0 dm V p p V γ − × = = × = Integrated activities 2.2 (a) and (b). The table below displays computed enthalpies of formation (semi-empirical, PM3 level, PC Spartan ProTM), enthalpies of combustion based on them (and on experimental enthalpies of formation of H2O(l) and CO2(g), –285.83 and –393.51 kJ mol –1 respectively), experimental enthalpies of combustion (Table 2.6), and the relative error in enthalpy of combustion. Compound ∆ f H O / kJ mol−1 ∆ c H O / kJ mol−1(calc.) ∆ c H O / kJ mol−1(expt.) % error CH4(g) –54.45 –910.72 –890 2.33 C2H6(g) –75.88 –1568.63 –1560 0.55 C3H8(g) –98.84 –2225.01 –2220 0.23 C4H10(g) –121.60 –2881.59 –2878 0.12 C5H12(g) –142.11 –3540.42 –3537 0.10 The combustion reactions can be expressed as: C n H 2n+2 (g) + 3n + 1 2 O 2 (g) → nCO 2 (g) + (n + 1) H 2 O(1). The enthalpy of combustion, in terms of enthalpies of reaction, is ∆ c H O = n∆ f H O (CO 2 ) + (n+1)∆ f H O (H 2 O) − ∆ f H O (C n H 2n+2 ), Where we have left out ∆ f H O (O 2 ) = 0. The % error is defined as: % error = ∆ c H O (calc) − ∆ c H O (expt.) ∆ c H O (expt.) × 100% The agreement is quite good. (c) If the enthalpy of combustion is related to the molar mass by ∆ c H O = k[M / (g mol−1)]n then one can take the natural log of both sides to obtain: ln ∆ c H O = ln k + n ln M / (g mol−1). Thus, if one plots ln ∆ c H O vs. ln [M / (g mol–1)], one ought to obtain a straight line with slope n and y-intercept ln |k|. Draw up the following table: 2:20 Compound M/(g mol–1) ∆ c H / kJ mol−1 ln M/(g mol–1) ln ∆ c H O / kJ mol−1 CH4(g) 16.04 –910.72 2.775 6.814 C2H6(g) 30.07 –1568.63 3.404 7.358 C3H8(g) 44.10 –2225.01 3.786 7.708 C4H10(g) 58.12 –2881.59 4.063 7.966 C5H12(g) 72.15 –3540.42 4.279 8.172 The plot is shown below in Fig I2.1. Figure I2.1 The linear least-squares fit equation is: ln | ∆ c H O / kJ mol−1 |= 4.30 + 0.903ln M / (g mol−1) R2 = 1.00 These compounds support the proposed relationships, with n = 0.903 and k = –e4.30 kJ mol–1 = –73.7 kJ mol–1. The agreement of these theoretical values of k and n with the experimental values obtained in Problem 2C.3 is rather good. 2:21 The heat lost in step 2 was more than the heat gained in step 1, resulting in a net loss of entropy. Or the ordering represented by confining the sample to a smaller volume in step 2 overcame the disordering represented by the temperature rise in step 1. A negative entropy change is allowed for a system as long as an increase in entropy elsewhere results in ∆S total > 0 . 3A.9(b) Since the masses are equal and the heat capacity is assumed constant, the final temperature will be the average of the two initial temperatures, f 1 (100 C 25 C) 62 5 C 2 T = + = . The heat capacity of each block is C = mCs where Cs is the specific heat capacity. So, 3 1 1s(individual) 10 0 10 g 0 449 J K g ( 37.5K) 168kJH mC T − −∆ = ∆ = . × × . × ± = ± These two enthalpy changes add up to zero: ∆H tot = 0 fs i ln ; 100 C 373 2 K 25 C 298 2 K 62 5 C 335 7 K T S mC T ∆ = = . ; = . ; . = . 3 1 1 11 335 7 (10 0 10 g) (0 449 J K g ) ln 532 J K 298 2 S − − − . ∆ = . × × . × = . 3 1 1 12 335 7 (10 0 10 g) (0 449 J K g ) ln 475J K 373 2 S − − − . ∆ = . × × . × = − . 1total 1 2 57 J KS S S −∆ = ∆ + ∆ = 3A.10(b) (i) 3 1 1f 1 3 i 1 1 21g 4.60 dm (gas) ln [3A.14] (8 314J K mol ) ln 39 95g mol 1.20 dm 5.873 J K 5.9 J K V S nR V − − − − − ∆ = = × . . = = 1(surroundings) (gas) 5.9 J K [reversible] S S −∆ = −∆ = − ∆S(total) = 0 (ii) 1(gas) 5.9 J K [ is a state function] S S−∆ = + ∆S(surroundings) = 0 [no change in surroundings] 1(total)= 5.9 J KS −∆ + (iii) q rev = 0 so ∆S(gas) = 0 ∆S(surroundings) = 0 [No heat is transfered to the surroundings] ∆S(total) = 0 3A.11(b) (i) ∆ vap S = ∆ vap H T b = 35.27 × 10 3 J mol−1 (64.1+ 273.15) K = +104.58J K−1 = 104.6 J K−1 (ii) If vaporization occurs reversibly, as is generally assumed ∆S sys + ∆S sur = 0 so ∆S sur = −104.6 J K−1 Comment. This calculation has been based on the assumption that the heat capacities remain constant over the range of temperatures involved and that the enthalpy of vaporization at 298.15 K given in Table 3A.2 can be applied to the vaporization at 373.15 K. Neither one of these assumptions are strictly valid. Therefore, the calculated value is only approximate. F12:3 3A.12(b) vapfusf f fp 2 p 2 p 2 i fus i vap i (H O,s)ln (H O,l)ln (H O,g)ln HHT T T S nC n nC n nC T T T T T ∆∆ ∆ = + + + + 1 15.0 g 0.832 mol 18.02 g mol n −= = 1 1 6.008 kJ/mol1 10.832 mol 38.02 J K mol ln 0.832 mol 273.15 K 1 1 0.832 mol 75.291 J K mol ln 40.657 kJ/mol 1 1 0.832 mol 0.832 mol 33.58 J K mol l 373.15 K 273.15 261.15 373.15 273.15 S − − − −∆ = × × + × − −+ × × − −+ × + × × n 378.15 373.15 1130.3 J KS −∆ = Comment. This calculation was based on the assumption that heat capacities were constant over the range of temperatures involved. This assumption is not strictly valid. Therefore the calculated value is only approximate. Problems 3A.2 The Otto cycle is represented in Fig. 3.1. Assume one mole of air. Figure 3A.1 cycle 2 w q η | | = | | [3A.8] w cycle = w 1 + w 3 = ∆U 1 + ∆U 3 [q 1 = q 3 = 0] = C V (T B − T A ) + C V (T D − T C ) q 2 = ∆U 2 = C V (T C − T B ) B A D C D A C B C B 1 T T T T T T T T T T η | − + − | − = = − | − | − We know that T A T B = 1/c V B V A and T D T C = 1/c V C V D [2E.2a] Since VB = VC and VA = VD, T A T B = T D T C , or T D = T A T C T B 3:4 Then A C A B A C B B 1 1 T T T T T T T T η − = − = − − or 1 B A 1 c V V η / = − Given that Cp,m = 7/2R, we have CV,m = 5/2R [2D.11] and c = 2 5 For ( )2 5A B 110, 1 0.47 10 V V η / = = − = ∆S 1 = ∆S 3 = ∆S sur,1 = ∆Ssur,3 = 0 [adiabatic reversible steps] ∆S 2 = C V ,m ln T C T B At constant volume T C T B = p C p B = 5.0 ( ) 1 1 12 5 (8 314J K mol ) (ln 5 0) +33J K2S − − −∆ = × . × . = ∆S sur,2 = −∆S 2 = −33J K−1 ∆S 4 = −∆S 2 T C T D = T B T A = −33J K −1 ∆S sur,4 = −∆S 4 = +33J K−1 3A.4 (a) As suggested, relate the work to the temperature-dependent coefficient of performance : dw = dq c c = C p dT T T h − T = C p T h dT T − dT Integrating yields w = C p T h dT TTi Tf ∫ + dTTi Tf ∫ = Cp Th ln T f T i − (T f − T i ) = C p T h ln T i T f − T i + T f (b) The heat capacity is Cp = (4.184 J K –1 g–1) × (250 g) = 1046 J K–1, so the work associated with cooling the water from 293 K to the freezing temperature is w cooling = 1046 J K−1 × 293 K × ln 293 K 273 K − 293 K + 273 K = 748 J The refrigerator must also remove the heat of fusion at the freezing temperature. For this isothermal process, the coefficient of performance does not change, so w freeze = q c c = ∆ fus H T c T h − T c = ∆ fus H T h − T c T c = 6.008 × 103 J mol−1 × 250 g 18.0 g mol−1 × 293 − 273 273 = 6113 J The total work is w total = w cooling + w freeze = (748 + 6113) J = 6.86 × 103 J = 6.86 kJ At the rate of 100 W = 100 J s–1, the refrigerator would freeze the water in t = 6.86 × 103 J 100 J s−1 = 68.6 s F12:5 3 3 1 1 1atm (130Torr) (30 6dm ) 760Torr 0 193mol (0 08206dm atm K mol ) (330 2 K) pV n RT − − × × . = = = . . × . This is a substantial fraction of the original amount of water and cannot be ignored. Consequently the calculation needs to be redone taking into account the fact