Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap24

Part 3: Change 24 Molecules in motion Solutions to exercises Discussion questions E24.1(b) Diffusion is the migration of particles (molecules) down a concentration gradient Diffusion can be interpreted at the molecular level as being the result of the random jostling of the molecules in a fluid The motion of the molecules is the result of a series of short jumps in random directions, a so-called random walk In the random walk model of diffusion, although a molecule may take many steps in a given time, it has only a small probability of being found far from its starting point because some of the steps lead it away from the starting point but others lead it back As a result, the net distance traveled increases only as the square root of the time There is no net flow of molecules unless there is a concentration gradient in the fluid, alse there are just as many molecules moving in one direction as another The rate at which the molecules spread out is proportional to the concentration gradient The constant of proportionality is called the diffusion coefficient On the molecular level in a gas, thermal conduction occurs because of random molecular motions in the presence of a temperature gradient Across any plane in the gas, there is a net flux of energy from the high temperature side, because molecules coming from that side carry a higher average energy per molecule across the plane than those coming from the low temperature side In solids, the situation is more complex as energy transport occurs through quantized elastic waves (phonons) and, in metals, also by electrons Conduction in liquids can occur by all the mechanisms mentioned At the molecular (ionic) level, electrical conduction in an electrolytic solution is the net migration of ions in any given direction When a gradient in electrical potential exists in a conductivity cell there will be a greater flow of positive ions in the direction of the negative electrode than in the direction of the positive electrode, hence there is a net flow of positive charge toward the region of low electrical potential Likewise a net flow of negative ions in the direction of the positive electrode will occur In metals, only negatively charged electrons contribute to the current To see the connection between the flux of momentum and the viscosity, consider a fluid in a state of Newtonian flow, which can be imagined as occurring by a series of layers moving past one another (Fig 24.11 of the text) The layer next to the wall of the vessel is stationary, and the velocity of successive layers varies linearly with distance, z, from the wall Molecules ceaselessly move between the layers and bring with them the x-component of linear momentum they possessed in their original layer A layer is retarded by molecules arriving from a more slowly moving layer because they have a low momentum in the x-direction A layer is accelerated by molecules arriving from a more rapidly moving layer We interpret the net retarding effect as the fluid’s viscosity E24.2(b) According to the Grotthuss mechanism, there is an effective motion of a proton that involves the rearrangement of bonds in a group of water molecules However, the actual mechanism is still highly contentious Attention now focuses on the H9 O4 + unit in which the nearly trigonal planar H3 O+ ion is linked to three strongly solvating H2 O molecules This cluster of atoms is itself hydrated, but the hydrogen bonds in the secondary sphere are weaker than in the primary sphere It is envisaged that the rate-determining step is the cleavage of one of the weaker hydrogen bonds of this secondary sphere (Fig 24.19a of the text) After this bond cleavage has taken place, and the released molecule has rotated through a few degrees (a process that takes about ps), there is a rapid adjustment of bond lengths and angles in the remaining cluster, to form an H5 O2 + cation of structure H2 O · · · H+ · · · OH2 (Fig 