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2003年第35届ICHO预备试题答案(英文)1-33



Problem 1: Proton – antiproton atom We can use the expression for hydrogen-like atoms to calculate the energy levels. 2πZ 2 e 4 ? En = ? where Z is the total number of charges in the nucleus

(= 1), e is (4πε 0 )2 h 2 n 2 the electron charge (= 1.6022 x 10 –19 C), μ is the reduced mass of the system with
1 ?1 ? = m1 + m? and m1 = m2 = mp = 1.6726 x 10-27 kg, thus μ = ? mp, ε0 is the 2 permittivity of vacuum (= 8.8542 x 10-12 C2 J-1 m-1) [note: (4πε0)-1 is Coulomb's constant], h is the Planck constant (= 6.626076 x 10-34 J s) and n is the principal quantum number of the system taking values 1, 2, …

(

)

?1

? e2 ? E1 ?15 For n = 1, E 1 = ? m p ? ?4hε ? ? = ?2.00129 x10 J and for n = 2, E 2 = 4 , hence 0 ? ? 3 ?E ≡ E 2 ? E1 = E1 = 1.50097 x10 ?15 J 4 ?E 1.50097 x10 ?15 J ΔE = h ν ,hence ν = = = 2.2652 s ?1 . ?34 h 6.626076x10 J s c 2.997925x10 8 m s ?1 c = λ ν, hence λ = c / ν = λ = = = 1.3234 x10 ?10 m = 1.3234 ?. ν 2.2652 s ?1 The Bohr radius is given by h 2 (4πε 0 ) 2h 2 ε 0 α= = = 5.76397 x10 ?14 m which is 1836/2 times smaller than the 2 2 2 4π ?e πm p e hydrogen radius due to the difference in reduced mass of the "atom". Problem 2: Annulene The number of π electrons is 18. Two electrons can occupy each state due to the Pauli exclusion principle. Each state above the lowest is two-fold degenerate. Based on these pieces of information, we can fill this table: total e- up N max number of e per state to this state 0 2 2 1 4 6 2 4 10 3 4 14 4 4 18 States up to N = 4 are fully occupied with 18 electrons. The lowest possible transition is from state N = 4 to state N = 5. The path forming a circular well is L = 18 x 1.4 ?. Hence 2 x (18x1.4 ? ) x 9.109389x10 kg The transition frequency is ν = ΔE/h = 5.1544 x 1014 s-1 and the corresponding wavelength is λ = c/ν = 581.6 nm.
2 ? 31

2

?E = E 5 ? E 4 = 5 ? 4
2

(

2

)

(6.6260755x10

?34

Js

)

2

= 3.415x10 ?19 J

Problem 3: Chemical bonding: The molecular ion O 2+ 2 1. 171.9 kcal/mol 2. No 42

3. Yes 4. 85.6 kcal/mol 5. (171.9 ? 85.6) kcal/mol = 86.3 kcal/mol or (86.3/NA) kcal/molecule = 1.43 x 10-23 kcal/molecule. 6. ~1.1 ? 7. ~1.6 ? Problem 4: Electrochemistry: Nicad batteries
0 Ec 2Ni(OH)2 (s) + 2OH? 2NiO(OH) (s) + 2H2O + 2e? = ?0.490 V 2 0 RT Ec = Ec ? ln ?OH ? ? ? 2F ? 0 Ea Cd (s) + 2OH? Cd(OH)2 (s) + 2e? = +0.809 V 1 0 RT Ea = Ea ln ? 2 F ?OH ? ? 2 ? ? disch arg e ? ? ? → 2Ni(OH)2 (s) + Cd(OH)2 (s) Cd (s) + 2NiO(OH) (s) + 2H2O ? ← ? ? ? ch arg e 0 0 E = Ea – Ec = Ea ? Ec = 0.809 V – (-0.490 V) = 1.299 V. 700 mAh = 0.700 A x 3600 s = 2520.0 C 2520.0 2520.0 C → moles of Cd = 0.013 moles of Cd → 0.013x112.4 = 1.47 g of 2 ? 96485 Cd.

Problem 5: Boiler Tank capacity: m = V ρ = 4 m3 x 0.73 g cm-3 = 2920 kg Heating power P = 116 kW m P 116 kJ s ?1 Consumption rate = = = 2.70 x 10 -3 kg s ?1 = 9.73 kg h ?1 7 ?1 ?h t 4.3x10 J kg m m 2920 kg = = 1.08 x 10 8 s = 300 h = 12.5 days Operation duration t = -3 ?1 m 2.70 x 10 kg s t CnH2n+2 + ?(3n+1) O2 → nCO2 + (n+1)H2O m CO 2 44n 22 = = This ratio does not depend heavily on n; for a 1 m C n H 2 n + 2 14n + 2 7+ n representative value of n = 10, the ratio takes the value of 3.1. m Cn H2n +2 Since = 9.73 kg h -1 , t m C n H 2 n + 2 m CO 2 m CO 2 = = 9.73 kg h -1 x3.1 = 30.2 kg h -1 t m Cn H 2n +2