that only a part, nl, of the vapor condenses into a liquid while the remainder (1.00 mol – nl) remains gaseous. The heat flow involving water, then, becomes q(H 2 O) = −n 1 ∆ vap H + n 1 C p,m (H2O, l)∆T (H2O) +(1.00 mol − n 1 )C p,m (H2O,g)∆T (H2O) Because nl depends on the equilibrium temperature through n 1 = 1.00 mol − pV RT , where p is the vapor pressure of water, we will have two unknowns (p and T) in the equation −q(H 2 O) = q(Cu) . There are two ways out of this dilemma: (1) p may be expressed as a function of T by use of the Clapeyron equation, or (2) by use of successive approximations. Redoing the calculation yields: θ = n l ∆ vap H + n l C p,m (H2O, l) × 100°C + (1.00 − nl )Cp,m (H2O,g) × 100°C mC s + nC p,m (H2O, l) + (1.00 − nl )Cp,m (H2O,g) With n 1 = (1.00 mol) − (0.193mol) = 0.807 mol (noting that Cp,m(H2O,g) = 33.6 J mol –1 K–1 [Table 2C.2]) θ = 47.2°C. At this temperature, the vapor pressure of water is 80.41 Torr, corresponding to n 1 = (1.00 mol) − (0.123mol) = 0.877 mol This leads to θ = 50.8°C. The successive approximations eventually converge to yield a value of θ = 49.9 C 323 1K= . for the final temperature. (At this temperature, the vapor pressure is 0.123 bar.) Using this value of the final temperature, the heat transferred and the various entropies are calculated as in part (a). q(Cu) = (2.00 × 103 g) × (0.385J K−1 g−1) × (49.9 K) = 38.4 kJ = −q(H 2 O) ∆S(H 2 O) = −n∆ vap H T b + nC p,m ln T f T i = −119.8J K−1 ∆S(Cu) = mC s ln T f T i = 129.2 J K−1 ∆S(total) = −119.8 J K−1 + 129.2 J K−1 = 9 J K−1 3A.10 ∆S depends on only the initial and final states, so we can use f m i ln [3A.20]p T S nC T, ∆ = Since q = nC p,m (Tf − Ti ), Tf = Ti + q nC p,m = T i + I 2 Rt nC p,m [q = ItV = I 2 Rt] That is, ∆S = nC p,m ln 1+ I 2 Rt nC p,mTi Since n = 500 g 63.5 g mol−1 = 7.87 mol 2 1 1 1 1 1 (1 00 A) (1000 ) (15 0 s) (7 87 mol) (24 4 J K mol ) ln 1 (7 87) (24 4 J K ) (293 K) (192 J K ) (ln1 27) 45.4 J K S − − − − − . × Ω × . ∆ = . × . × + . × . × = × . = + 3:8 [1 J = 1 AVs = 1 A 2Ω s] For the second experiment, no change in state occurs for the copper, hence, ∆S(copper) = 0 . However, for the water, considered as a large heat sink 2 2 1(1 00 A) (1000 ) (15 0 s)(water) 51.2 J K 293 K q I Rt S T T −. × Ω × .∆ = = = = + 3A.12 Let us write Newton’s law of cooling as follows: d ( ) d s T A T T t = − − Where A is a constant characteristic of the system and TS is the temperature of the surroundings. The negative sign appears because we assume T > TS. Separating variables S d d T A t T T = − − , and integrating, we obtain Sln( )T T At K− = − + , where K is a constant of integration. Let Ti be the initial temperature of the system when t = 0, then Siln( )K T T−= Introducing this expression for K gives S S S S i i ln or ( )e At T T At T T T T T T −− − − = − = + i d d d ln ( ln ) d d d S T C C T t t T t = = From the above expression for T, we obtain S Siln ln ln( )T T At T T−= − . Substituting ln t we obtain Si d ln( ) d S CA T T t −= − , where now Ti can be interpreted as any temperature T during the course of the cooling process. 3B The measurement of entropy Solutions to exercises 3B.1(b) Use Sm = R ln s, where s is the number of orientations of about equal energy that the molecule can adopt. Draw up the following table: n: 0 1 2 3 4 5 6 o m p a b c o m p s 1 6 6 6 3 6 6 2 6 6 3 6 1 Sm/R 0 1.8 1.8 1.8 1.1 1.8 1.8 0.7 1.8 1.8 1.1 1.8 0 where a is the 1,2,3 isomer, b the 1,2,4 isomer, and c the 1,3,5 isomer. F12:9 3B.2(b) (i) ∆ r S O = S m O (Zn2+ ,aq) + S m O (Cu,s) − S m O (Zn,s) − S m O (Cu2+ ,aq) = −112.1+ 33.15 − 41.63+ 99.6 J K −1 mol−1 = −21.0 J K−1 mol−1 (ii) ∆ r S O = 12S m O (CO 2 ,g) + 11S m O (H 2 O, l) − S m O (C 12 H 22 O 11 ,s) − 12S m O (O 2 ,g) = (12 × 213.74) + (11× 69.91) − 360.2 − (12 × 205.14) J K −1 mol−1 = +512.0 J K−1 mol−1 Solutions to problems 3B.2 ,mm m 0 d ( ) (0) [3A.19] T pC TS T S T = + ∫ From the data, draw up the following table T / K 10 15 20 25 30 50 Cp ,m T / (J K–2 mol–1) 0.28 0.47 0.540 0.564 0.550 0.428 T / K 70 100 150 200 250 298 Cp ,m T / (J K–2 mol–1) 0.333 0.245 0.169 0.129 0.105 0.089 Plot Cp,m / T against T (Fig. 3B.1). This has been done on two scales. The region 0 to 10 K has been constructed using Cp,m = aT 3, fitted to the point at T = 10 K, at which Cp,m = 2.8 J K –1 mol–1, so a = 2.8 × 10 −3 J K−4 mol−1 . The area can be determined (primitively) by counting squares. Area A = 38.28 J K–1 mol–1. Area B up to 0°C = 25.60 J K–1 mol–1; area B up to 25°C = 27.80 J K–1 mol–1. Hence Figure 3B.1 (a) S m (273K) = S m (0) + 63.88J K−1 mol−1 (b) S m (298 K) = S m (0) + 66.08J K−1 mol−1 3B.4 S m (T ) = S m (0) + 0 T ∫ C p,m dT T [3A.19] 3:10 Taking Cp,m constant yields S m O (200 K) = S m O (100 K) + C p,m ln (200 K / 100 K) = [29.79 + 24.44 ln(200 / 100 K)] J K−1 mol−1 = 46.60 J K−1 mol−1 The difference is slight. 3B.8 S = k ln W [also see Exercises 3B.1(a) and (b)] so S = k ln 4N = Nk ln 4 = (5 × 108 ) × (1.38×10−23J K−1) × ln 4 = 9.57 ×10−15J K−1 Question. Is this a large residual entropy? The answer depends on what comparison is made. Multiply the answer by Avogadro’s number to obtain the molar residual entropy, 5.76×109 J K–1 mol–1, surely a large number—but then DNA is a macromolecule. The residual entropy per mole of base pairs may be a more reasonable quantity to compare to molar residual entropies of small molecules. To obtain that answer, divide the molecule’s entropy by the number of base pairs before multiplying by NA. The result is 11.5 J K–1 mol–1, a quantity more in line with examples discussed in Exercises 3B.1(a) and (b). 3C Concentrating on the system Answers to discussion questions 3C.2 All of the thermodynamic properties of a system that we have encountered, U, H, S, A, and G can be used as the criteria for the spontaneity of a process under specific conditions. The criteria are derived directly from the fundamental relation of thermodynamics which is a combination of the first and second laws, namely ext non-pVd d d d 0U p V w T S− − + + ≥ The inequality sign gives the criteria for the spontaneity of a process, the equality gives the criteria for equilibrium. The specific conditions we are interested in and the criteria that follow from inserting these conditions into the fundamental relation are the following: (1) Constant U and V, no work at all ,d 0U VS ≥ (2) Constant S and V, no work at all ,d 0S VU ≤ (3) Constant S and p, no work at all ,d 0S pH ≤ (4) Constant T d dTA w≤ (5) Constant T and V, only non-pV work , non-pVd dT VA w≤ (6) Constant T and V, no work at all ,d 0T VA ≤ (7) Constant T and p, p = pext , non-pVd dT pG w≤ (8) Constant T and p, no non-pV work ,d 0T pG ≤ Exercises F12:13 3C.1(b) (i) ∆ r H O = ∆ f H O (Zn2+ ,aq) − ∆ f H O (Cu2+ ,aq) = −153.89 − 64.77 kJ mol−1 = −218.66 kJ mol−1 ∆ r G O = −218.66 kJ mol−1 − (298.15K) × (−21.0 J K−1 mol−1) = −212.40 kJ mol−1 (ii) ∆ r H O = ∆ c H O = −5645kJ mol−1 ∆ r G O = −5645kJ mol−1 − (298.15K) × (512.0 J K−1 mol−1) = −5798 kJ mol−1 3C.2(b) 3 2 3 2CO(g) CH CH OH(l) CH CH COOH(l)+ → O O O r f f Products Reactants 1 1 1 1 [2C.5] 510.7 kJ mol ( 277.69 kJ mol ) ( 110 53kJ mol ) 122.5kJ mol H H Hν ν − − − − ∆ = ∆ − ∆ = − − − − − . = − ∑ ∑ O O O r m m Products Reactants 1 1 -1 1 1 1 1 1 [3B.2] 191.0 J K mol 160.7 J K mol 197 67 J K mol 167.4 J K mol S S Sν ν − − − − − − − ∆ = − = − − . = − ∑ ∑ O O O r r r 1 1 1 1 122.5kJ mol (298K) ( 167.4 J K mol ) 72.6 kJ mol G H T S − − − − ∆ = ∆ − ∆ = − − × − = − 3C.3(b) C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(l) ∆ r G O = 3∆ f G O (CO 2 ,g) + 4∆ f G O (H 2 O, l) − ∆ f G O (C 3 H 8 ,g) − 0 = 3(−394.36 kJ mol−1) + 4(−237.13kJ mol−1) − 1(−23.49 kJ mol−1) = −2108.11kJ mol−1 The maximum non-expansion work is 2108.11kJ mol−1 since w add = ∆G 3C.4(b) (a) ∆ r G O = ∆ f G O (Zn2+ ,aq) − ∆ f G O (Cu2+ ,aq) = −147.06 − 65.49 kJ mol-1 = −212.55 kJ mol−1 (b) ∆ r G O = 12∆ f G O (CO 2 ,g) + 11∆ f G O (H 2 O, l) − ∆ f G O (C 12 H 22 O 11 ,s) − 12∆ f G O (O 2 ,g) = 12 × (−394.36) + 11× (−237.13) − (−1543) − 12 × 0 kJ mol −1 = −5798 kJ mol−1 Comment. In each case these values of ∆ r G O agree closely with the calculated values in Exercise 3C.1(b). 