24.19b) Shortly after this reorganization has occurred, a new H9 O4 + cluster forms as other molecules rotate into a position where they can become members of a secondary hydration sphere, INSTRUCTOR’S MANUAL 386 but now the positive charge is located one molecule to the right of its initial location (Fig 24.19c) According to this model, there is no coordinated motion of a proton along a chain of molecules, simply a very rapid hopping between neighbouring sites, with a low activation energy The model is consistent with the observation that the molar conductivity of protons increases as the pressure is raised, for increasing pressure ruptures the hydrogen bonds in water E24.3(b) Because the drift speed governs the rate at which charge is transported, we might expect the conductivity to decrease with increasing solution viscosity and ion size Experiments confirm these predictions for bulky ions, but not for small ions For example, the molar conductivities of the alkali metal ions increase from Li+ to Cs+ (Table 24.6) even though the ionic radii increase The paradox is resolved when we realize that the radius a in the Stokes formula is the hydrodynamic radius (or “Stokes radius”) of the ion, its effective radius in the solution taking into account all the H2 O molecules it carries in its hydration sphere Small ions give rise to stronger electric fields than large ones, so small ions are more extensively solvated than big ions Thus, an ion of small ionic radius may have a large hydrodynamic radius because it drags many solvent molecules through the solution as it migrates The hydrating H2 O molecules are often very labile, however, and NMR and isotope studies have shown that the exchange between the coordination sphere of the ion and the bulk solvent is very rapid The proton, although it is very small, has a very high molar conductivity (Table 24.6)! Proton and O-NMR show that the times characteristic of protons hopping from one molecule to the next are about 1.5 ps, which is comparable to the time that inelastic neutron scattering shows it takes a water molecule to reorientate through about rad (1–2 ps) 17 Numerical exercises E24.4(b) (a) The mean speed of a gas molecule is c= so 8RT 1/2 πM c(He) = c(Hg) M(Hg) 1/2 = M(He) 200.59 1/2 = 7.079 4.003 (b) The mean kinetic energy of a gas molecule is 21 mc2 , where c is the root mean square speed c= 3RT 1/2 M So 21 mc2 is independent of mass, and the ratio of mean kinetic energies of He and Hg is E24.5(b) (a) The mean speed can be calculated from the formula derived in Example 24.1 c= RT 1/2 = πM × (8.314 J K −1 mol−1 ) × (298 K) π × (28.02 × 10−3 kg mol−1 ) 1/2 = 4.75 × 102 m s−1 kT [24.14] (b) The mean free path is calculated from λ = 1/2 σp with σ = πd = π × (3.95 × 10−10 m)2 = 4.90 × 10−19 m2 (1.381 × 10−23 J K−1 ) × (298 K) Then, λ = atm 1.013×105 Pa 21/2 × (4.90 × 10−19 m2 ) × (1 × 10−9 Torr) × 760 atm Torr × = × 104 m MOLECULES IN MOTION E24.6(b) 387 (c) The collision frequency could be calculated from eqn 31, but is most easily obtained from eqn 32, 4.75 × 102 m s−1 c = × 10−2 s−1 since λ and c have already been calculated z = = λ 4.46 × 104 m Thus there are 100 s between collisions, which is a very long time compared to the usual timescale of molecular events The mean free path is much larger than the dimensions of the pumping apparatus used to generate the very low pressure kT p = 1/2 [24.14] σλ σ = πd , p = σ 1/2 d= = π 0.36 nm2 π 1/2 = 0.34 nm (1.381 × 10−23 J K−1 ) × (298 K) (21/2 ) × (0.36 × 10−18 m2 ) × (0.34 × 10−9 m) = 2.4 × 107 Pa This pressure corresponds to about 240 atm, which is comparable to the pressure in a compressed gas cylinder in which argon gas is normally stored E24.7(b) The mean free path is kT (1.381 × 10−23 J K−1 ) × (217 K) λ = 1/2 = = 4.1 × 10−7 m σp 21/2 [0.43 × (10−9 m)2 ] × (12.1 × 103 Pa atm−1 ) E24.8(b) Obtain data from Exercise 24.7(b) The expression for z obtained in Exercise 24.8(a) is z = 1/2 16 σp π mkT Substituting σ = 0.43 nm2 , p = 12.1 × 103 