(

)

Problem 6: Ammonium nitrate Mixing is endothermic and the process is adiabatic, thus heat has to be provided by the solution itself. Since water is at its freezing point, it will tend to freeze, but the solution created will experience a depression of freezing point due to the presence of

43

dissolved ions. The large amount of heat required for solvation will necessitate some freezing of water. A Hess cycle of 3 steps will be considered. A. mixing at 0 °C with ΔH1 > 0 B. lowering of the temperature of the mixture to its final temperature with ΔH2 < 0 C. freezing of some water ms with ΔH3 < 0 The final temperature is given by θ 2 = ?K f 2
n m ? ms

where Kf is the cryoscopy

constant, 2 is the number of particles per formula weight for NH4NO3, n the number 80 g of moles of NH4NO3 n = = 1 mol , m = 1000 g. 80 g mol ?1 ΔH1 = Δhs n = -25.69 kJ mol-1 x 1 mol = -25.69 kJ m ?H 2 = c P θ 2 where cP is the molar heat capacity of water and M its molecular M mass (18 g mol-1). m ?H 3 = ? ?h f s where Δhf is the molar enthalpy of fusion. M ΔH1 + ΔH2 + ΔH3 = ΔHtotal = 0 because no heat is allowed to be exchanged between the system and its surroundings. Substituting for θ2 and solving for ms yields the following expression ? m ?h s nM ? 2nK f c P m m ?h s nM ? ms = + ± ? ? + ? 2 ?h 2 ? 2 ?h f 2 ?h f f ? ? We discount the solution derived from the + sign as unphysical (ms > m) and arrive at the result ms = 28.52 g ice. Hence θ2 = -3.83 °C. If we made the simplification that we expect ms << m, then θ2 is immediately calculated as –3.72 °C, which yields a value for ms = 29.9g. If this result is used to improve the θ2 value using the exact expression, we get θ2 = -3.83 °C. Then, ms can be further improved to 28.5 g. The process is spontaneous, irreversible, one where separation of components is possible, adiabatic, isobaric, isenthalpic, nearly isoenergetic. The equation ΔG = ΔH - T ΔS can be used here because T varies less than 2%. ΔG < 0 because the process is spontaneous and ΔH = 0, hence ΔS > 0. This is also to be expected from stability criteria under the constraint ΔH = 0. Problem 7: Carbon dioxide A rather accurate phase diagram for CO2 is shown here using data from LandoltB?rnstein New Series (Pc,Tc) IV/20B, p. 22 which 6000 has been based on Dykyi, J. Repas, M.: solid liquid 4000 Saturated Vapor (P3,T3) Pressure of Organic Compounds, Bratislava, 2000 gas Czech.: Slovakian Academy of Science, 0 1979.
100 150 200 250 Temperature (K) 300
2

Pressure (kPa)

44

A qualitative one can be drawn based on the triple and the critical points. Since the room temperature is well above the triple point, there is no way there can be any solid CO2 in the fire extinguisher. At 298 K, the CO2 vapor pressure is 63.1 bar. This value can be estimated by drawing a straight line between the triple and the critical points. It can be calculated also based on the empirical Antoine equation, viz., P B log =A? with A = 6.46212, B = 748.28 and C = -16.9. kPa C+T K Problem 8: Iron crystal (a) Let R be the atomic radius of iron and a =2.87 ? the length of the unit cell edge. Then, as atoms touch each other along the body diagonal and from a Pythagorean theorem in the cube: a 3 =4R ? R = (a 3 )/4 ? R = 1.24 ? The Avogadro number (NA) can be calculated from the density (ρ) formula. The latter is obtained by finding the number of atoms per unit cell, multiplying this number by the mass of each atom mole of atoms (g / mol) ( ) and, N A (atoms / mol) eventually, dividing the result by the volume of the unit cell (a3). Note that each bcc unit cell contains two whole spheres, that is 2 Fe atoms. 2 × (55.847 / N A ) g ? ρ(g/cm3) = a 3 (cm 3 ) 2 × 55.847 g NA = ? NA ≈ 6.01×1023 g 7.86 × (2.87 × 10 ?8 ) 3 cm 3 3 cm (b) By applying the Pythagorean theorem in the cube, one finds: a2 + a2 = (4R)2 ? a = 2R 2 ? R = a 2 ? R = 1.27 ? (slightly different from the 4 value found above for bcc structure, because of the different packing, having an influence on the atomic radius or at least its estimation). As for the density, recalling that each fcc unit cell contains four whole spheres, that is 4 Fe atoms, once again one has: 4 × (55.847 / N A ) g ρ(g/cm3) = = a 3 (cm 3 )