3C.5(b) The formation reaction of glycine is 2 2 2 2 2 1 5 2 2 2C(gr) O (g) N (g) H (g) NH CH COOH(s)+ + + → The combustion reaction is 72 2 2 2 2 22 5 1 2 2 NH CH COOH(s) O (g) 2CO (g) H O(1) N (g)+ → + + 3:14 O O Oc f 2 f 2 f 2 2 5 2 2 (CO ,g) (H O,1) (NH CH COOH(s))H H H H∆ = ∆ + ∆ − ∆ O O O f 2 2 f 2 f 2 c 2 2 1 1 1 1 5 2 5 2 (NH CH COOH(s)) 2 (CO ,g) (H O,1) (NH CH COOH(s)) 2 393.51 kJ mol ( ) ( 285.83 kJ mol ) ( 969 kJ mol ) 532.6 kJ mol H H H H − − − − ∆ = ∆ + ∆ − ∆ = − × + × − − − = − O O O O O O f m 2 2 m m 2 m 2 m 2 1 1 1 1 1 1 1 1 1 1 1 1 1 5 2 2 1 5 2 2 (NH CH COOH(s)) 2 (C,gr) (O ,g) (N ,g) (H ,g) 103.5J K mol 2 5.740 J K mol (205.138J K mol ) 191.61J K mol (130.684J K mol ) 535.63J K mol S S S S S S − − − − − − − − − − − − × × × × ∆ = − × − − − = − × − − − = − O O O f f f 1 1 1 1 532.6 kJ mol (298.15 K) ( 535.63 J K mol ) 373kJ mol G H T S − − − − ∆ = ∆ − ∆ = − − × − = − Solutions to problems 3C.2 Begin with the partition function of an oscillator [See Chapter 15]. V 1 , 1 e x q x hcv T θ β ωβ−= = = =− The molar internal energy, molar entropy, and molar Helmholtz energy are obtained from the partition function as follows: U − U (0) = − N q ∂q ∂β V = −N (1− e− x ) d dβ (1− e− x )−1 = Nωe− x 1− e− x = Nω ex − 1 S = U − U (0) T + nR ln q = Nkxe− x 1− e− x − Nk ln(1− e− x ) = Nk x ex − 1 − ln(1− e− x ) A − A(0) = G − G(0) = −nRT ln q = NkT ln(1− e− x ) The functions are plotted in Fig. 3C.1. F12:15 (b) At 1000 K, τ = 1000 298 = 3.356 , so ∆ r G O (1000 K) = {(3.356) × 2 × (−16.45) + (1− 3.356) × 2 × (−46.11)}kJ mol−1 = +107 kJ mol−1 3D.4 T ∂S ∂V = V ∂p ∂T [Table 3D.1] (a) For a van der Waals gas p = nRT V − nb − n2a V 2 = RT V m − b − a V m 2 Hence, T ∂S ∂V = V ∂p ∂T = R V m − b (b) For a Dieterici gas p = RTe− a/RTVm Vm −b T ∂S ∂V = ∂p ∂T V = R 1+ a RV m T e− a/ RVmT V m − b For an isothermal expansion, ∆S = dS Vi Vf ∫ = T ∂S ∂V dVVi Vf ∫ so we can simply compare T ∂S ∂V expressions for the three gases. For a perfect gas, p = nRT V = RT V m so T ∂S ∂V = V ∂p ∂T = R V m T ∂S ∂V is certainly greater for a van der Waals gas than for a perfect gas, for the denominator is smaller for the van der Waals gas. To compare the van der Waals gas to the Dieterici gas, we assume that both have the same parameter b. (That is reasonable, for b is an excluded volume in both equations of state.) In that case, T ,Die ∂S ∂V = R 1+ a RV m T e− a/ RVmT V m − b = T ,vdW ∂S ∂V 1+ a RV m T e− a/ RVmT Now notice that the additional factor in T ,Die ∂S ∂V has the form (1+x)e –x, where x > 0. This factor is always less than 1. Clearly (1+x)e–x < 1 for large x, for then the exponential dominates. But (1+x)e–x < 1 even for small x, as can be seen by using the power series expansion for the exponential: (1+x)(1– x+x2/2+...) = 1 – x2/2 + ... So T ,Die ∂S ∂V < T ,vdW ∂S ∂V To summarize, for isothermal expansions: ∆S vdW > ∆S Die and ∆S vdW > ∆S perfect The comparison between a perfect gas and a Dieterici gas depends on particular values of the constants a and b and on the physical conditions. 3D.6 (a) ( ) ( ) ( )1 1 p T T V V V T V p α κ ∂ ∂= × ; = − × ∂ ∂ 3:18 (1) T ∂S ∂V = V ∂p ∂T [Maxwell relation] ( ) ( ) ( ) ( ) ( ) [Euler chain relation 2] [reciprocal identity 2] 1 1 p TV p T p T T p pV Mathematical Background T VT V T Mathematical Background V p V V T V V p α κ ∂ ∂ ∂= − , ∂ ∂∂ ∂ ∂ = − , ∂ ∂ ∂ ∂ = − = + ∂ ∂ ( ) p S V T S p ∂ ∂= ∂ ∂ [Maxwell relation] ( ) [Euler chain] [reciprocal]T p TS p S pT T S S p Sp T ∂ ∂ ∂ ∂ ∂= − = − ∂ ∂ ∂∂ ∂ First treat the numerator: ( ) [Maxwell relation] p T S V V Tp α ∂ ∂= − = − ∂∂ As for the denominator, at constant p dS = p ∂S ∂T dT and dS = dq rev T = dH T = C p dT T [dq p = dH ] Therefore, ( ) p p CS T T ∂ = ∂ and ( ) Cp p TVV S α∂ = ∂ (2) V ∂p ∂S = − S ∂T ∂V [Maxwell relation] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 [Euler chain] [reciprocal] [Maxwell relation] [Euler chain relation] [recipr T S V V T pTV VV V V p V VT S VT SS VV TT S p Vp TVT S U S U U T U T V U T S V U p T ∂ ∂∂ − = = ∂∂ ∂∂ ∂∂ ∂ ∂ ∂∂ − ∂∂∂ = = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂− ∂ ∂ = ∂ ∂ ∂ ∂ ocal identity, twice] T V V T U T C S α κ ∂ = = ∂ (b) ( ) ( )J V U V T UC V T µ ∂ ∂= = ∂ ∂ ( ) ( ) ( ) ( ) J 1 [Euler chain relation] [reciprocal identity] [3D.6] V U V T T V T UC V T V U pU p T V T µ ∂ ∂ −= = ∂ ∂ ∂ ∂ ∂ ∂= − = − ∂ ∂ F12:19 V ∂p ∂T = −1 p ∂T ∂V T ∂V ∂p [Euler chain] = − p ∂V ∂T T ∂V ∂p = α κ T Therefore, µ J C V = p − αT κ T 3D.8 [3D.6]T V p T p T π ∂ = − ∂ ln ln [Chapter 15] T T A Q A kT Q p kT V V ∂ ∂ = − = − = ∂ ∂ then ! N T q NkT q Q p N q V ∂ = = ∂ Substitute this expression for p into eqn. 3D.6. We obtain after differentiating p with respect to T at constant V 2 T T NkT q q T V π ∂ ∂ = ∂ ∂ 3D.10 The Gibbs–Helmholtz equation is ∂ ∂T ∆G T = − ∆H T 2 so for a small temperature change O O r 2 2 O O O r r 2 r 1 r 2 1 and G H G G H T T T T TT T ∆ ∆ ∆ ∆ ∆ ∆ = ∆ = − ∆ so d ∆ r G T O ∫ = − ∆ r H Od T T 2∫ and ∆ r G 190 O T 190 = ∆ r G 220 O T 220 + ∆ r H O 1 T 190 − 1 T 220 ∆ r G 190 Ο = ∆ r G 220 Ο T190 T 220 + ∆ r H O 1− T 190 T 220 For the monohydrate ∆ r G 190 O = (46.2 kJ mol−1) × 190 K 220 K + (127 kJ mol−1) × 1− 190 K 220 K , ∆ r G 190 O = 57.2 kJ mol−1 For the dihydrate ∆ r G 190 Ο = (69.4 kJ mol−1) × 190 K 220 K + (188kJ mol−1) × 1− 190 K 220 K , ∆ r G 190 Ο = 85.6 kJ mol−1 For the trihydrate ∆ r G 190 O = (93.2 kJ mol−1) × 190 K 220 K + (237 kJ mol−1) × 1− 190 K 220 K , ∆ r G 190 O = 112.8kJ mol−1 3:20 F = C – P + 2 . Restricting to pure substances (C=1) and rearranging for phases gives P = 3 – F . Areas in the phase diagram have two degrees of freedom; one can vary pressure and temperature independently (within limits) and stay within the area. Thus, F = 2 and P = 1 in areas. Lines have one degree of freedom; one can vary pressure or temperature, but to stay on the line the value of the other is determined by the line. Thus, F = 1 and P = 2 on lines. Points on the phase diagram have zero degrees of freedom; one can vary neither pressure nor temperature and on a given point. Thus, F = 0 and P = 3 on points. (a) is in an area, so there is a single phase. (b) and (c) are points, so there are three phases present. (d) is on a line, so there are two phases present. 