Pa, m = (28.02 u), and T = 217 K we obtain z = × (0.43 × 10−18 m2 ) × (12.1 × 103 Pa) [π × (28.02) × (1.6605 × 10−27 kg) × (1.381 × 10−23 J K−1 ) × (217 K)]1/2 = 9.9 × 108 s−1 E24.9(b) The mean free path is kT (1.381 × 10−23 J K−1 ) × (25 + 273) K 5.50 × 10−3 m Pa λ = 1/2 = = p σp 21/2 [0.52 × (10−9 m)2 ]p (a) λ = (b) λ = (c) λ = 5.50 × 10−3 m Pa (15 atm) × (1.013 × 105 Pa atm−1 ) 5.50 × 10−3 m Pa (1.0 bar) × (105 Pa bar −1 ) 5.50 × 10−3 m Pa (1.0 Torr) × 1.013×105 Pa atm−1 760 Torr atm−1 = 3.7 × 10−9 m = 5.5 × 10−8 m = 4.1 × 10−5 m INSTRUCTOR’S MANUAL 388 E24.10(b) The fraction F of molecules in the speed range from 200 to 250 m s−1 is 250 m s−1 F = 200 m s−1 f (v)dv where f (v) is the Maxwell distribution This can be approximated by F ≈ f (v) v = 4π 3/2 −Mv M v exp 2πRT 2RT v, with f (v) evaluated in the middle of the range F ≈ 4π 44.0 × 10−3 kg mol−1 3/2 × (225 m s−1 )2 2π(8.3145 J K −1 mol−1 ) × (300 K) × exp −(44.0 × 10−3 kg mol−1 ) × (225 m s−1 )2 2(8.3145 J K−1 mol−1 ) × (300 K) × (50 m s−1 ), F ≈ 9.6 × 10−2 Comment The approximation we have employed, taking f (v) to be nearly constant over a narrow range of speeds, may not be accurate enough, for that range of speeds includes about 10 per cent of the molecules Numerical exercises E24.11(b) The number of collisions is N = ZW At = = pAt (2πmkT )1/2 (111 Pa) × (3.5 × 10−3 m) × (4.0 × 10−3 m) × (10 s) {2π × (4.00 u) × (1.66 × 10−27 kg u−1 ) × (1.381 × 10−23 J K−1 ) × (1500 K)}1/2 = 1.1 × 1021 E24.12(b) The mass of the sample in the effusion cell decreases by the mass of the gas which effuses out of it That mass is the molecular mass times the number of molecules that effuse out m = mN = mZW At = mpAt m 1/2 = pAt = pAt 2π kT (2πmkT )1/2 = (0.224 Pa) × π × 21 × 3.00 × 10−3 m × × (24.00 h) × (3600 s h−1 ) 300 × 10−3 kg mol−1 2π × (8.3145 J K −1 mol−1 ) × (450 K) = 4.89 × 10−4 kg 1/2 M 2π RT 1/2 MOLECULES IN MOTION 389 E24.13(b) The flux is J = −κ dT dT = − λCV ,m v [X] dz dz where the minus sign indicates flow toward lower temperature and λ= √ So J = − 2N σ , 2CV ,m 3σ NA =− × v = 8kT 1/2 = πm 8RT 1/2 , πM and [M] = n/V = N/NA RT 1/2 dT πM dz × (28.832 − 8.3145) J K −1 mol−1 × [0.27 × (10−9 m)2 ] × (6.022 × 1023 mol−1 ) (8.3145 J K−1 mol−1 ) × (260 K) 1/2 × (3.5 K m−1 ) π × (2.016 × 10−3 kg mol−1 ) = 0.17 J m−2 s−1 E24.14(b) The thermal conductivity is κ = 2CV ,m λCV ,m v [X] = 3σ NA RT 1/2 πM so σ = 2CV ,m 3κNA RT 1/2 πM = (0.240 mJ cm−2 s−1 ) × (K cm−1 )−1 = 0.240 × 10−1 J m−1 s−1 K −1 so σ = × (29.125 − 8.3145) J K −1 mol−1 × (0.240 × 10−1 J m−1 s−1 K −1 ) × (6.022 × 1023 mol−1 ) × (8.3145 J K−1 mol−1 ) × (298 K) 1/2 π × (28.013 × 10−3 kg mol−1 ) = 1.61 × 10−19 m2 E24.15(b) Assuming the space between sheets is filled with air, the flux is J = −κ dT = [(0.241 × 10−3 J cm−2 s−1 ) × (K cm−1 )−1 ] × dz [50 − (−10)] K 10.0 cm = 1.45 × 10−3 J cm−2 s−1 So the rate of energy transfer and energy loss is J A = (1.45 × 10−3 J cm−2 s−1 ) × (1.50 m2 ) × (100 cm m−1 )2 = 22 J s−1 INSTRUCTOR’S MANUAL 390 E24.16(b) The time dependence of the pressure of a gas effusing without replenishment is √ p = p0 e−t/τ where τ ∝ m The time t it takes for the pressure to go from any initial pressure p0 to a prescribed fraction of that pressurefp0 is t = τ ln fp0 = τ ln f p0 √ so the time is proportional to τ and therefore also to m Therefore, the ratio of times it takes two different gases to go from the same initial pressure to the same final pressure is related to their molar masses as follows M1 1/2 M2 t1 = t2 and t2 t1 M = M1 82.3 s = 554 g mol−1 18.5 s E24.17(b) The time dependence of the pressure of a gas effusion without replenishment is So Mfluorocarbon = (28.01 g mol−1 ) × p = p0 e−t/τ where τ = = V A0 so t = τ ln p0 /p 2πm 1/2 V = kT A0 2π M 1/2 RT 22.0 m3 π × (0.50 × 10−3 m)2 2π × (28.0 × 10−3 kg mol−1 ) × 1/2 = 2.4 × 105 s (8.3145 J K−1 mol−1 ) × (293 K) 122 kPa = 1.5 × 104 s 105 kPa E24.18(b) The coefficient of viscosity is so t = (8.6 × 105 s) ln η = 13 λmN v = 3σ mkT 1/2 π σ = so 3η mkT 1/2 