45

4 × 55.847 g = 8.02 g/cm3 3 ?8 3 6.023 × 10 × (3.59 × 10 ) cm The higher value of γ-Fe density, as compared with α-Fe, points at the fact that the fcc structure is denser than bcc. fcc represents the, so called, cubic close packed structure which, together with the hexagonal close packed, are the most efficient ways of packing together equal sized spheres in three dimensions. (c) and (d) The unit cells below are illustrated by using reduced size spheres. Note that, in hard spheres packing model the represented atoms must be in contact one to each other.
23

According to the left figure, a perfectly fitted interstitial atom centered at (?, 0, ?) in an α-Fe cell, would have a radius of:

Rinterstitial = ? a - RFe, where a = 2.87 A and RFe = 1.24 ? [(see question (a)]. Therefore: Rinterstitial (α-Fe) ≈ 0.20 ?
Similarly, according to the figure in right, a perfectly fitted interstitial atom centered at (?, ?, ?) in an γ-Fe cell, would have a radius of:
! !

!

Rinterstitial = ? a - RFe, where a =3.59 A and RFe= 1.27 A Therefore: Rinterstitial (γ-Fe) ≈ 0.53 ?
(e) 1 nm = 10 ?. Thus: For α-Fe: R carbon 0.77 A = = 3.85 ! R int erstitial 0.20 A
! !

[(see question (b)].

R carbon 0.77 A = 1.45 For γ-Fe: = ! R int erstitial 0.53 A Therefore, the carbon atom is roughly four times too large to fit next to the nearest iron atoms in α-Fe without strain, while it is only 1.5 times oversize to fit in the γ-Fe structure. The above estimations explain well the low solubility of carbon in α-Fe (< 0.1 %). (f) The wavelength (λ) of the X-rays will be calculated from Bragg's law, assuming first order diffraction: 2d sinθ = λ, where θ is the angle of diffraction equal to 32.6o and d is the interplanar spacing of the (200) set of parallel lattice planes, that is, the perpendicular distance between any pair of adjacent planes in the set. The (200) planes are shown

46

shaded in the figure. Let a be the length of the cubic unit cell edge. Then, from previous data for α-Fe: a = ! a = 1.44 ? Therefore, 2.87 A , so the distance between adjacent (200) planes is: d = 2 from Bragg's law: λ = 2d sinθ = 2×1.44×sin(32.6o) ? λ ≈ 1.55 ? This value corresponds to the Kα1 transition of iron. Problem 9: Cyclodextrine a) V = a x c . b = a b c sin(β) = 7474 ?3, Vm= 7474 ?3 /4 = 1868 ?3 b) FW = 1535.4 g mol-1, ρ = FW 4/(V NA), hence ρ = 1.3646 g cm-3 Problem 10: Infrared spectroscopy 1. The number of vibrational modes is given by 3N-6 for non-linear and 3N-5 for linear molecules, where N is the number of atoms in the molecule. Hence CO: 1, H2O: 3, C6H6: 30, C60: 174 2. The fact that the molecules have a vibrational infra red absorption implies that the molecules have a permanent dipole moment, hence these diatomic molecules must be heteronuclear. For a simple harmonic oscillator-type diatomic molecule, the eigen frequency is given 1 k by the equation ν = , where k is the force constant and μ the reduced mass of 2π ? the molecule. In the absence of any further information, nothing can be said about the reduce masses or the force constants (though bond strength is not related to k, but more likely to bond dissociation energy). The eigen frequency is equal to the frequency of the absorbed photons because the vibrational energy is given by Ev = (v+?) h ν and the energy for the resonance transition is ΔE = Ev=1 – Ev=0 = h ν. Problem 11: Radioactivity and chemical reactivity 1. 2. 3. 4. 5. γ-rays are a form of high energy electromagnetic radiation. There are no stable (non-radioactive) isotopes beyond Bi. There are numerous light isotopes that are radioactive. Xe compounds such as XeF2 are commercially available. Cs is the element with the lowest ionization potential (3.89 eV).

Problem 12: Carbon dating a) Let N0 be the 14C/12C ratio in living systems and N the same ratio found in a sample coming from a system that died t years ago. Then, the following relation between them is true: N = N 0 e ? λt , where λ (= ln 2 / t?) the disintegration constant for 14C. The above equation becomes N ? ln N0 t N ln 0.25 t= = ? ? ln = ?5700 y = 11400 y λ ln 2 ln 2 N 0 b) The β decay scheme is based on the nuclear reaction n → p + β + ν e , where p is a proton and ν e an electron antineutrino. In the case of 14C we have 14 C → 14N + β- + ν hence C becomes a (common) 14N atom. c) If an organic molecule contains 14C, the consequence of its disintegration can be grave for the structure of the molecule, causing great damage to the molecule in 47