4A.2(b) For pure substances (one-component systems), the chemical potential is the molar Gibbs energy: dG = (µ2 – µ1)dn so ∆G = (µ2 – µ1)n = (–8.3 kJ mol–1)(0.15×10–3 mol) = +1.2×10–3 kJ = 1.2 J. 4A.3(b) Use the phase rule (eqn 4A.1) F = C – P + 2 to solve for the number of phases: P = C – F + 2 = 4 – F + 2 = 6 – F ≤ 6 . The maximum number of phases in equilibrium occurs when the number of degrees of freedom is at a minimum, namely zero; that number is six. 4B Phase diagrams of pure substances Answers to discussion questions 4B.2 See Topic 4B.1(b). The mathematical reason can be seen in eqn 4B.2, ∂µ ∂p T = V m Because Vm > 0 for all pure substances, the slope of the change in chemical potential with respect to change in pressure is positive: chemical potential increases with increasing pressure. See also the answer to Discussion question 4A.2, which addresses why the chemical potential changes even in incompressible substances. 4B.4 See Topic 4B.3 for classification of phase transitions. First-order phase transitions show discontinuities in the first derivative of the Gibbs energy with respect to temperature. They are recognized by finite discontinuities in plots of H, U, S, and V against temperature and by an infinite discontinuity in Cp. Second-order phase transitions show discontinuities in the second derivatives of the Gibbs energy with respect to temperature, but the first derivatives are continuous. The second-order transitions are recognized by kinks in plots of H, U, S, and V against temperature, but most easily by a finite discontinuity in a plot of Cp against temperature. A λ-transition shows characteristics of both first and second-order transitions and, hence, is difficult to classify by the Ehrenfest scheme. It resembles a first-order transition in a plot of Cp against T, but appears to be a higher-order transition with respect to other properties. At the molecular level first-order transitions are associated with discontinuous changes in the interaction energies between the atoms or molecules constituting the system and in the volume they occupy. One kind of second-order transition may involve only a continuous change in the arrangement of the atoms from one crystal structure (symmetry) to another while preserving their orderly arrangement. In one kind of λ-transition, called an order- disorder transition, randomness is introduced into the atomic arrangement. See Figures 4B.9 through 4B.12 of the text. Solutions to Exercises 4B.1(b) The difference between the definition of normal and standard transition temperatures is the pressure at which the transition takes place: normal refers to exactly 1 atm (101325 Pa), 2 while standard refers to exactly 1 bar (exactly 105 Pa). At the standard boiling temperature and pressure, the liquid and gas phases are in equilibrium, so their chemical potentials are equal: µliquid(Tstd,pstd) = µgas(Tstd,pstd) The same can be said at the normal boiling temperature and pressure: µliquid(Tnorm,pnorm) = µgas(Tnorm,pnorm) Equations 4B.1 and 4B.2 show how the chemical potential changes with temperature and pressure, so for small changes we can write dµ = ∂µ ∂T p dT + ∂µ ∂p T dp = −S m dT + V m dp Assuming that the differences between standard and normal boiling point are small enough, we can equate the differences in the chemical potentials of the two phases: ∆µgas = –Sm,gas∆T + Vm,gas∆p = –Sm,liquid∆T + Vm,liquid∆p = ∆µliquid , where ∆p is defined as pnorm–pstd. Rearrange to isolate ∆T: (Sm,liquid–Sm,gas)∆T = (Vm,liquid–Vm,gas)∆p , (–∆vapS)∆T = (Vm,liquid–Vm,gas)∆p ≈ –Vm,gas∆p Use the ideal gas law to find the molar volume of the gas. Also, we need to find ∆vapS or to use Trouton’s rule (eqn 3A.17): ∆T ≈ V m,gas ∆p ∆ vap S = RT ∆p p∆ vap S = RT b 2∆p p∆ vap H = (8.3145 J K−1 mol−1)(373 K)2 (1325 Pa) (105 Pa)(40.656 × 103 J) = 0.38 K That is, the normal boiling temperature is 0.38 K higher than the standard boiling temperature. 4B.2(b) Equation 4B.1 shows how the chemical potential changes with temperature dµ = ∂µ ∂T p dT = −S m dT so ∆µ = − S m dT∫ = −Sm∆T = −53 J K−1 mol−1 × (1000 − 100) K = 4.8 × 104 J mol−1 = 48 kJ mol−1 4B.3(b) Equation 4B.2 shows how the chemical potential changes with pressure dµ = ∂µ ∂p T dp = V m dp = M ρ dp so ∆µ = M ρ d p∫ = M ρ ∆p = 78.11 g mol−1 0.879 g cm−3 × (10 × 106 − 100 × 103) Pa × 1 m3 106 cm3 = 8.8 × 102 J mol−1 = 0.088 kJ mol−1 Note: we assumed that the sample is incompressible. 4B.4(b) The effect on vapour pressure of a change in applied external pressure on a liquid is given by eqn 4B.3: p = p *e Vm ( l)∆P/ RT . For liquid naphthalene, the molar volume is V m = M ρ = 118.16 g mol−1 0.962 g cm−3 = 122.8 cm3 mol−1 so V m (l)∆P RT = 122.8 cm3 mol−1 × (15 × 106 − 1.0 × 105 ) Pa 8.3145 J K−1 mol−1 × 368 K × 1 m3 106 cm3 = 0.598 and p = p * e Vm ( l)∆P/ RT = (2.0 kPa)e0.598 = 3.6 kPa . 4B.5(b) Use the Clapeyron equation (eqn 4B.5a) dp dT = ∆ trs S ∆ trs V 3 Assume that ∆fusS and ∆fusT are independent of temperature: ∆ fus S = ∆ fus V × dp dT ≈ ∆ fus V × ∆p ∆T ∆ fus S = (152.6cm3 mol−1 − 142.0cm3 mol−1) × (1.2 × 106 Pa − 1.01× 105 Pa) 429.26 K − 427.15K = (10.6cm3 mol−1) × 1m3 106 cm3 × (5.21× 105 Pa K−1) = 5.52 Pa m3 K−1 mol−1 = +5.5J K−1 mol−1 At the melting temperature ∆ fus H = T f ∆ fus S = (427.15K) × (5.52 J K−1 mol−1) = +2.4 kJ mol−1 4B.6(b) On the assumption that the vapour is a perfect gas and that ∆vapH is independent of temperature, we may write [4B.11] p = p∗e− χ , χ = ∆ vap H R × 1 T − 1 T ∗ , ln p∗ p = χ 1 T = 1 T ∗ + R ∆ vap H ln p∗ p = 1 293.2 K + 8.3145J K−1 mol−1 32.7 × 103 J mol−1 × ln 58.0 66.0 = 3.378 × 10−3 K−1 Hence T = 1 3.378 × 10−3 K−1 = 296 K = 23°C 4B.7(b) Integrating the Clausius-Clapeyron eqation (4B.10) yields an expression for ln p: d ln p∫ = ∆ vap H RT 2∫ dT so ln p = constant − ∆ vap H RT Therefore, ∆ vap H = 3036.8 K × R = 8.3145 J K−1 mol−1 × (3036.8 K) = +25.25kJ mol−1 4B.8(b) (i) The indefinitely integrated form of eqn 4B.10 is used as in Exercise 4B.7(b). ln p = constant − ∆ vap H RT , or log p = constant − ∆ vap H 2.303 RT Thus ∆ vap H = 1625K × R × 2.303 = 1625K × 8.3145 J K−1 mol−1 × 2.303 = 31.11kJ mol−1 (ii) The normal boiling point corresponds to p = 1.000 atm = 760 Torr, so log 760 = 8.750 − 1625 K T and T = 1625 K 8.750 − log 760 = 276.9 K 4B.9(b) ∆T ≈ ∆ fus V ∆ fus S × ∆p [4B.5a and Exercise 4B.5(a)] ≈ T f ∆ fus V ∆ fus H × ∆p = T f M ∆p ∆ fus H × ∆ 1 ρ [V m = M / ρ] Normal freezing point is Tf = (273.15 – 3.65) K = 269.50 K at a pressure of 1 atm, which is about 0.1 MPa. Thus, to the nearest MPa, ∆p = 100 MPa = 1.00×108 Pa 4 Figure 4B.1 The points are plotted in Figure 4B.1. The slope is –4569 K, so −∆ vap H R = −4569 K, or ∆ vap H = +38.0 kJ mol-1 The normal boiling point occurs at p = 1 atm = 101.3 kPa, or at ln(p/kPa) = 4.618, which from the figure corresponds to 1000 K/T = 2.80. Therefore, Tb = 357 K (84°C) The accepted value is 83°C. 4B.10 The slope of the solid–vapour coexistence curve is given by dp dT = ∆ sub H T ∆ sub V [analogous to 4B.9] so ∆ sub H = T ∆ sub V dp dT Figure 4B.2 The slope can be obtained by differentiating an equation fit to the coexistence curve (Figure 4B.2). Fit the data to an exponential function or take natural logarithms of the pressures and make a linear fit to the transformed data. The fit equation is p/Pa = 2.659×10–10 e0.1687T/K 7 so dp dT = (2.659 × 10−10 Pa) × (0.1687 K−1) × e0.1687T /K = 4.41Pa K−1 at 150 K. The change in volume is essentially the volume of the vapour V m = RT p = (8.3145 J K−1 mol−1) × (150 K) (2.659 × 10−10 Pa) × e0.1687×150 = 47.7 m3 So ∆ sub H Ο = (150 K) × (47.7 m3) × 4.41Pa K−1 = 3.16 × 104 J mol−1 = 31.6 kJ mol−1 4B.12 dH = CpdT + V dp implies d∆H = ∆CpdT + ∆V dp , where ∆ signifies a difference between phases. Along a phase boundary dp and dT are related by dp dT = ∆H T ∆V [4B.6 or 4B.9] Therefore, d∆H = ∆C p + ∆V × ∆H T ∆V dT = ∆C p + ∆H T dT and d∆H dT = ∆C p + ∆H T Then, since d dT ∆H T = 1 T d∆H dT − ∆H T 2 = 1 T d∆H dT − ∆H T substituting the first result gives d dT ∆H T = ∆C p T Therefore, ( ) dd d In p pC TH C TT T ∆∆ = = ∆ 4B.14 Equation 4B.3 gives the vapour pressure of a liquid under an additional applied pressure ∆P: p = p *eVm ( l)∆P/ RT The applied pressure is the hydrostatic pressure of the liquid overlying the depth d: ∆P = ρgd The molar volume of the liquid is Vm(l) = M / ρ Substituting into eqn. 4B.3 yields p = p*eMgd/RT For a 10-m column of water at 25°C, Mgd RT = (18.02 × 10−3 kg mol−1) × (9.81 m s−2 ) × (10 m) (8.3145 J K−1 mol−1) × (298 K) = 7.1× 10−4 so p p* = e7.1×10 −4 ≈ 1+ 7.1× 10−4 That is, the fractional increase in vapor pressure is 7.1×10–4 or 0.071 per cent. 4B.16 In each phase the slopes of curves of chemical potential plotted against temperature are p ∂µ ∂T = −S m [4.1] The curvatures of the graphs are given by p ∂2µ ∂T 2 = − p ∂S m ∂T To evaluate this derivative, consider dS at constant p: dS = dq rev T = dH T = C p dT T so p ∂2µ ∂T 2 = − p ∂S m ∂T = − C p,m T Since Cp,m is necessarily positive, the curvatures in all states of matter are necessarily negative. Cp,m is often largest for the liquid state, though not always. In any event, it is the 8 ratio Cp,m/T that determines the magnitude of the curvature, so no general answer can be given for the state with the greatest curvature. It depends upon the substance. 4B.18 S = S(T,p) dS = p ∂S ∂T dT + T ∂S ∂p dp p ∂S ∂T = C p T [Problem 4B.16] T ∂S ∂p = − p ∂V ∂T [Table 3D.1] = −αV m dq rev = T dS = C p dT − T p ∂V ∂T dp C S = S ∂q ∂T = C p − TVα S ∂p ∂T = C p − αV × ∆ trs H ∆ trs V [4B.6] Integrated activities 4.2 (a) The phase diagram is shown in Figure I4.1. Figure I4.1 (b) The standard melting point is the temperature at which solid and liquid are in equilibrium at 1 bar. That temperature can be found by solving the equation of the solid–liquid coexistence curve for the temperature: 1 = p3/bar + 1000(5.60+11.727x)x . Put the equation into standard form: 11727x2 + 5600x + (4.362×10–7 –1) = 0 The quadratic formula yields { } ( ) ( ) 2 1/21/22 11727 4 11727 5600 5600 1 15600 (5600) 4 11727 (–1) 2 11727 2 x ×− ± +− ± − × × = = × × The square root is rewritten to make it clear that the square root is of the form 1+ a{ }1 2 , with a = 1 ; thus the numerator is approximately −1+ 1+ 12 a( )= 12 a , and the whole expression reduces to x ≈ 1/5600 = 1.79×10–4 . Thus, the melting point is T = (1+x)T3 = (1.000179) × (178.15 K) = 178.18 K. (c) The standard boiling point is the temperature at which the liquid and vapour are in equilibrium at 1 bar. That temperature can be found by solving the equation of the liquid– vapour coexistence curve for the temperature. This equation is too complicated to solve 9 The total volume is also equal to V = VAnA + VBnB [5A.3] . Therefore, VA = V −VBnB nA = 1001.21 cm 3 − (−1.4 cm3) × (0.050 mol) 55.49 mol = 18.04 cm3 mol−1 Question. VA is essentially the same as the molar volume of pure water, but clearly VB is not even approximately the molar volume of pure solid MgSO4. What meaning can be ascribed to a negative partial molar volume? 5A.3(b) Use the Gibbs-Duhem equation [5A.13], replacing infinitesimal changes in chemical potential (dµJ) with small finite changes (δµJ) δµB ≈ − nA nB δµA = − 0.22nB nB × (−15 J mol−1) = +3.3 J mol−1 5A.4(b) The Gibbs energy of mixing perfect gases is ∆mixG = nRT(xA ln xA + xB ln xB) [5A.16] = pV(xA ln xA + xB ln xB) [perfect gas law] Because the compartments are of equal size, each contains half of the gas; therefore, ( )mix 3 3 3 3 6 3 1 1 1 1( ) ln ln ln 2 2 2 2 2 1m (100 10 Pa) (250cm ) ln 2 17.3 Pa m = 17.3 J 10 cm G pV pV∆ = × + = − = − × × × = − − ∆mixS = −nR(xA ln xA + xB ln xB ) [5A.17] = −∆mixG T = +17.3 J 273 K = +0.635 J K−1 5A.5(b) ∆ mix S = −nR x J J ∑ ln xJ [5A.17] We need mole fractions: xJ = nJ nJ J ∑ Since we have mass percentages, 100.0 g is a convenient sample size. The amounts of each component are nN2 = 75.52 g × 1 mol 2 ×14.007 g = 2.696 mol nO2 = 23.15 g × 1 mol 2 ×15.999 g = 0.7235 mol nAr = 1.28 g × 1 mol 39.95 g = 0.0320 mol nCO2 = 0.046 g × 1 mol (12.011+ 2 ×15.999) g = 0.00105 mol The mole fractions are xN2 = nN2 nN2 + nO2 + nAr + nCO2 = 2.696 mol (2.696 + 0.7235+ 0.0320 + 0.00105) mol = 0.7809 Similarly, xO2 = 0.2096 , xAr = 0.00928, and xCO2 = 0.00030 . Once we have mole fractions, the convenient sample size is for a total of one mole of gas: ∆mixS = −R xJ ln xJ J ∑ = −R{(0.7809ln0.7809) + (0.2096ln0.2096) +(0.00928ln0.00928) + (0.00030ln0.00030)} = 0.5665R = +4.710 J K−1 mol−1 From the data in Exercise 5A.5(a), the entropy of mixing was ∆mixS = −R xJ ln xJ J ∑ = −R{(0.781ln0.781) + (0.210ln0.210) + (0.0094ln0.0094)} = 0.565R = +4.70 J K−1 mol−1 2 So the difference is 1 1mix mixΔ (b) – Δ (a) 0.0015 0.012 J K molS S R − −= = + Comment. We can readily see that the data in this exercise (b) includes the CO2 term, which contributes –R(0.00030 ln 0.00030) = 0.0025R to the entropy of mixing—more than the total difference. The fact that the mole fractions of the other components are slightly smaller in part (b) to make room for the small amount of CO2 partly offsets the direct CO2 term itself. 5A.6(b) Let 12 refer to 1,2-dimethylbenzene and 13 to 1,3-dimethylbenzene. Because the two components are structurally similar, we assume Raoult’s Law [5A.21] applies. ptotal = p12 + p13 = x12p12* + x13p13* = (0.500)(20 + 18) kPa = 19 kPa. The mole fractions in the vapor phase are the ratios of partial to total pressure: x12,vap = p12 ptotal = xliq,12 p12 * ptotal = (0.500)(20 kPa) 19 kPa = 0.53 and xvap,13 = 0.47 5A.7(b) Total volume V = nAVA + nBVB = n(xAVA + xBVB), where n = nA + nB Total mass m = nAMA + nBMB = n{xAMA + (1–xA)MB} So n = m x A M A + (1− x A )M B = 1.000 × 10 3 g (0.3713) × (241.1g mol−1) + (1− 0.3713) × (198.2 g mol−1) = 4.670 mol and V = n(x A V A + x B V B ) = (4.670 mol) ×{(0.3713) × (188.2) + (1− 0.3713) × (176.14)} cm3 mol−1 = 843.5 cm3 5A.8(b) Let W denote water and E ethanol. The total volume of the solution is V = nWVW + nEVE We are given VE, we need to determine nW and nE in order to solve for VW, for V W = V − n E V E n W Take 100 cm3 of solution as a convenient sample. The mass of this sample is m = ρV = (0.9687 g cm–3) × (100 cm3) = 96.87 g . 