π = 1.66 µP = 166 × 10−7 kg m−1 s−1 so σ = × (166 × 10−7 kg m−1 s−1 ) × (28.01 × 10−3 kg mol−1 ) × (1.381 × 10−23 J K−1 ) × (273 K) π × (6.022 × 1023 mol−1 ) = 3.00 × 10−19 m2 E24.19(b) The rate of fluid flow through a tube is described by )π r (p2 − pout dV = in dt 16lηp0 so pin = 1/2 16lηp0 dV + p out π r dt Several of the parameters need to be converted to MKS units r = 21 (15 × 10−3 m) = 7.5 × 10−3 m dV and = 8.70 cm3 × (10−2 m cm−1 )3 s−1 = 8.70 × 10−6 m3 s−1 dt 1/2 MOLECULES IN MOTION 391 Also, we have the viscosity at 293 K from the table According to the T 1/2 temperature dependence, the viscosity at 300 K ought to be η(300 K) = η(293 K) × 300 K 1/2 = (176 × 10−7 kg m−1 s−1 ) × 293 K 300 1/2 293 = 1.78 × 10−7 kg m−1 s−1 pin = 16(10.5 m) × (178 × 10−7 kg m−1 s−1 ) × (1.00 × 105 Pa) π × (7.5 × 10−3 m)4 1/2 ×(8.70 × 10 −6 −1 m s ) + (1.00 × 10 Pa) = 1.00 × 105 Pa Comment For the exercise as stated the answer is not sensitive to the viscosity The flow rate is so low that the inlet pressure would equal the outlet pressure (to the precision of the data) whether the viscosity were that of N2 at 300 K or 293 K—or even liquid water at 293 K! E24.20(b) The coefficient of viscosity is η = 13 λmN v = 3σ = mkT 1/2 π 3[0.88 × (10−9 m)2 ] × (78.12 × 10−3 kg mol−1 ) × (1.381 × 10−23 J K−1 )T π × (6.022 × 1023 mol−1 ) = 5.72 × 10−7 × (T /K)1/2 kg m−1 s−1 (a) At 273 K η = (5.72 × 10−7 ) × (273)1/2 kg m−1 s−1 = 0.95 × 10−5 kg m−1 s−1 (b) At 298 K η = (5.72 × 10−7 ) × (298)1/2 kg m−1 s−1 = 0.99 × 10−5 kg m−1 s−1 (c) At 1000 K η = (5.72 × 10−7 ) × (1000)1/2 kg m−1 s−1 = 1.81 × 10−5 kg m−1 s−1 E24.21(b) The thermal conductivity is 2CV ,m κ = 13 λCV ,m v [X] = 3σ NA (a) RT 1/2 πM × [(20.786 − 8.3145) J K −1 mol−1 ] κ= 3[0.24 × (10−9 m)2 ] × (6.022 × 1023 mol−1 ) × (8.3145 J K −1 mol−1 ) × (300 K) π(20.18 × 10−3 kg mol−1 ) = 0.0114 J m−1 s−1 K −1 1/2 1/2 INSTRUCTOR’S MANUAL 392 The flux is J = −κ (305 − 295) K 0.15 m dT = (0.0114 J m−1 s−1 K −1 ) × dz = 0.76 J m−2 s−1 so the rate of energy loss is J A = (0.76 J m−2 s−1 ) × (0.15 m)2 = 0.017 J s−1 (b) × [(29.125 − 8.3145) J K −1 mol−1 ] κ= 3[0.43 × (10−9 m)2 ] × (6.022 × 1023 mol−1 ) 8.3145 J K −1 mol−1 ) × (300 K) × 1/2 π(28.013 × 10−3 kg mol−1 ) = 9.0 × 10−3 J m−1 s−1 K −1 The flux is J = −κ (305 − 295) K 0.15 m dT = (9.0 × 10−3 J m−1 s−1 K −1 ) × dz = 0.60 J m−2 s−1 so the rate of energy loss is J A = (0.60 J m−2 s−1 ) × (0.15 m)2 = 0.014 J s−1 E24.22(b) The rate of fluid flow through a tube is described by )π r (p2 − pout dV = in dt 16lηp0 so the rate is inversely proportional to the viscosity, and the time required for a given volume of gas to flow through the same tube under identical pressure conditions is directly proportional to the viscosity t1 η1 = t2 η2 η1 t2 t1 (208 µP) × (18.0 s) = 52.0 µP = 52.0 × 10−7 kg m−1 s−1 ηCFC = 72.0 s The coefficient of viscosity is so η2 = η = 13 λmN v = 3σ × mkT 1/2 = π 3π d × mkT 1/2 π so the molecular diameter is 1/2 × 3πη d = mkT 1/4 π 1/2 = 3π(52.0 × 10−7 kg m−1 s−1 ) × (200 × 10−3 kg mol−1 ) × (1.381 × 10−23 J K−1 ) × (298 K) π × (6.022 × 1023 mol−1 ) = 9.23 × 10−10 m = 923 pm 1/4 MOLECULES IN MOTION E24.23(b) 393 RT 1/2 πM 2CV ,m κ = 13 λCV ,m v [X] = 3σ NA × (29.125 − 8.3145) J K −1 mol−1 = 3[0.43 × (10−9 m)2 ] × (6.022 × 1023 mol−1 ) × (8.3145 J K−1 mol−1 ) × (300 K) 1/2 π × (28.013 × 10−3 kg mol−1 ) = 9.0 × 10−3 J m−1 s−1 K −1 E24.24(b) The diffusion constant is 2(RT )3/2 D = 13 λ v = 3σpNA (π M)1/2 = = 2[(8.3145 J K −1 mol−1 ) × (298 K)]3/2 3[0.43 × (10−9 m)2 ]p(6.022 × 1023 mol−1 ) × π(28.013 × 10−3 kg mol−1 ) 1/2 1.07 m2 s−1 p/Pa The flux due to diffusion is J = −D d[X] d = −D dx dx n V =− D RT dp dx where the minus sign indicates flow from high pressure to low So for a pressure gradient of 0.10 atm cm−1 J = D/(m2 s−1 ) (8.3145 J K−1 mol−1 ) × (298 K) × (0.20 × 105 Pa m−1 ) = (8.1 mol m−2 s−1 ) × (D/(m2 s−1 )) (a) D= 1.07 m2 s−1 = 0.107 m2 s−1 10.0 and J = (8.1 mol m−2 s−1 ) × (0.107) = 0.87 mol m−2 s−1 (b) D= 1.07 m2 s−1 = 1.07 × 10−5 m2 s−1 