the vicinity of the 14C atom. At least the chemical bond is raptured since 14N is a chemically different atom than 14C. Free radicals may also be created. d) The total carbon inside a human body of 75 Kg is 75 kg x 0.185 = 13.9 kg. The total radioactivity (R) is R = 0.277 Bq / g x13.9 kg = 3850 Bq The amount of 14C present is estimated from the total radioactivity as follow: dN R≡? = λN dt Then t 5700 y 60 x 60 x 24 x 365.25 s A N = = A ? = 3850s ?1 = 1.00 x1015 atoms = 1.66 nmol λ ln 2 0.693 y Problem 13: Uranium (ΔΑ = -4, ΔΖ= -2) a) alpha decay: X(A,Z) → X(A-4, Z-2) + 4He2+(2p+2n) (ΔΑ = 0, ΔΖ = +1) beta decay: X(A,Z) → X(A, Z+1) + β + ν e Since changes in the mass number (ΔA) are due to the emission of alpha particles only, in each series we have: total alpha particles emitted = ΔAtotal/4 Alpha emission also changes the atomic number (Z) (ΔZ = -2), so the total decrease in Z due to the total alpha particles emitted would be twice their total number. But Z of the final (stable) element of the radioactive series is higher than the expected Z based on alpha emission. This difference in Z is due to the number of beta particle emitted. Thus 238 U → 206Pb, α = ΔA/4 = (238-206)/4 = 32/4 = 8, β = 2α - ΔΖ = 18 -(92-82) = 6 235 U → 207Pb, α = ΔA/4 = (235-207)/4 = 28/4 = 7, β = 2α - ΔΖ = 14-(92-82) = 4 b) This occurs when an alpha decay (ΔZ = -2) is followed by two successive beta decays (ΔZ = +2). c) For each radioisotope of uranium we can write 235 Ν = 235Ν0 exp(-λ235 t) and 238Ν = 238Ν0 exp(-λ238 t) where N the number of nuclei at time t, N0 at time t = 0 and λ = ln 2 / t? = 0.693 / t? the disintegration constant. At t = 0, 235N0 = 238N0, then exp(? λ 238 t ) 238 N 99.3 = = = 142 exp(? λ 235 t ) 235 N 0.7 Thus λ235 t - λ238 t = ln 142 = 4.95 λ238 = 0.693/t? = 0.693/4.5x109 = 1.54x10-10 y-1 λ235 = 0.693/t? = 0.693/7.1x108 = 9.76x10-10 y-1 4.95x1010 y 4.95 t= = = 6.0 x10 9 y 8.22 (9.76 ? 1.54)x10 ?10 y ?1 d) The energy released by the complete fission of 1g 235U is E = (1/235) x 6.022 x 1023 x 200 MeV = 5.13 x1023 MeV and the energy released upon combustion of 1 g C is E = (1/12) x 6.022 x 1023 x 4.1 eV = 2.06 x 1023 eV = 2.06x1017 MeV Hence the amount of carbon that would release the same amount of energy as the fission of 1 g 235U is m = (5.13x1023)/(2.06x1017) = 2.49 x 103 kg C

48

Problem 14: Lead iodide 1. The graph obtained is one of two straight lines, meeting at a peak of about 2.50 g Pb(NO3)2. Mass of PbI2(s) in g 5.00

_ _ _ _ _ *
1.00

4.00

Data according to the reaction 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) Mass of Pb(NO)3 (g) 0.500 1.000 1.500 4.000 (1.000 g KI) 3.000 (2.000 g KI) Mass of PbI2 (g) 0,696 1.392 2.088 1.389 2,778

3.00

.
* *

2.00

.
_ _ _
2.00

1.00

2. The total quantity of reactant is limited to 5.000 g. If either reactant is in excess, the amount in excess will be “wasted”, because it cannot be used to form product. Thus, we obtain the maximum amount of product when neither reactant is in excess; there is a stoichiometric amount of each. The balanced chemical equation for this reaction, 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) shows that stoichiometric quantities are two moles of KI (166.00 g/mol) for each mole of Pb(NO3)2 (331.21 g/mol). If we have 5.000 g total, we can let the mass of KI equal x g, so that the mass of Pb(NO3)2 = (5.000 – x) g. Then we have 1 mol KI x amount KI = x g KI × = 166.00 g KI 166.00 1 mol Pb(NO3) 2 5.000 - x amount Pb(NO3)2 = (5.000 – x) g Pb(NO3)2 × = 331.21 g Pb(NO3)2 331.21 At the point of stoichiometric balance, amount KI = 2 × amount Pb(NO3)2 x 5.000 - x =2× or 331.21x = 10.00 × 166.00 – 332.00x 166.00 331.21 1660.0 1 mol KI = 2,503 g KI × x= = 0.01508 mol KI 331.21 + 332.00 166.00 g KI 1 mol Pb(NO3) 2 = 0.007539 mol Pb(NO3)2 5.000 – x = 2.497 g Pb(NO3)2 × 331.21 g Pb(NO3) 2 To determine the proportions precisely, we use the balanced chemical equation. 49