80 per cent of this mass water and 20 per cent ethanol, so the moles of each component are n W = (0.80) × (96.87 g) 18.02 −1g mol = 4.3 mol and n E = (0.20) × (96.87 g) 46.07 −1 g mol = 0.42 mol−1 . V W = V − n E V E n W = 100 cm3 − (0.42 mol) × (52.2 cm3 mol−1) 4.3 mol = 18 cm3 mol−1 5A.9(b) Henry’s law is [5A.23] pB = xBKB, so check whether pB / xB is equal to a constant (KB) x 0.010 0.015 0.020 p/kPa 82.0 122.0 166.1 (p/kPa) / x 8.2×103 8.1×103 8.3×103 Hence, KB = p / x = 8.2×10 3 kPa (average value). 5A.10(b) Refer to Brief Illustration 5A.4 and use the Henry’s Law constant from Table 5A.1. Henry’s law in terms of molal concentration is pB = bBKB. So the molal solubility of methane in benzene at 25°C in equilibrium with 1.0 bar of methane is bCH4 = pCH4 KCH4 = 100 kPa 44.4 ×103 kPa kg mol−1 = 2.25 ×10−3 mol kg−1 To find the molar solubility, we assume that the density of the solution is the same as that of pure benzene, given at a nearby temperature (20°C) in Table 0.1: [CH4] = bCH4 ρbenzene = 2.25 ×10 −3 mol kg−1 × 0.879 kg dm−3 = 2.0 ×10−3 mol dm−3 5A.11(b) With concentrations expressed in molalities, Henry’s law [5A.23] becomes pB = bBKB. 3 Solving for bB, the molality, we have b B = p B K = x B p total K , where ptotal = 1 atm = 101.3 kPa For N2, K = 1.56×10 5 kPa kg mol–1 [Table 5A.1] b = 0.78 × 101.3 kPa 1.56 × 105 kPa kg mol−1 = 5.1× 10−4 mol kg−1 For O2, K = 7.92×10 4 kPa kg mol–1 [Table 5A.1] b = 0.21× 101.3 kPa 7.92 × 104 kPa kg mol−1 = 2.7 × 10−4 mol kg−1 5A.12(b) As in Exercise 5A.11(b), we have b B = p B K = 2.0 × 101.3 kPa 3.01× 103 kPa kg mol−1 = 0.067 mol kg−1 Hence, the molality of the solution is about 0.067 mol kg–1. Since molalities and molar concentrations (molarities) for dilute aqueous solutions are numerically approximately equal, the molar concentration is about 0.067 mol dm–3. Solutions to problems 5A.2 C = 1; hence, according to the phase rule (eqn 4A.1) F = C – P + 2 = 3 – P Since the tube is sealed there will always be some gaseous compound in equilibrium with the condensed phases. Thus when liquid begins to form upon melting, P = 3 (s, l, and g) and F = 0, corresponding to a definite melting temperature. At the transition to a normal liquid, P = 3 (l, l′, and g) as well, so again F = 0. 5A.4 Letting B stand for CuSO4(aq), the partial molar volume of the dissolved salt is V B = ∂V ∂n B nA [5A.1] We will determine VB by plotting V against nB while holding nA constant. We can find the volume from the density: ρ = m A + m B V so V = m A + m B ρ . The data include the composition of the solution expressed as mass percent. (That is, m(CuSO4)/g, the mass in grams of B dissolved in 100 g solution, is numerically equal to w, defined as mass of B over total solution mass expressed as a percent). For our plot, we need nB per fixed amount of A. Let us choose that fixed quantity to be mA = 1 kg exactly, so nB is numerically equal to the molal concentration. So n B = m B M B such that m B m A + m B × 100 = w . Solve for mB: m B = wm A 100 − w . Draw up the following table of values of mB, nB, and V at each data point, using mA = 1000 g. W 5 10 15 20 ρ /(g cm–3) 1.051 1.107 1.167 1.23 mB/g 52.6 111.1 176.5 250.0 nB/mol 0.330 0.696 1.106 1.566 V/cm3 1001.6 1003.7 1008.1 1016.3 VB/(cm 3 mol–1) 2.91 8.21 14.13 20.78 A plot V against nB is shown in Figure 5A.1. 4 n = 2.00 mol and xhexane = xheptane = 0.500 Therefore, ∆ mix G = (2.00 mol) × (8.3145 J K−1 mol−1) × (298 K) × 2 × (0.500 ln 0.500) = −3.43 × 103 J = −3.43 kJ and ∆ mix S = −∆ mix G T = +3.43 × 103 298 K = +11.5 J K−1 For an ideal solution, ∆mixH = 0, just as it is for a mixture of perfect gases [5A.18]. It can be demonstrated from ∆mix H = ∆mixG + T∆mixS = ∆mixG + T −∆mixG T = 0 5B.6(b) (i) Benzene and ethylbenzene form nearly ideal solutions, so. ∆mixS = –nRT(xA ln xA + xB ln xB) [5A.17] We need to differentiate eqn 5A.17 with respect to xA and look for the value of xA at which the derivative is zero. Since xB = 1 –xA, we need to differentiate ∆mixS = –nRT{xA ln xA + (1–xA)ln(1–xA)} This gives using d ln x dx = 1 x d∆ mix S dx A = −nR{ln x A + 1− ln(1− x A ) − 1} = −nR ln x A 1− x A which is zero when x A = 1 2 . Hence, the maximum entropy of mixing occurs for the preparation of a mixture that contains equal mole fractions of the two components. (ii) Because entropy of mixing is maximized when nE = nB (changing to notation specific to Benzene and Ethylbenzene) m E M E = m B M B This makes the mass ratio m B m E = M B M E = 78.11 g mol −1 106.17 g mol−1 = 0.7357 5B.7(b) The ideal solubility in terms of mole fraction is given by eqn 5B.15: ln x Pb = ∆ fus H R × 1 T f − 1 T = 5.2 × 103 J mol−1 8.3145 J K−1 mol−1 × 1 600.K − 1 553K = −0.089 Therefore, xPb = e –0.089 = 0.92 . x Pb = n Pb n Bi + n Pb implying that n Pb = n Bi x Pb 1− x Pb = m Bi M Bi × x Pb 1− x Pb Hence the amount of lead that dissolves in 1 kg of bismuth is n Pb = 1000 g 209 g mol−1 × 0.92 1− 0.92 = 52 mol or, in mass units, mPb = nPb×MPb = 52 mol × 207 g mol –1 = 1.1×104 g = 11 kg. Comment. A mixture of 11 kg of lead and 1 kg of bismuth would normally be regarded as a solution of bismuth in lead, not the other way around. It is unlikely that such a mixture could be regarded as an ideal dilute solution of lead in bismuth. Under such circumstances eqn 5B.15 ought to be considered suggestive at best, rather than quantitative. 5B.8(b) The best value of the molar mass is obtained from values of the data extrapolated to zero concentration, since it is under this condition that the van’t Hoff equation (5B.16) applies. 7 ΠV = nBRT so Π = mRT MV = cRT M where c = m/V . But the osmotic pressure is also equal to the hydrostatic pressure Π = ρgh [1A.1] so h = RT ρgM c Figure 5B.1 Hence, plot h against c and identify the slope as RT ρgM . Figure 5B.1 shows the plot of the data. The slope of the line is 1.78 cm /(g dm–3), so RT ρgM = 1.78 cm g dm−3 = 1.78 cm dm3 g−1 = 1.78 × 10−2 m4 kg−1 Therefore, M = RT (ρg) × (1.78 × 10−2 m4 kg−1) = (8.3145 J K−1 mol−1) × (293K) (1.000 × 103 −3kg m ) × (9.81 m s−2 ) × (1.78 × 10−2 m4 kg−1) = 14.0 kg mol-1 5B.9(b) In an ideal dilute solution the solvent (CCl4, A) obeys Raoult’s law [5A.21] and the solute (Br2, B) obeys Henry’s law [5A.23]; hence pA = xA p * = (0.934) × (23 kPa) = 21.5 kPa pB = xBKB = (0.066) × (73 kPa) = 4.8 kPa ptotal = (21.5 + 4.8) kPa = 26.3 kPa The composition of the vapour in equilibrium with the liquid is yA = pA ptotal = 21.5 kPa 23.3 kPa = 0.82 and yB = pB ptotal = 4.8 kPa 23.3 kPa = 0.18 5B.10(b) Let subscript 12 denote the 1,2 isomer and 13 the 1,3 isomer. Assume that the structurally similar liquids obey Raoult’s law [5A.21]. The partial pressures of the two liquids sum to 19 kPa. p13 + p12 = p = x13p13 * + x12p12 * = x13p13 * + (1–x13)p12 * Solve for x13: x13 = p − p12 * p13 * − p12 * = (19 − 20) kPa (18 − 20) kPa = 0.5 and x12 = 1 – 0.5 = 0.5 . The vapour phase mole fractions are given by eqn 1A.8: 8 y13 = p13 p = x13 p13 * p = (0.5) ×18 kPa 19 kPa = 0.47 and y12 = x12 p12 * p = (0.5) × 20.0 kPa 50.7 kPa = 0.53 . 