资源描述
38th International Chemistry Olympiad * Solutions for Preparatory ProblemsProblem 1: “A brief history” of life in the universe1-1.T = 1010 / (1)1/2 = 1010 K (10 billion degrees)1-2.T = 1010 / (180)1/2 = 0.7 x 109 109 K (1 billion degrees)1-3t = 1010/(3x103)2 s = 1013 s = 3 x 105 yr1-4.t = (1010/103)2 s = 1014 s = 3 x 106 yr1-5. 100 K1-6.10 K 1-7.a - ( f ) - ( d ) - ( h ) - ( i ) - ( c ) - ( g ) - ( j ) - ( e ) - ( b ) Problem 2: Hydrogen in outer space2-1. (8 x 8.3 J K-1 mol-1 x 2.7 K)/(3.14)(10-3 kg mol-1)1/2 = 240 m s-12-2. volume of cylinder = (2)1/2 (3.14)(10-8 cm)2(2.4 x 104 cm s-1) = 1.1 x 10-11 cm3 s-12-3. collision/sec = (volume of cylinder) x (atoms/unit volume)= (1.1 x 10-11 cm3 s-1)(10-6 cm-3) = 1.1 x 10-17 s-1time between collisions = 1/(1.1 x 10-17 s-1) = 9 x 1016 s = about 3 billion yr2-4. (240 m s-1)(9 x 1016 s) = 2.2 x 1019 m (about 2,000 light yr)2-5. Speed is proportional to the square root of the temperature.(240 m s-1)(40/2.7)1/2 = 920 m s-12-6. volume of cylinder = (2)1/2 (3.14)(10-8 cm)2(9.2 x 104 cm s-1) = 4.1 x 10-11 cm3 s-1 collision/sec = (volume swept per second) x (atoms/unit volume) = (4.1 x 10-11 cm3 s-1)(1 cm-3) = 4.1 x 10-11 s-1time between collisions = 1/(4.1 x 10-11 s-1) = 2.4 x 1010 s = about 800 yrmean free path = (920 m s-1)(2.4 x 1010 s) = 2.2 x 1013 m (intergalactic space)/(interstellar space) = (2.2 x 1019 m)/(2.2 x 1013 m) = about a million2-7.very smallProblem 3: Spectroscopy of interstellar molecules3-1.100l = 2.9 x 10-3 m K l = 2.9 x 10-5 mE(photon) = hc/l = (6.63 x 10-34 J s) (3.0 x 108 m/s) /(2.9 x 10-5 m)= 6.9 x 10-21 J3-2.J: 0 1m = (12x16/28)(1.66 x 10-27 kg) = 1.14 x 10-26 kgI = mR2 = (1.14 x 10-26 kg)(1.13 x 10-10 m)2 = 1.45 x 10-46 kg m2E(01) = 2 h2/8p2I = 2(6.63 x 10-34 J s)2/8p2(1.45 x 10-46 kg m2)= 7.68 x 10-23 JE(photon) of Problem 3-1 = 6.9 x 10-21 J E(01) = 7.68 x 10-23 J Rotational excitation by the background radiation is feasible. 3-3.E(02) = 6 h2/8p2I = hc/ll = 8p2cI /6hI = mR2 = (1/2) x 1.66 x 10-27 kg(0.74 x 10-10 m)2 = 4.55 x 10-48 kg m2l = 8p2Ic /6h = 8p2 x 4.55 x 10-48 kg m2 x 3 x 108 m/s/(6 x 6.63 x 10-34 J s) = 2.71 x 10-5 mT = 2.9 x 10-3 m K /l = 2.9 x 10-3 m K / 2.71 x 10-5 m = 107 KObservation of hydrogen rotational spectra is feasible at 100 K.Problem 4: Ideal gas law at the core of the sun4-1.protons: (158 g/cm3 x 0.36)/(1.0 g/mole) = 57 mol/cm3helium nuclei: (158 g/cm3 x 0.64)/(4.0 g/mole) = 25 mol/cm3electrons: 57 + (25 x 2) = 107 mol/cm3Total: 189 mol/cm34-2.volume of a hydrogen molecule = 2 (4/3) p r3= 2 x (4/3) p x (0.53 x 10-8 cm)3 = 1.2 x 10-24 cm3hydrogen gas: V/n = RT/p = (0.082 atm L K-1 mol-1) x 300 K / 1 atm = 24.6 L/mole = 4.1 x 10-23 L/molecule = 4.1 x 10-20 cm3/molecule1.2 x 10-24 cm3 / 4.1 x 10-20 cm3 = 3 x 10-5 = 0.003 %liquid hydrogen: (2 g/mole) / (0.09 g/cm3) / (6 x 1023 molecule /mole)= 3.7 x 10-23 cm3(1.2 x 10-24 cm3)/(3.7 x 10-23 cm3) = 0.03 = 3 % solar plasma: neglect volume of electrons(4/3)(p)(1.4 x 10-13 cm)3 (1 x 57mol/cm3 + 4 x 25 mol/cm3)(6 x 1023 mol-1)= 1.1 x 10-12 = 1 x 10-10 %Volume occupied is extremely small and ideal gas law is applicable.4-3. From 4-1, we know there are 189 moles of particles/cm3.T = pV/nR = (2.5 x 1011)(1 x 10-3)/(189)(0.082) = 1.6 x 107K Problem 5: Atmosphere of the planets5-1. U Pb + 8He + 6e5-2. After almost one half-life, the molar ratio between Pb-206 and U-238 is 1. Mass ratio: Pb-206/U-238 = 206/238 = 0.87 5-3.