100 × 103 and J = (8.1 mol m−2 s−1 ) × (1.07 × 10−5 ) = 8.7 × 10−5 mol m−2 s−1 (c) D= 1.07 m2 s−1 = 7.13 × 10−8 m2 s−1 15.0 × 106 and J = (8.1 mol m−2 s−1 ) × (7.13 × 10−8 ) = 5.8 × 10−7 mol m−2 s−1 E24.25(b) Molar ionic conductivity is related to mobility by λ = zuF = (1) × (4.24 × 10−8 m2 s−1 V−1 ) × (96 485 C mol−1 ) = 4.09 × 10−3 S m2 mol−1 E24.26(b) The drift speed is given by s = uE = u φ (4.01 × 10−8 m2 s−1 V−1 ) × (12.0 V) = = 4.81 × 10−5 m s−1 l 1.00 × 10−2 m INSTRUCTOR’S MANUAL 394 E24.27(b) The limiting transport number for Cl− in aqueous NaCl at 25◦ C is ◦ t− = u− 7.91 = 0.604 = u+ + u − 5.19 + 7.91 (The mobilities are in 10−8 m2 s−1 V−1 ) E24.28(b) The limiting molar conductivity of a dissolved salt is the sum of that of its ions, so ◦ m (MgI2 ) = λ(Mg2+ ) + 2λ(I− ) = ◦ m (Mg(C2 H3 O2 )2 ) + ◦ m (NaI) − ◦ m (NaC2 H3 O2 ) = (18.78 + 2(12.69) − 2(9.10)) mS m2 mol−1 = 25.96 mS m2 mol−1 E24.29(b) Molar ionic conductivity is related to mobility by λ = zuF F− : u= so u = λ zF 5.54 × 10−3 S m2 mol−1 Cl− : u= Br − : u= (1) × (96 485 C mol−1 ) = 5.74 × 10−8 m2 V−1 s−1 7.635 × 10−3 S m2 mol−1 (1) × (96 485 C mol−1 ) = 7.913 × 10−8 m2 V−1 s−1 7.81 × 10−3 S m2 mol−1 = 8.09 × 10−8 m2 V−1 s−1 (1) × (96 485 C mol−1 ) E24.30(b) The diffusion constant is related to the mobility by D = (4.24 × 10−8 m2 s−1 V−1 ) × (8.3145 J K −1 mol−1 ) × (298 K) uRT = zF (1) × (96 485 C mol−1 ) = 1.09 × 10−9 m2 s−1 E24.31(b) The mean square displacement for diffusion in one dimension is x = 2Dt In fact, this is also the mean square displacement in any direction in two- or three-dimensional diffusion from a concentrated source In three dimensions r = x + y + z2 so r = x + y + z2 = x = 6Dt So the time it takes to travel a distance t= r is r2 (1.0 × 10−2 m)2 = 4.1 × 103 s = 6D 6(4.05 × 10−9 m2 s−1 ) E24.32(b) The diffusion constant is related to the viscosity of the medium and the size of the diffusing molecule as follows D= kT 6πηa so a= kT (1.381 × 10−23 J K−1 ) × (298 K) = 6πηD 6π(1.00 × 10−3 kg m−1 s−1 ) × (1.055 × 10−9 m2 s−1 ) a = 2.07 × 10−10 m = 207 pm MOLECULES IN MOTION 395 E24.33(b) The Einstein–Smoluchowski equation related the diffusion constant to the unit jump distance and time λ2 λ2 so τ = 2τ 2D If the jump distance is about one molecular diameter, or two effective molecular radii, then the jump distance can be obtained by use of the Stokes–Einstein equation D= D= and τ = kT kT = 6πηa 3πηλ so λ= kT 3π ηD [(1.381 × 10−23 J K−1 ) × (298 K)]2 (kT )2 = 18(π η)2 D 18[π(0.387 × 10−3 kg m−1 s−1 )]2 × (3.17 × 10−9 m2 s−1 )3 = 2.00 × 10−11 s = 20 ps E24.34(b) The mean square displacement is (from Exercise 24.31(b)) r = 6Dt so t = (1.0 × 10−6 m)2 r2 = 1.7 × 10−2 s = 6D 6(1.0 × 10−11 m2 s−1 ) Solutions to problems Solutions to numerical problems P24.3 X = Ni Xi [See Problem 24.2] N i {1.80 m + × (1.82 m) + · · · + 1.98 m} 1.89 m 53 (1.80 m)2 + × (1.82 m)2 + · · · + (1.98 m)2 = 3.57 m2 (b) h2 = 53 (a) h = h2 = 1.89 m P24.4 κ = 13 λcCV ,m [A] [24.28] 8kT 1/2 c= [24.7] ∝ T 1/2 πm Hence, κ ∝ T 1/2 CV ,m , so P24.7 κ = κ T T 1/2 × CV ,m CV ,m At 300 K, CV ,m ≈ 23 R + R = 25 R At 10 K, CV ,m ≈ 23 R [rotation not excited] 300 1/2 κ × = 9.1 = Therefore, 10 κ The atomic current is the number of atoms emerging from the slit per second, which is ZW A with A = × 10−7 m2 We use p [24.15] ZW = (2πmkT )1/2 p/Pa = [(2π) × (M/g mol−1 ) × (1.6605 × 10−27 kg) × (1.381 × 10−23 J K−1 ) × (380 K)]1/2 = (1.35 × 1023 m−2 s−1 ) × p/Pa (M/g mol−1 )1/2 INSTRUCTOR’S MANUAL 396 (a) Cadmium: ZW A = (1.35 × 1023 m−2 s−1 ) × (1 × 10−7 m2 ) × (b) Mercury: ZW A = (1.35 × 1023 m−2 s−1 ) × (1 × 10−7 m2 ) × P24.10 κ c= c≈ m [24.98] ≈ κ ◦ m 0.13 (112.4)1/2 = × 1014 s−1 152 (200.6)1/2 = × 1017 s−1 [c small, conductivity of water allowed for in the data] 1.887 × 10−6 S cm−1 138.3 S cm2 mol−1 [Exercise 24.28(a)] ≈ 1.36 × 10−8 mol cm−3 = solubilit y = 1.36 × 10−5 M P24.12 t (H+ ) = u(H+ ) u(H+ ) + u(Cl− ) [24.61] = 3.623 = 0.82 3.623 + 0.791 When a third ion is present we use t (H+ ) = I (H+ ) [24.58] I (H+ ) + I (Na+ ) + I (Cl− ) For each I , I = zuνcF AE = constant × cu Hence, when NaCl is added t (H+ ) = = P24.14 t+ = c(H+ )u(H+ ) c(H+ )u(H+ ) + c(Na+ )u(Na+ ) + c(Cl− )u(Cl− ) (1.0 × 10−3 ) × (3.623) = 0.0028 (1.0 × 10−3 ) × (3.623) + (1.0) × (0.519) + (1.001) × (0.791) zcAF I × x t [Problem 24.13] The density of the solution is 0.682 g cm−3 ; the concentration c is related to the molality m by c/(mol L−1 ) = ρ/(kg L−1 ) × m/(mol kg−1 ) which holds for dilute solutions such as these A = πr = π × (2.073 × 10−3 m)2 = 1.350 × 10−5 m2 czAF (1.350 × 10−5 m2 ) × (9.6485 × 104 C mol−1 ) × c = (0.1042 m2 mol−1 ) × c = I t (5.000 × 10−3 A) × (2500 s) = (0.1042 m2 mol−1 ) × ρ × m = (0.1042 m2 mol−1 ) × (682 kg m−3 ) × m = (71.06 kg m−1 mol−1 ) × m = (0.07106 kg mm−1 mol−1 ) × m and so t+ = (0.07106 kg mm−1 mol−1 ) × x × m In the first solution t+ = (0.07106 kg mm−1 mol−1 ) × (286.9 mm) × (0.01365 mol kg−1 ) = 0.278 In the second solution t+ = (0.07106 kg mm−1 mol−1 ) × (92.03 mm) × (0.04255 mol kg−1 ) = 0.278 MOLECULES IN MOTION 397 Therefore, t (H+ ) = 0.28, a value much less than in pure water where t (H+ ) = 0.63 Hence, the mobility is much less relative to its counterion, NH− P24.17 If diffusion is analogous to viscosity [Section 24.5, eqn 24.36] in that it is also an activation energy controlled process, then we expect D ∝ e−Ea /RT Therefore, if the diffusion constant is D at T and D at T , Ea = − R ln D D T =− − T1 (8.314 J K−1 mol−1 ) × ln 2.89 2.05 298 K − 2731 K = 9.3 kJ mol−1 That is, the activation energy for diffusion is 9.3 kJ mol−1 x = 2Dt [24.91], P24.19 Hence, η = D= kT [24.83] 6π aη kT kT t 1.381 × 10−23 J K−1 ) × (298.15 K) × t = = 6πDa 3πa x (3π ) × (2.12 × 10−7 m) × x t x2 = (2.06 × 10−15 J m−1 ) × and therefore η/(kg m−1 s−1 ) = We draw up the following table t/s 108 x /cm2 103 η/(kg m−1 s−1 ) 2.06 × 10−11 (t/s) ( x /cm2 ) 30 60 90 120 88.2 0.701 113.4 1.09 128 1.45 144 1.72 Hence, the mean value is 1.2 × 10−3 kg m−1 s−1 P24.21 The viscosity of a perfect gas is η = 13 N mλc = mkT 1/2 π mc √ = 3σ 3σ so σ = 3η mkT 1/2 π The mass is m= (a) 17.03 × 10−3 kg mol−1 6.022 × 1023 mol−1 σ = = 2.828 × 10−26 kg 3(9.08 × 10−6 kg m−1 s−1 ) × (2.828 × 10−26 kg) × (1.381 × 10−23 J K−1 ) × (270 K) π = 4.25 × 10 −19 m = πd so d= 4.25 × 10−19 m2 π 1/2 1/2 = 3.68 × 10−10 m INSTRUCTOR’S MANUAL 398 (b) σ = 3(17.49 × 10−6 kg m−1 s−1 ) × 1/2 (2.828 × 10−26 kg) × (1.381 × 10−23 J K−1 ) × (490 K) π = 2.97 × 10−19 m2 = πd so 2.97 × 10−19 m2 π d= 1/2 = 3.07 × 10−10 m Comment The change in diameter with temperature can be interpreted in two ways First, it shows the approximate nature of the concept of molecular diameter, with different values resulting from measurements of different quantities Second, it is consistent with the idea that, at higher temperatures, more forceful collisions contract a molecule’s perimeter P24.22 The diffusion constant of an ion in solution is related to the mobility of the ion and to its radius in separate relations kT uRT = zF 6πηa D= a= so a= zF k ze = 6π ηuR 6π ηu (1) × (1.602 × 10−19 C) 6π(0.93 × 10−3 kg m−1 s−1 ) × (1.1 × 10−8 m2 V−1 s−1 ) = 8.3 × 10−10 m = 830 pm Solutions to theoretical problems P24.25 Write the mean velocity initially as a; then in the emerging beam vx = K a vx f (vx ) dvx where K is a constant which ensures that the distribution in the emergent beam is also normalized That is, a m 1/2 a −mv /2kT x 1=K f (vx ) dvx = K e dvx 2πkT 0 This integral cannot be evaluated analytically but it can be related to the error function by defining x2 = mvx2 2kT which gives dvx = 1= K = 2kT 1/2 dx Then m m 1/2 2πkT 2kT 1/2 b −x e dx m [b = (m/2kT )1/2 × a] b K e−x dx = 21 Kerf(b) π 1/2 where erf (z) is the error function [Table 12.2]: erf(z) = z π 1/2 e−x dx 2 Therefore, K = erf(b) The mean velocity of the emerging beam is vx = K m 1/2 a m 1/2 vx e−mvx /2kT dvx = K 2πkT 2π kT −kT m a d −mv /2kT x (e dvx dv x MOLECULES IN MOTION 399 kT 1/2 −ma /2kT (e − 1) 2mπ = −K 2kT 1/2 mπ This expression for