5B.11(b) The partial vapour pressures are given by Raoult’s law [5A.21]: pA = xApA * and pB = xBpB * = (1–xB)pB * . Eqn 1A.8 relates these vapour pressures to the vapour-phase mole fractions: y A = p A p total = x A p A * x A p A * + (1− x A ) p B * Solve for xA: x A p A * + (1− x A ) p B * = x A p A * y A x A p A * − p B * − p A * y A = − pB * x A = p B * p B * + p A * y A − p A * = 82.1 kPa 82.1+ 68.8 0.621 − 68.8 kPa = 0.662 and xB = 1 – xA = 1 – 0.662 = 0.338 . The total vapour pressure is ptotal = xApA * + xBpB * = 0.662 × 68.8 kPa + 0.338 × 82.1 kPa = 73.3 kPa . 5B.12(b) (i) If the solution is ideal, then the partial vapour pressures are given by Raoult’s law [5A.21]: pA° = xApA * = 0.4217 × 110.1 kPa = 46.4 kPa and pB° = xBpB * = (1–xB)pB * = (1–0.4217) × 76.5 kPa = 44.2 kPa . (Note the use of the symbol ° to emphasize that these are idealized quantities; we do not yet know if they are the actual partial vapour pressures.) At the normal boiling temperature, the partial vapour pressures must add up to 1 atm (101.3 kPa). These ideal partial vapour pressures add up to only 90.7 kPa, so the solution is not ideal. (ii) We actually do not have enough information to compute the initial composition of the vapour above the solution. The activities and activity coefficients are defined by the actual partial vapour pressures. We know only that the actual vapour pressures must sum to 101.3 kPa. We can make a further assumption that the proportions of the vapours are the same as given by Raoult’s law. That is, we assume that y A = y A ° = p A ° p A ° + p B ° = 46.4 kPa (46.4 + 44.2) kPa = 0.512 and y B = y B ° = p B ° p A ° + p B ° = 44.2 kPa (46.4 + 44.2) kPa = 0.488 . By Eqn. 1A.8, the actual partial vapour pressures would then be pA = yAptotal = 0.512 × 101.3 kPa = 51.9 kPa and pB = yBptotal = 0.488 × 101.3 kPa = 49.4 kPa . To find the activity coefficients, note that γ A = p A x A p A * = p A p A ° = 51.9 kPa 46.4 kPa = 1.117 and γ B = 49.4 kPa 44.2 kPa = 1.117 Comment. Assuming that the actual proportions of the vapours are the same as the ideal proportions begs the question (i.e., arrives at the answer by assumption rather than calculation). The assumption is not unreasonable, though. It is equivalent to assuming that the activity coefficients of the two components are equal (when in principle they could be different). The facts that the difference between ideal and actual total pressure is relatively small (on the order of 10%), that non-ideal behavior is due to the interaction of the two components, and that the two components are present in comparable quantities combine to suggest that the error we make in making this assumption is fairly small. 9 5B.10 By the van’t Hoff equation [5B.16] Π = [B]RT = cRT M Division by the standard acceleration of free fall, g, gives Π g = c(R / g)T M (a) This expression may be written in the form ′Π = c ′R T M which has the same form as the van’t Hoff equation, but the unit of osmotic pressure (Π′) is now force / area length / time2 = (mass length) / (area time2 ) length / time2 = mass area This ratio can be specified in g cm–2. Likewise, the constant of proportionality (R′) would have the units of R/g. energy K−1 mol−1 length / time2 = (mass length2 / 2time ) K −1 mol−1 length / time2 = mass length K−1 mol−1 This result may be specified in g cm K–1 mol–1 . ′R = R g = 8.314 47 J K −1 mol−1 9.80665m s−2 = 0.847840 kg m K−1 mol−1 10 3 g kg × 10 2 cm m = 84784.0 g cm K−1 mol−1 In the following we will drop the primes giving Π = cRT M and use the Π units of g cm–2 and the R units g cm K–1 mol–1. (b) By extrapolating the low concentration plot of Π / c versus c (Figure 5B.3(a)) to c = 0 we find the intercept 230 g cm–2/(g cm–3). In this limit the van’t Hoff equation is valid so Figure 5B.3(a) RT M = intercept or M = RT intercept 12 M = RT intercept = (84784.0 g cm K−1 mol−1) × (298.15K) (230 g cm−2 ) / (g cm−3) = 1.1×105 g mol−1 (c) The plot of Π / c versus c for the full concentration range (Figure 5B.3(b)) is very nonlinear. We may conclude that the solvent is good. This may be due to the nonpolar nature of both solvent and solute. Figure 5B.3(b) (d) The virial analogue to the van’t Hoff equation (eqn. 5B.18) rearranges to Π / c = (RT / M )(1+ ′B c + ′C c 2 ) Since RT / M has been determined in part (b) by extrapolation to c = 0, it is best to determine the second and third virial coefficients with the linear regression fit (Π / c) / (RT / M )−1 c = ′B + ′C c R = 0.9791. B′ = 21.4 cm3 g–1; standard deviation = 2.4 cm3 g–1 . C′ = 211 cm6 g–2; standard deviation = 15 cm6 g–2 . (e) Using 1/4 for g and neglecting terms beyond the second power, we may write ( ) ( ) ( )1 2 1 2 11 2RT B cc MΠ / / ′= + We can solve for B′ , then g(B′)2 = C′ . ( ) ( ) 1 2 1 2 11 2 c B c RT M Π / / ′− = RT / M has been determined above as 230 g cm–2/(g cm–3). We may analytically solve for B′ from one of the data points, say, Π / c = 430 g cm −2 / g cm−3 at c = 0.033 g cm–3. 430 g cm−2 / g cm−3 230 g cm−2 / g cm−3 1/2 − 1 = 1 2 ′B × (0.033 g cm−3) 13 ′B = 2 × (1.367 − 1) 0.033 g cm−3 = 22 cm3 g−1 ′C = g( ′B ) 2 = 0.25 × (22 cm3 g−1)2 = 123 cm6 g−2 Better values of B′ and C′ can be obtained by plotting ( ) ( )1 2 1 2RTc MΠ / // against c. This plot is shown in Figure 5B.3(c). The slope is 14 cm3 g–1. B′ =2 × slope = 28 cm3 g–1. C′ is then 196 cm6 g−2 . The intercept of this plot should theoretically be 1.00, but it is in fact 0.916 with a standard deviation of 0.066. The overall consistency of the values of the parameters confirms that g is roughly 1/4 as assumed. Figure 5B.3(c) 5B.12 The Gibbs energy of mixing an ideal solution is [5A.16] ∆mixG = nRT(xA ln xA + xB ln xB) The molar Gibbs energy of mixing is plotted against composition for several temperatures in Fig. 5B.4. The legend shows the temperature in kelvins. Figure 5B.4 The composition at which the temperature dependence is strongest is the composition at which the function has its largest magnitude, namely xA = xB = 0.5. 5B.14 The theoretical solubility [5B.15] is 14 5C.3(b) Refer to the figure given with the exercise. At the lowest temperature shown on the phase diagram, there are two liquid phases, a water-rich phase (xB = 0.07) and a methylpropanol- rich phase (xB = 0.88); the latter phase is about 10 times as abundant as the former (lever rule). On heating, the compositions of the two phases change, the water-rich phase increasing significantly in methylpropanol and the methylpropanol-rich phase more gradually increasing in water. (Note how the composition of the left side of the diagram changes more with temperature than the right.) The relative proportions of the phases continue to be given by the lever rule. Just before the isopleth intersects the phase boundary, the methylpropanol-rich phase (xB = 0.84) is in equilibrium with a vanishingly small water- rich phase (xB = 0.3). Then the phases merge, and the single-phase region is encountered with xB = 0.3. 