_

_

3.00

4.00

5.00

Mass of Pb(NO3) 2(aq) in g

maximum PbI2 mass = 2.503 g KI × 3.476 g PbI2 Problem 15: Octahedral complexes d1 : tg1eg0 d2 : tg2eg0 d3 : tg3eg0 d4 : tg4eg0 (Δ > P) or tg3eg1 (Δ < P) d5 : tg5eg0 (Δ > P) or tg3eg2 (Δ < P) d6 : tg6eg0 (Δ > P) or tg4eg2 (Δ < P) d7 : tg6eg1 (Δ > P) or tg5eg2 (Δ < P) d8 : tg6eg2 d9 : tg6eg3

1 mol PbI 2 461.0 g PbI 2 1 mol KI × × = 166.00 g KI 2 mol KI 1 mol PbI 2

Problem 16: Isomerism in Inorganic Chemistry 1. [1s22s22p63s23p6] 3d74s2 Ar ↑↓ ↑ ↑ ↑ ↑ Co3+ [Ar]3d6 2 3 2. dative covalent by the ligand into an empty metal orbital. d sp hybridization
27Co

↑↓ ↑↓

↑↓ ↑

↑↓ 3d

↑↓

↑↓

↑↓ 4s

↑↓ ↑↓ 4.
Cl

↑↓ 4p ↑↓ 4p fac
Cl

↑↓ ↑↓ ↑↓ ↑↓ mer
Cl Cl Cl H3N Cl Rh NH3

↑ ↑ ↑ ↑↓ 3d 4s outer sphere paramagnetic complex 3. trans cis
Cl NH3 H3N H3N Cl Co NH3 H3N H3N NH3 Co NH3

↑↓ 4d

↑↓

↑↓

Cl

H3N H3N

Rh

Cl

NH3

NH3

5.

2 enantiomers ± Co(en)33+
N N N N N N Co N N N Co N N N

Α

Β

50

Problem 17: Tetrahedral and square complexes 3d ↑↓ ↑↓ ↑↓ ↑ ↑ sp3: tetrahedral/paramagnetic 3d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ dsp2: square planar/diamagnetic Problem 18: Copper enzyme 1. Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1, Cu(I): 1s2 2s2 2p6 3s2 3p6 3d10, a. Cu(II): 1s2 2s2 2p6 3s2 3p6 3d9 2. Oxidised PC 3. A=ε.c.l ? c = 0.700 / (4500 x 1) = 1.56 x 10-4 mol dm-3. 5 cm3 of the solution contain 1.56 x 10-4 x 5 x 10-3 x 10500 x 1000 = 8.2 mg PC. #Cu atoms = 1.56 x 10 –4 x 5 x 10-3 x 6.0221 x 1023 = 4.7 x 1017 4. Electronic configurations : Zn(II): 1s2 2s2 2p6 3s2 3p6 3d10, Cd(II): 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10, Co(II): 1s2 2s2 2p6 3s2 3p6 3d7, Ni(II): 1s2 2s2 2p6 3s2 3p6 3d8. Redox inactive are the Zn(II) and Cd(II) reconstituted Blue Copper Proteins. Problem 19: Palladium nanoclusters 1. N = N o ? ρ ?V = 307 Pd(0) atoms per nanocluster AtomicWeight
3

4s ↑↓ 4s ↑↓

↑↓

4p ↑↓ 4p ↑↓

↑↓

↑↓

4 ?D? V = the volume of a nanocluster = π ? ? 3 ?2? 2 According to the equation y=10n +2, the number of Pd(0) atoms in a 4 full-shells nanocluster is N = 1+12+42+92+162 = 309, hence it is a full shell cluster. 2. From Fig. 4 the H2 uptake is ?PH 2 = 4.15 ? 2.05 = 2.10 atm in 184 min. ?PH 2 V = ?n H 2 RT , hence ?n H 2 = 0.029 mol where V = 400-55=345 cm3 Initially n C6H12 = Vρ 5 cm 3 x 0.81 g cm -3 = = 0.048 mol M (6 x12.0107 + 12 x1.00794 )g mol ?1 reacted moles 0.029 (i) Conversion = = = 0.60 = 60% initial moles 0.048 162 (ii) The catalytically active Pd(0) atoms are = 0.524 = 52.4% of the total Pd(0) 309 nH 0.029mol amount. So: TON = 2 = = 1106 and n Pd 0.524x 50x10 ? 6 mol TOF 1106 TOF = = = 6.0 min ?1 t 184 min 3. The spectral regions (δ / ppm) and the respective relative integrals in the 1H-NMR spectrum of hex-1-ene (Fig. 5a and Table) are assigned as follows: 1 2 3 4 5 1 2 3 4 5 0.88-0.96 1.15-1.32 1.99-2.08 5.65-5.79 4.85-4.98 CH3(CH2)2CH2CH CH2 (3) (4) (2) (1) (2)