the average magnitude of the one-dimensional velocity in the x direction may be obtained from Now use a = vx initial = vx = ∞ vx f (vx )dvx = = ∞ vx m 1/2 −mv /2kT x e dvx 2π kT m 1/2 2π kT 2kT m = 2kT 1/2 mπ It may also be obtained very quickly by setting a = ∞ in the expression for vx in the emergent beam with erf(b) = erf(∞) = Substituting a = erf 2kT 1/2 into vx in the emergent beam e−ma /2kT = e−1/π and erf(b) = mπ π 1/2 2kT 1/2 − e−1/π × mπ erf π 11/2 From tables of the error function (expanded version of Table 12.2), or from readily available software, or by interpolating Table 12.2 Therefore, vx = erf π 1/2 = erf(0.56) = 0.57 and e−1/π = 0.73 Therefore, vx = 0.47 vx initial P24.27 The most probable speed, c∗ , was evaluated in Problem 24.23 and is 2kT 1/2 m c∗ = v(most probable) = Consider a range of speeds v around c∗ and nc∗ , then with v = c∗ ∗2 ∗2 f (nc∗ ) (nc∗ )2 e−mn c /2kT [24.4] = n2 e−(n −1)mc /2kT = n2 e(1−n ) = ∗2 ∗ f (c ) c∗2 e−mc /2kT f (4c∗ ) f (3c∗ ) −3 −8 = × e = 3.02 × 10 = 16 × e−15 = 4.9 × 10−6 f (c∗ ) f (c∗ ) The current Ij carried by an ion j is proportional to its concentration cj , mobility uj , and charge number |zj | [Justification 24.9] Therefore Therefore, P24.28 Ij = Acj uj zj where A is a constant The total current passing through a solution is I= j Ij = A j cj uj z j INSTRUCTOR’S MANUAL 400 The transport number of the ion j is therefore tj = Ij Acj uj zj = = A j cj u j z j I c j uj z j j c j u j zj If there are two cations in the mixture t cuz = t c u z P24.29 cu c u = ∂ 2c ∂c = D [24.84] ∂t ∂x a or c = 1/2 e−bx /t t then ∂c =− ∂t ∂c = ∂x a t 1/2 =− = 2Dt e−bx /t + t 3/2 −2bx t × × n0 e−x /4Dt [24.88] A(π Dt)1/2 with c = a × ∂ 2c 2b =− t ∂x if z = z a t 1/2 c+ a t 1/2 × bx t2 bx c e−bx /t = − + c 2t t e−bx /t e−bx /t + bx Dt a t 1/2 × 2bx −bx /t 2b e =− c+ t t 2bx c t c ∂c as required D ∂t Initially the material is concentrated at x = Note that c = for x > when t = on account of the very strong exponential factor (e−bx /t → more strongly than 1/2 → ∞) When t x = 0, e−x /4Dt = We confirm the correct behaviour by noting that x = and x = at t = [24.90], and so all the material must be at x = at t = P24.31 Draw up the following table based on the third and last equations of Justification 24.12 N P (6λ)Exact P (6λ)Approx 0.004 N P (6λ)Exact P (6λ)Approx 30 0.0806 0.0799 0.016 0.162 40 0.0807 0.0804 0.0313 0.0297 60 0.0763 0.0763 10 0.0439 0.0417 20 0.0739 0.0725 100 0.0666 0.0666 The points are plotted in Fig 24.1 The discrepancy is less than 0.1 per cent when N > 60 MOLECULES IN MOTION 401 0.10 0.05 0 20 40 60 80 100 Figure 24.1 Solutions to applications P24.33 The work required for a mass, m, to go from a distance r from the centre of a planet of mass m to infinity is ∞ w= r F dr where F is the force of gravity and is given by Newton’s law of universal gravitation, which is Gmm r2 F = G is the gravitational constant (not to be confused with g) Then w = ∞ r Gmm Gmm dr = r r2 Since according to Newton’s second law of motion, F = mg, we may make the identification g= Gm r2 Thus, w = grm This is the kinetic energy that the particle must have in order to escape the planet’s gravitational attraction at a distance r from the planet’s centre; hence w = 21 mv = mgr ve = (2g Rp )1/2 [Rp = radius of planet] which is the escape velocity (a) ve = [(2) × (9.81 m s−2 ) × (6.37 × 106 m)]1/2 = 11.2 km s−1 (b) g(Mars) = R(Earth)2 m(Mars) × × g(Earth) = (0.108) × m(Earth) R(Mars)2 6.37 × (9.81 m s−2 ) 3.38 = 3.76 m s−2 Hence, ve = [(2) × (3.76 m s−2 ) × (3.38 × 106 m)]1/2 = 5.0 km s−1 INSTRUCTOR’S MANUAL 402 πMc2 8RT 1/2 ,T = πM 8R and we can draw up the following table Since c = 10−3 T /K Earth Mars H2 11.9 2.4 He 23.7 4.8 O2 190 38 [c = 11.2 km s−1 ] [c = 5.0 km s−1 ] In order to calculate the proportion of molecules that have speeds exceeding the escape velocity, ve , we must integrate the Maxwell distribution [24.4] from ve to infinity ∞ P = ve f (v)dv = ∞ m M m 3/2 −mv /2kT = v e dv R k 2π kT 4π ve This integral