5C.4(b) The feature that indicates incongruent melting (Topic 5C.4(c)) is circled in Figure 5C.3. The incongruent melting point is marked as T1. The composition of the eutectic is xB ≈ 0.58 and its melting point is labeled T2 . Figure 5C.3 5C.5(b) The cooling curves corresponding to the phase diagram in Figure 5C.4(a) are shown in Figure 5C.4(b). Note the breaks (abrupt change in slope) at temperatures corresponding to points a1, b1, and b2. Also note the eutectic halts at a2 and b3. Figure 5C.4 5C.6(b) Refer to Figure 5C.5. Dotted horizontal lines have been drawn at the relevant temperatures. 17 Figure 5C.5 (i) At 500°C, the phase diagram shows a single liquid phase at all compositions, so B is soluble in A in all proportions. (ii) At 390°C, solid B exists in equilibrium with a liquid whose composition is circled and labeled x1 on Figure 5.11. That composition is xB = x1 = 0.63. (iii) At point x2, two phases coexist: solid AB2 and a liquid mixture of A and B with mole fraction xB = x2 = 0.41. Although the liquid does not contain any AB2 units, we can think of the liquid as a mixture of dissociated AB2 in A. Call the amount (moles) of the compound nc and that of free A na. Thus, the amount of A (regardless of whether free or in the compound) is nA = na + nc , and the amount of B is nB = 2nc . The mole fraction of B is x B = x 2 = n B n A + n B = 2n c (n a + n c ) + 2n c = 2n c n a + 3n c Rearrange this relationship, collecting terms in nc on one side and na on the other: nax2 = nc(2–3x2) . The mole ratio of compound to free A is given by n c n a = x 2 2 − 3x 2 = 0.41 2 − 3× 0.41 = 0.53 . 5C.7(b) The phase diagram is shown in Figure 5C.6. Point symbols are plotted at the given data points. The lines are schematic at best. 18 Figure 5C.6 At 860°C, a solid solution with x(ZrF4) = 0.27 appears. The solid solution continues to form, and its ZrF4 content increases until it reaches x(ZrF4) = 0.40 at 830°C. At that temperature and below, the entire sample is solid. 5C.8(b) The phase diagram for this system (Figure 5C.7) is very similar to that for the system methyl ethyl ether and diborane of Exercise 5C.7(a). The regions of the diagram contain analogous substances. The mixture in this Exercise has a diborane mole fraction of 0.80. Follow this isopleth down to see that crystallization begins at about 123 K. The liquid in equilibrium with the solid becomes progressively richer in diborane until the liquid composition reaches 0.90 at 104 K. Below that temperature the system is a mixture of solid compound and solid diborane. Figure 5C.7 5C.9(b) The cooling curves are sketched in Figure 5C.8. Note the breaks and halts. The breaks correspond to changes in the rate of cooling due to the freezing out of a solid which releases its heat of fusion and thus slows down the cooling process. The halts correspond to the existence of three phases and hence no variance until one of the phases disappears. Figure 5C.8 19 (100) × 24.3 24.3+127 = 16 and in Mg2Cu it is (100) × 48.6 48.6 + 63.5 = 43 . The initial point is a1, corresponding to a single-phase liquid system. At a2 (at 720°C) MgCu2 begins to come out of solution and the liquid becomes richer in Mg, moving toward e2. At a3 there is solid MgCu2 + liquid of composition e2 (33 per cent by mass of Mg). This solution freezes without further change. The cooling curve will resemble that shown in Figure 5C.12(b). Figure 5C.12 5C.8 The data are plotted in Figure 5C.13. At 360°C, K2FeCl4(s) appears. The solution becomes richer in FeCl2 until the temperature reaches 351°C, at which point KFeCl3(s) also appears. Below 351°C the system is a mixture of K2FeCl4(s) and KFeCl3(s). Figure 5C.13 5C.10 Equation 5C.5 is 22 p = pA * pB * pA * + ( pB * − pA * )yA First divide both sides by pA * to express the pressure in units of pA * . Next, divide both numerator and denominator by pB * to see if the right hand side can be expressed as a function of the ratio pA * / pB * rather than of each vapor pressure separately: p / pA * = 1 pA * / pB * + (1− pA * / pB * )yA The plot of p / pA * vs. yA at several values of the vapor pressure ratio is shown in Figure 5C.4 of the main text. 5C.12 Equation 5C.7 is The simplest way to construct a plot of ξ vs. xA is to isolate ξ: A plot based on this equation is shown in Figure 5C.14(a). Figure 5C.14(a) (a) The graphical method described in section 5C.3(b) and illustrated in the main text’s Figure 5C.19 is also shown below in Figure 5C.14(b). Here the left-hand side of eqn 5C.7 is plotted as the bold curve, and the lighter lines are the right-hand side for ξ = 1, 2, 3, and 5. Small squares are placed where the curve intersects one of the lines. Note that the curve intersects every line at xA = ½, the composition at which H E is maximized. For values of ξ ≤ 2, that is the only point of intersection; for values of ξ > 2, there are two additional points of intersection arranged at equal distance from xA = ½. 23 Figure 5C.14(b) The root xA = ½ is unlike the other roots of eqn 5C.7 in several respects. The graphical approach shows that it is a root for all values of ξ. That fact can be confirmed by inspection by substituting xA = ½ into eqn 5C.7, leading to 0 = 0 for any finite value of ξ. For ξ > 2, that root of eqn 5C.7 is a maximum in the Gibbs energy, not a minimum, as can be seen in Figure 5C.18 of the main text. However, in the equation obtained by isolating ξ, xA = ½ leads only to ξ = 2. That equation yields an indeterminate form for xA = ½, but application of L’Hospital’s rule yields 1 1 1 2 2 2 A 1 1 A A A A A A A ln 1 ln ln(1 ) ( ) 2 2 lim lim lim 2 2 1 2 1 2 2x x x x x x x x x x x − − → → → − − − − − + = = = = − − (b) One method of numerical solution is illustrated by the following cells from a spreadsheet. Set up one column to represent xA, one for the left-hand side of eqn 5C.7, and one for the right-hand side (with variable ). From the cells shown here, it is apparent that when xA = 0.9980 or 0.9985, but when xA = 0.9990. Therefore, the value of xA when the two sides were equal lies somewhere between 0.9985 and 0.9990, or, to three decimal places, at 0.999. Therefore, a root of eqn 5C.7 when ξ = 7 is xA = 0.999. x ln(x/(1-x)) 7(2x–1) 0.998 6.213 6.972 0.9985 6.501 6.979 0.999 6.907 6.986 5D Phase diagrams of ternary systems Answers to discussion question 5D.2 The lever rule [5C.6] applies in a ternary system, but with an important caveat. The tie lines along which the rule applies are experimentally determined, not necessarily horizontal lines or lines parallel to any edge of the triangular diagram. Thus the lever rule applies, but as a practical matter it can be used only in the vicinity of plotted tie lines. (By contrast, recall that the lever rule in a binary phase diagram could be applied within a two-phase region simply by drawing a horizontal line to the appropriate phase boundaries.) See Topic 5D.2(a) and Figure 5D.4 of the main text. Solutions to exercises 5D.1(b) The ordered triples (xA, xB, xC) are plotted in Figure 5D.1. The vertices of the triangular phase diagram are labeled for the component that is pure at that vertex. For example, the top of the diagram is pure A: (1, 0, 0). As a reminder, at the edge opposite a labeled vertex, that 24