51

The integral ratios of the second 1H-NMR spectrum (Fig. 5b and Table) suggest that both hex-1-ene and hexane are present. The differences in the integral values of the spectral regions 0.88-0.96 ppm and 1.12-1.37 ppm must be due to the presence of hexane. The relative integrals of second spectrum are converted as shown in the table below: δ / ppm relative integral Solution of the reaction 0.88-0.96 3+6 (Fig. 5b) 1.12-1.37 4+8 1.99-2.08 2 4.85-4.98 2 5.65-5.79 1 So, the spectral regions (δ / ppm) and the respective relative integrals in the 1H-NMR spectrum corresponding to hexane are assigned as follows: 1’ 2’ 1' 2' 1' 0.88-0.96 1.15-1.32 CH3(CH2)4CH3 (6) (8) Finally, comparing the integral values per proton for the hex-1-ene and hexane the % conversion of hex-1-ene to hexane, after 30 min is 50% Problem 20: Drug kinetics
k1 As ? Ab ? ?→ ?→ products

(1)

d[A]s = k 1[A]s (2) dt Integration of Eq. 2 gives [A]s = [A]o exp(-k1t), where [A]o the concentration of the drug in the stomach at zero time. ? 1.000 0.875 0.750 0.625 0.500 0.375 0.250 0.125 0.000 0.0 0.5 1.0 1.5 2.0 2.5 3.0 time (hours)
2

[A]s/[A]0

3.5

4.0

4.5

?1? = 0.25 = ? ? Since ? of the initial amount remains after ? 2? 2 one hour, (?) = 1/16 = 0.625 will remain after 2 hours, which corresponds to 4 half lives. That is 6.25% of [A]s is left.

[A]0 ? [A]s [A]0

[A]s = 0.75 ? [A]0

52

Problem 21: Br2 + CH4 reaction mechanism 1 The rate of formation of CH3Br is given by the equation: d[CH 3 Br] v=+ = k 3 [CH 3 ][Br2 ] (1) dt The “steady state” approximations for CH3 and Br are given by the equations: d[CH 3 ] = k 2 [Br][CH 4 ] ? [CH 3 ](k 3 [Br2 ] + k 4 [HBr]) = 0 (2) dt d[Br ] 2 = 2k 1 [Br2 ][M ] ? k 2 [Br ][CH 4 ] + [CH 3 ](k 3 [Br2 ] + k 4 [HBr ]) ? 2k 5 [Br ] [M ] = 0 (3) dt From equation (2): [CH 3 ]st = k 2 [Br ][CH 4 ] (4) k 3 [Br2 ] + k 4 [HBr ] From equations (3) and (4): ? k1 ?2 ? [Br]st = ? [ ] Br 2 ?k ? (5) ? 5 ? By combining equations (1), (4) and (5) the expression for the rate of formation of CH3Br as a function of the concentrations of the stable species that are involved in the reaction is given by equation (6): ? k1 v=? ?k ? 5 2 Start of the reaction Steady state condition Near to the end of the reaction 3 Start of the reaction [Br2] >> [HBr] and since k3 ≈ k4: k3 [Br2] >> k4[HBr], so k4[HBr] / k3 [Br2] << 1 Steady state condition Near to the end of the reaction [Br2] << [HBr] and since k3 ≈ k4: k3 [Br2] << k4[HBr], so k4[HBr] / k3 [Br2] >> 1 II I III ?2 [Br2 ]2 [CH 4 ] ? ? k 2 ? k [HBr ] 4 ? +1 k 3 [Br2 ]
1 1 1

(6)

53

Problem 22: Buffer solutions 1. The equilibrium, which governs the concentration of H+ within the solution is HCOO? + H+ HCOOH! [ H + ][ HCOO ? ] = 2.1× 10 ? 4 Hence K a = [ HCOOH] and since [HCOOH] ≈ 0.200 M and [HCOO?] ≈ 0.150 M 0.200 [H + ] = 2.1 × 10 -4 × = 2.8 × 10 ? 4 M 0.150 and pH = 3.55. 2. Since sodium hydroxide reacts with formic acid: HCOOH + OH? → HCOO? + H2O the concentration of formic acid in the solution is reduced to [HCOOH] = 0.200 M – 0.0100 M = 0.190 M and the concentration of formate is increased to [HCOO?] = 0.150 M + 0.0100 M = 0.160 M 0.190 Therefore: [H + ] = 2.1 × 10 -4 × = 2.5 × 10 ? 4 M 0.160 and pH = 3.60 Note that the addition of sodium hydroxide, which is a strong base, causes a very small increase of the pH of the solution. 3. Let V the volume of the solution of sodium hydroxide. Therefore, the final volume of the solution will be (100.0 + V) mL and the number of mmol of CH3COOH and OH? which are mixed are 100.0 mL × 0.150 mmol/mL = 15.00 mmol and V mL × 0.200 mmol/mL = 0.200×V mmol, respectively. From the reaction: CH3COOH + OH? → CH3COO? + H2O it is obvious that the amount of acetate produced is 0.200×V mmol and the amount of acetic acid which remains unreacted is (15.00 – 0.200×V) mmol. Hence, the concentration of each constituent in the buffer solution is: (15.00 - 0.200 × V) [CH 3 COOH] = M 100.0 + V 0.200 × V and [CH 3 COO ? ] = M 100.0 + V From the acid dissociation constant expression of acetic acid [CH 3 COO ? ][H + ] Ka = = 1.8 × 10 ?5 [CH 3 COOH] [CH 3 COO ? ] Ka = + [CH 3 COOH] [H ] 0.200 × V M 1.8 × 10 ?5 100.0 + V = and (15.00 - 0.200 × V) × ?5 M 1.0 10 100.0 + V from which V = 48.21 cm-3. 4. a 5. (i) c, (ii) b 6. (i) b, (ii) d 7. (i) c, (ii) c it can be derived 54