cannot be evaluated analytically and must be expressed in terms of the error function We proceed as follows m Defining β = and y = βv gives v = β −1/2 y, v = β −1 y , ve = β −1/2 ye , 2kT ye = β 1/2 ve , dv = β −1/2 dy ∞ 2 β 3/2 −1 −1/2 ∞ β β y e−y dy = 1/2 y e−y dy π π β 1/2 ve β 1/2 ve P = 4π = and ∞ π 1/2 y e−y dy − β 1/2 ve y e−y dy The first integral can be evaluated analytically; the second cannot ∞ y e−y dy = P =1− β 1/2 ve π 1/2 π 1/2 , hence ye−y (2y dy) = − β 1/2 ve π 1/2 y d(−e−y ) This integral may be evaluated by parts P =1− y(−e π 1/2 −y ) β 1/2 ve − β 1/2 ve (−e−y ) dy β 1/2 ve 2 β 1/2 β 1/2 −βve2 P =1+2 ve e − 1/2 e−y dy = + ve e−βve − erf(β 1/2 ve ) π π π = erfc(β 1/2 ve ) + From β = β 1/2 ve e−βve π m M = and ve = (2gRp )1/2 2kT 2RT β 1/2 ve = MgRp 1/2 RT [erfc(z) = − erf(z)] MOLECULES IN MOTION 403 For H2 on Earth at 240 K β 1/2 ve = (0.002016 kg mol−1 ) × (9.807 m s−2 ) × (6.37 × 106 m) 1/2 = 7.94 (8.314 J K−1 mol−1 ) × (240 K) P = erfc(7.94) + 7.94 π 1/2 e−(7.94) = (2.9 × 10−29 ) + (3.7 × 10−27 ) = 3.7 × 10−27 at 1500 K β 1/2 ve = (0.002016 kg mol−1 ) × (9.807 m s−2 ) × (6.37 × 106 m) 1/2 = 3.18 (8.314 J K−1 mol−1 ) × (1500 K) P = erfc(3.18) + 3.18 π 1/2 e−(3.18) = (6.9 × 10−6 ) + (1.46 × 10−4 ) = 1.5 × 10−4 For H2 on Mars at 240 K β 1/2 ve = (0.002016 kg mol−1 ) × (3.76 m s−2 ) × (3.38 × 106 m) 1/2 = 3.58 (8.314 J K−1 mol−1 ) × (240 K) P = erfc(3.58) + 3.58 π 1/2 e−(3.58) = (4.13 × 10−7 ) + (1.10 × 10−5 ) = 1.1 × 10−5 at 1500 K, β 1/2 ve = 1.43 P = erfc(1.43) + (1.128) × (1.43) × e−(1.43) = 0.0431 + 0.209 = 0.25 For He on Earth at 240 K β 1/2 ve = (0.004003 kg mol−1 ) × (9.807 m s−2 ) × (6.37 × 106 m) 1/2 = 11.19 (8.314 J K−1 mol−1 ) × (240 K) P = erfc(11.2) + (1.128) × (11.2) × e−(11.2) = + (4 × 10−54 ) = × 10−54 at 1500 K, β 1/2 ve = 4.48 P = erfc(4.48) + (1.128) × (4.48) × e−(4.48) = (2.36 × 10−10 ) + (9.71 × 10−9 ) = 1.0 × 10−8 For He on Mars at 240 K β 1/2 ve = (0.004003 kg mol−1 ) × (3.76 m s−2 ) × (3.38 × 106 m) (8.314 J K−1 mol−1 ) × (240 K) 1/2 = 5.05 P = erfc(5.05) + (1.128) × (5.05) × e−(5.05) = (9.21 × 10−13 ) + (4.79 × 10−11 ) = 4.9 × 10−11 at 1500 K, β 1/2 ve = 2.02 P = erfc(2.02) + (1.128) × (2.02) × e−(2.02) = (4.28 × 10−3 ) + (0.0401) = 0.044 For O2 on Earth it is clear that P ≈ at both temperatures INSTRUCTOR’S MANUAL 404 For O2 on Mars at 240 K, β 1/2 ve = 14.3 P = erfc(14.3) + (1.128) × (14.3) × e−(14.3) = + (2.5 × 10−88 ) = 2.5 × 10−88 ≈ at 1500 K, β 1/2 ve = 5.71 P = erfc(5.71) + (1.128) × (5.71) × e−(5.71) = (6.7 × 10−6 ) + (4.46 × 10−14 ) = 4.5 × 10−14 Based on these numbers alone, it would appear that H2 and He would be depleted from the atmosphere of both Earth and Mars only after many (millions?) years; that the rate on Mars, though still slow, would be many orders of magnitude larger than on Earth; that O2 would be retained on Earth indefinitely; and that the rate of O2 depletion on Mars would be very slow (billions of years?), though not totally negligible The temperatures of both planets may have been higher in past times than they are now In the analysis of the data, we must remember that the proportions, P , are not rates of depletion, though the rates should be roughly proportional to P The results of the calculations are summarized in the following table 240 K P (Earth) P (Mars) H2 He O2 H2 1500 K He 3.7 × 10−27 1.1 × 10−5 × 10−54 4.9 × 10−11 0 1.5 × 10−4 0.25 1.0 × 10−8 0.044 O2 4.5 × 10−14 ... INSTRUCTOR S MANUAL 386 but now the positive charge is located one molecule to the right of its initial location (Fig 24.19c) According to this model, there is no coordinated motion of a proton... Pa (1.0 Torr) × 1.013×105 Pa atm−1 760 Torr atm−1 = 3.7 × 10−9 m = 5.5 × 10−8 m = 4.1 × 10−5 m INSTRUCTOR S MANUAL 388 E24.10(b) The fraction F of molecules in the speed range from 200 to 250... is given by Newton’s law of universal gravitation, which is Gmm r2 F = G is the gravitational constant (not to be confused with g) Then w = ∞ r Gmm Gmm dr = r r2 Since according to Newton’s second