(1/2)mve2 = GMm/R ve2 = (2GM/R) = (2)(6.67x10-11 N m2 kg-2)(5.98 x 1024 kg)/(6.37 x 106 m) ve = 1.12 x 104 m s-1 5-4.hydrogen atom: (8RT/p M)1/2 = (8)(8.3145 kg m2 s-2 mol-1 K-1)(298 K)/(3.14)(1.008 x 10-3 kg mol-1)1/2 = 2500 m s-1 (22% of the escape velocity)nitrogen molecule: 2500 m s-1 x (1/28)1/2 = 470 m s-1 (4% of the escape velocity)The fraction with speed exceeding the escape velocity is much greater for hydrogen atoms than for nitrogen molecules.5-5. a. Jupiter: large mass, low temperature, H/He retained at high pressureb. Venus: lost light elements, rich in carbon dioxide, high pressure c. Mars: small mass, rich in carbon dioxide, low pressure d. Earth: lost light elements, carbon dioxide converted to oxygen throughphotosynthesise. Pluto: very small mass, lost light elements, very low atmospheric pressure 5-6.5-7. He (4K) H2 (20K) N2 (77K) O2 (90K) CH4 (112K) Dispersion force is greater for larger molecules.Nitrogen with the triple bond has a smaller bond length than oxygen.Nitrogen also has less lone pair electrons to be involved in dispersion. Problem 6: Discovery of the noble gases6-1. In 1816 Prout published a hypothesis that all matter is composed ultimately of hydrogen. (Later, Harlow Shapley, an eminent astronomer, said that if God did create the world by a word, the word would have been hydrogen.) Prout cited as evidence the fact that the specific gravities of gaseous elements appeared to be whole-number multiples of the value for hydrogen.6-2.28 NH3 + 21 O2 + 78 N2 + Ar 92 N2 + 42 H2O + Ar6-3.(92)(2)(14.0067) + 39.948/93 = 28.1426-4.78 N2 + 21 O2 + Ar + 42 Cu 78 N2 + 42 CuO + Ar6-5.(78)(2)(14.0067) + 39.948/79 = 28.1646-6.28.164/28.142 = 1.0008(about 0.1%)6-7.4 NH3 + 3 O2 2 N2 + 6 H2O Molecular weight of pure nitrogen = (2)(14.0067) = 28.01328.164/28.013 = 1.0054The discrepancy would increase about 7-fold (0.0054/0.0008).6-8.40/29 = 1.46-9.5R/3R = 1.67translational6-10.volume of air = 1000 m3 = 106 liter(106)/22.4 = 4.5 x 104 mol of airweight of argon = (4.5 x 104)(0.01)(40) = 1.8 x 104 g = 18 kg6-11.helium- sunneon- newargon- lazykrypton- hidden xenon- stranger Problem 7: Solubility of salts7-1. AgCl(s) Ag+(aq) + Cl-(aq)Ksp = Ag+Cl- = x2 = 1.810-10 Ag+ = Cl- = 1.3410-5 M AgBr(s) Ag+(aq) + Br-(aq) Ksp = Ag+Br- = x2 = 3.310-13 Ag+ = Br- = 5.7410-7 M 7-2. In this hypothetical case, Ag+ = Cl- = 1.3410-5 M just as in 7-1. Cl-(aq)/Cl(total) = Cl-(aq)/(Cl-(aq)+AgCl(s) = (1.310-5 M)(0.200 L)/1.0010-4 mol = 0.027 = 2.7% 7-3. Similarly, Ag+ = Br- = 5.7410-7 M just as in 7-1. Br-(aq)/Br(total) = Br-(aq)/(Br-(aq)+AgBr(s) = (5.710-7 M)(0.200 L)/1.0010-4 mol = 1.110-3 = 0.11% 7-4. Assume that 1.0010-4 mol of AgCl is precipitated, and 1.0010-6 mol of Ag+ ion remains in solution. Then a portion of AgCl dissolves. Ag+ = 5.010-6 + x , Cl- = x Ksp = Ag+Cl- = (5.010-6 + x)(x) = 1.810-10 Cl- = 1.110-5 M (slightly decreased) Ag+ = 1.610-5 M (slightly increased) Cl-(aq)/Cl(total) = Cl-(aq)/(Cl-(aq)+AgCl(s) = (1.110-5 M)(0.200 L)/1.0010-4 mol = 0.022 = 2.2% Similarly, Ag+ = 5.010-6 + x , Br- = x Ksp = Ag+Br- = (5.010-6 + x)(x) = 3.310-13 x 16400 cal/mol (40.0 cal/molK) T5.7 (cal/molK) T 1200 cal/molT BCl3 BBr311-4. Like neutralization that occurs between HCl and NaOH, the reaction producing stable acid-base adduct is expected to be exothermic. The enthalpy change will be the largest for the strongest Lewis acid, BF3.11-5. BF3 BCl3 BBr3H3 = H1 + H2;- 31.7-39.5-44.5(kcal/mol) The actual order of acidity is opposite from prediction based on the electronegativity of the halides.11-6. A = BF3 H2OB = B(OH)3, C = 3 HX (3 HCl or 3 HBr)Strong Lewis acids such as BCl3 and BBr3 can activate O-H bonds in H2O molecule to produce B(OH)3 by releasing HX. Dative -bonding with lone pair electrons of O, which have a similar energy level, can stabilize B(OH)3 as explained in 11-7.11-7. Empty pz-orbital in boron can accept a dative p-bond from the lone pair electrons of fluorine, which satisfies the octet rule for boron and shortens the boron-fluorine bond distance.Since resonance structures of this kind are not possible in the adduct compounds, effective resonance will reduce the tendency for pyridine adduct formation.The ability to form dative p-bonding appears to decrease sharply in the heavier elements due to the energy differences between B and X. Resonance of this dative p-bonding should be of lesser importance in the chloride and least importance in the bromide. These resonance structures having dative p-bonding are sufficiently large so as to reverse the trend expected from the relative inductive effects and the steric effects from the adduct formations.Problem 12: Solubility equilibrium in a buffer solution12-1. 440 mL H2S in 100 mL of water = 4.4 L H2S in 1 L of water = 0.20 M 12-2. For approximation, the concentration of all anions in (5) except Cl-, which is 0.02, can be crossed out. Thus (5) becomesH+ + 2Fe2+ = Cl- = 0.020(6)Combine (2) and (3): H+2 S2-/H2S = 1.24 x 10-21 Since H2S = 0.2, one gets H+2 S2- = 2.48 x 10-22 (7)Combine (1) and (7):H+2 (8.0 x 10-19/Fe2+) = 2.48 x 10-22 H+2 = 0.031 Fe2+(8)Combine (6) and (8): 64.5 H+2 + H+ 0.02 = 0H+ = 0.0115pH = 1.94 Fe2+ = 0.0043 (43% remains in solution)Check: HS- = (9.5 x 10-8)H2S/H+ = 2.2 x 10-6 Cl- = 0.02S2- = (1.3 x 10-14)HS-/H+ = 2.5 x 10-18OH- = 8.7 x 10-13Eq. (8) shows that 10-fold decrease in H+ increases Fe2+ 100-fold.12-3.From H+2 = 0.031 Fe2+, H+ = (0.031)(1 x 10-8)1/2 = 1.76 x 10-5pH = 4.7512-4.original HOAc = 0.10 M x 100 mL = 10 mmolHenderson-Hasselbalch eq for the HOAc-OAc- buffer at pH 4.75pH = 4.75 = pK + log OAc-/HOAc = 4.74 + log OAc-/HOAcinitial Fe2+ = 0.01 M x 100 mL = 1 mmolH+ produced upon precipitation of 1 mmol Fe2+ = 2 mmolOAc- consumed by H+ produced = 2 mmollog OAc-/HOAc = 4.75 - 4.74 = 0.01Let x = original mmol OAc- (x - 2)/(10 + 2) = 100.01 = 1.02, x = 14.3 mmol OAc- = 14.3 mmol/100 mL = 0.143 M12-5.pH = 4.74 + log (0.143/0.10) = 4.90Problem 13: Redox potential, Gibbs free energy, and solubility13-1.Ag+(aq) + e Ag(s)E = 0.7996 VG = Gf(Ag(s) + Gf(e) Gf(Ag+(aq) = Gf(Ag+(aq) = FETherefore, Gf(Ag+(aq) = FE = 77.15 kJ/mol13-2.Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq)G = Gf(Ag(NH3)2+(aq) Gf(Ag+(aq) 2 Gf(NH3(aq)= 17.12 kJ 77.15 kJ (2)(26.50) kJ = 41.27 kJln Kf = = 16.65Kf = = e16.65 = 1.7 x 10713-3.AgBr(s) Ag+(aq) + Br(aq)E = (0.0713 0.7996) V = 0.7283 Vln K sp = = = 28.17K sp = Ag+ Br = e28.347 = 4.89 x 101313-4.Let us assume Ag+ Ag(NH3)2+.AgBr(s) Ag+(aq) + Br(aq)K sp = 4.89 x 1013Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq)Kf = 1.7 x 107AgBr(s) + 2 NH3(aq) Ag(NH3)2+(aq) + Br(aq) K = K sp Kf = 8.31 x 106Initial0.100 0 0Change2S+S+SEquilibrium0.100 2S S SK = = 8.31 x 106 = 2.88 x 103S = Ag(NH3)2+ = Br = 2.9 x 104 MAg+ = K sp/Br = 1.7 x 1010 M Ag(NH3)2+Thus, the solubility of AgBr is 2.9 x 104 M13-5.Br = KSP / Ag+ = 4.89 x 1013 / 0.0600 = 8.15 x 1012 E = E + ln = 1.721 + log10 = 1.065 V13-6.In order to estimate the solubility of Br2(aq), we need to calculate the Gibbs free energy of the reaction:Br2(l) Br2(aq)G?From (e), Br2(l) + 2 e 2 Br(aq)E1 = 1.065 V, G1 = 2FE1 = 2.130F VLet us first calculate E2 for the half-cell reaction:Br2(aq) + 2 e 2 Br(aq)E2, G2 = 2FE2From the Latimer diagram,BrO3(aq) + 6 H3O+(aq) + 6 e Br(aq) + 9 H2O(l)E3 = 1.441 VBrO3(aq) + 5 H3O+(aq) + 4 e HOBr + 7 H2O(l)E4 = 1.491 V2 HOBr + 2 H3O+(aq) + 2 e Br2(aq) + 4 H2O(l)E5 = 1.584 VThen, 2 BrO3(aq) + 12 H3O+(aq) + 10 e Br2(aq) + 18 H2O(l)E6 = (2 x 4E4 + 2E5)/10 = 1.5096 VSimilarly, Br2(aq) + 2 e 2 Br(aq)E2 = (2 x 6E3 10E6)/2 = 1.098 V(Note that 6 x E3 = 4 x E4 + 1 x E5 + 1 x E2)Then, G2 = 2E2 = 2.196F VFinally,G = G1 G2 = 0.066F V = 6368 J/molTherefore,Br2(aq) = K = = e2.569 = 0.077 (M)Problem 14: Measuring the ozone level in air14-1.3I- I3- + 2e-O3 + 2H+ + 2e- O2 + H2O3I- + O3 + 2H+ I3- +O2 + H2O14-2. 14-3. The absorbance is given byA = - log T = -log(Isample/ Iblank) = log (Rsample/Rblank) = log (19.4 k/12.1 k) = 0.205I3- = A /b = 0.205/(240,000 M-1cm-1)(1.1 cm) = 7.7610-7 MNumber of moles O3 = Vsample I3- = (0.01 L)(7.7610-7 mol/L) = 7.7610-9 mol14-4.The number of moles of air sampled = PV/RT = P(tsampling F)/RT= (750 torr)(30 min)(0.250 L/min)/(62.4 torrL mol-1K-1)(298 K) = 0.302 molThe concentration of O3 in ppb = (7.76 10-9 mol / 0.302 mol) 109 = 25.7Problem 15: Lifesaving chemistry of the airbag15-1.15-2.moles of N2 = PV/RT= (1.25 atm)(15L)/(0.08206 L atm K-1mol-1)(323K) = 0.7072 moles of sodium azide generate 3.2 moles of nitrogen.Weight of sodium azide needed to generate 0.707 moles of nitrogen= (2)(0.707/3.2)(65g) = 29 g15-3.4 C3H5(NO3)3 6 N2 + O2 + 12 CO2 + 10 H2OPb(N3)2 Pb + 3 N2In all three reactions, the reactants are solid or liquid with small volume. A large volume of nitrogen gas is produced. Nitroglycerin produces other gases.The nitrogen molecule has a triple bond and is very stable. Thus, the reactions are highly exothermic, so that gases produced expand rapidly. 15-4.2 NaN3+ H2SO4 2 HN3 + Na2SO415-5.NaN3 = 60g / (65g/mol) = 0.923 mol H2SO4 = 3 mol/L 0.1 L = 0.3 molHN3 =
展开阅读全文