8. (i) b, (ii) c Problem 23: Titrations of weak acids The titration reaction is CH3COOH + OHCH3COO- + H2O a) Initial pH The pH of the solution before the titration begins, is calculated by the acid dissociation constant and the initial concentration of CH3COOH: CH3COO- + H+ CH3COOH From the acid dissociation constant expression: [CH3COO - ][H + ] Ka = = 1.8 × 10 -5 [CH3COOH] the concentration of H+ can be calculated: [H + ] = 1.8 × 10 -5 × 0.1000 = 1.34 × 10 ?3 M and pH = 2.87 b) pH after the addition of 10.00 cm3 of titrant The solution contains acetic acid and sodium acetate. Therefore it is a buffer solution. The concentration of each constituent is calculated: (50.00 cm 3 × 0.1000 M) - (10.00 cm 3 × 0.1000 M) [CH3COOH] = = 0.0667 M 60.00 cm 3 10.00 cm 3 × 0.1000 M [CH3COO - ] = = 0.01667 M 60.00 cm 3 These concentrations are then substituted into the dissociation constant expression of acetic acid for calculating the concentration of [H+]: 0.0667 [H + ] = 1.8 × 10 ?5 × = 7.20 × 10 ?5 M and pH = 4.14 0.01667 c) pH at the equivalence point At the equivalence point, all acetic acid has been converted to sodium acetate and the pH is calculated from the hydrolysis of acetate ions: CH3COO- + H2O CH3COOH + OHThe volume of titrant required for the equivalence point (Vep) is calculated: 50.00 cm 3 × 0.1000 M Vep = = 50.00 cm 3 0.1000 M and the total volume of solution is 100.0 cm3. Therefore, at this point of the titration [CH3COOH] = [OH-] and 50.00 cm 3 × 0.1000 M [CH3COO - ] = ? [OH - ] ≈ 0.0500 M 3 100.0 cm - 2 w [OH ] K 1.00 × 10 ?14 = = = 5.56 × 10 ?10 ?5 0.0500 M Ka 1.8 × 10 [OH - ] = 0.0500 × 5.56 × 10 ?10 = 5.27 × 10 ?6 M pOH = 5.28 and thus pH = 14-5.28 = 8.72 d) pH after the addition of 50.10 cm3 of titrant At this stage, all acetic acid has been converted to sodium acetate and the pH of the solution is calculated by the excess of sodium hydroxide, which has been added: (50.10 cm 3 × 0.1000 M) - (50.00 cm 3 × 0.1000 M) [OH ] = = 1.0 × 10 ? 4 M 3 100.1 cm Therefore pOH = 4.00 and pH = 10.00

55

14 12 10 pH 8 6 4 2 0 0 20 40 60
3

80

Volume of NaOH, cm

Titration curve of 0.1000 M acetic acid with 0.1000 M NaOH e) Selection of indicator Since the pH at the equivalence point is 8.72, the appropriate acid base indicator is phenolphthalein. 2. (i) b, (ii) c, (iii) a, (iv) b, (v) c, (vi) d Problem 24: Separation by extraction 1 Starting with an amount W0 of S in phase 1, after the extraction this amount is distributed between the two phases as follows: W0 = (CS)1V1 + (CS)2V2 Since D = (CS)2 / (CS)1, we have W0 = (CS)1V1 + D(CS)1V2 = (DV2 + V1)(CS)1 Therefore, after removing phase 2, the remaining amount of S in phase 1 is: V1 W1 = (CS)1V1 = W0 DV2 +V1 By repeating extraction with a fresh portion of volume V2 of phase 2, the amount W1 of S is similarly distributed. After removing phase 2, the remaining amount of S in phase 1 is: ? ? V1 V1 = W0 ? W2 = (CS)1V1 = W1 ? DV +V ? ? DV2 +V1 2 1 ? ? and so on. Therefore after n extractions with a fresh portion of volume V2 of phase 2, the remaining amount of S in phase 1 will be: ? ? V1 Wn = W0 ? ? DV +V ? ? 2 1 ? ? 2. (a) The remaining fraction of S after 1 extraction with 100 mL of chloroform is calculated using Equation 1.4: ? ? W 50 ? = 0.135, f1 = 1 = ? ? W0 ? 3.2 × 100 + 50 ? ? therefore the percentage of S extracted is 100 ?13.5 = 86.5% (b) The remaining fraction of S after 4 extractions with 25 mL of chloroform each time is similarly calculated: ? ? W 50 ? = 0.022, f4 = 4 = ? ? W0 ? 3.2 × 25 + 50 ? ?
4 1 n 2

56

therefore the percentage of S extracted is 100 ?2.2 = 97.8%. The result is indicative of the fact that successive extractions with smaller individual volumes of extractant are more efficient than a single extraction with all the volume of the extractant. 3. Using Equation 1-4 we have: ? ? 100.0 n 0.01 = ? ? 9.5 × 25.0 + 100.0 ? ? or 0.01 = 0.2963 hence n = log(0.01)/log(0.2963) = ? ? 3.78, therefore at least 4 extractions are required. 4. The equilibria involved are represented schematically as follows:
n

We have (subscripts w and o denote concentrations in aqueous and organic phase, respectively) (C HA ) o [HA]o D= = (C HA ) w [HA]w + [A ? ] w

KD = [HA]o / [HA]w and Ka = [H+]w [A?]w / [HA]w Combining all three equations we finally obtain: K D [H + ] w D= + (1.5) [H ] w + K a Last equation predicts that if [H+]w >> Ka (strongly acidic aqueous phase), then D ≈ KD (i.e. D acquires the maximum possible value) and the acid is extracted (or prefers to stay) in the organic phase. On the other hand, if [H+]w << Ka (strongly alkaline aqueous phase), we have D ≈ KD[H+]w / Ka, and because of the small value of D the acid is then extracted (or prefers to stay) in the aqueous phase. In this way, by regulating the pH of the aqueous phase, the course of extraction is shifted towards the desired direction. 5. (a) By using the previously derived Equation 1.5, we obtain the following plots of the D/KD ratio vs. pH.

(b) From these plots it is clear that at the pH region 7-8 the distribution ratio for benzoic acid will be practically 0, whereas that of phenol will acquire the maximum possible value. Therefore, phenol can be efficiently separated from an aqueous solution of both compounds by extraction with diethylether, provided that the pH of this solution has been adjusted in the range 7 to 8 (e.g. by the presence of excess of NaHCO3). 57

6. (a) The equilibria involved are represented schematically as follows:

We have the expressions (C OxH ) o [OxH]o D= = + (C OxH ) w [OxH 2 ] w + [OxH]w + [Ox ? ] w KD = [OxH]o / [OxH]w = 720 [OxH]w [H + ] w K1 = = 1×10?5 + [OxH 2 ] w [Ox ? ] w [H + ] w K2 = = 2×10?10 [OxH]w Combining all four equations, we have the sought-for expression [OxH]o D= + + [OxH ] w [H ] w K [OxH]w + [OxH]w + 2 + K1 [H ] w [OxH]o KD 1 ? + = = + [OxH]w [H ] w [H ] w K K2 + 1 + +2 +1+ K1 K1 [H ] w [H + ] w (b) Using last equation we obtain the following D-pH plot:

(c) We calculate the 1st and 2nd derivative of the denominator, i.e., [H + ] w K + + 1 + +2 F([H ]w) = K1 [H ] w K 1 whereupon we have the 1st derivative F'([H+]w) = ? +2 2 K 1 [H ] w 2K2 and for the 2nd derivative F''([H+]w) = [H + ]3 w

58

Since always F''([H+]w) > 0, then when F'([H+]w) = 0, F([H+]w) is minimum under these conditions. Consequently, the distribution ratio is maximum when K 1 ? +2 2 = 0 or [H+]w = K 1 K 2 = (1 × 10 ?5 )(2 × 10 ?10 ) = 4.5×10?8 M K 1 [H ] or pH = 7.35 Problem 25: Mass spectroscopy 1 The ionic fragment SiCl2+ will be represented by the following peaks: 28 35 M = 98 Si Cl2+ 29 35 M+1 = 99 Si Cl2+ M+2 = 100 28Si 35Cl 37Cl+ + 30Si 35Cl2+ M+3 = 101 29Si 35Cl 37Cl+ M+4 = 102 30Si 35Cl2+ + 28Si 35Cl 37Cl+ M+5 = 103 29Si 37Cl2+ M+6 = 104 30Si 37Cl2+ Therefore, the correct answer is 7. 2 The expected peaks and the corresponding probabilities are: 10 35 m/z = 45 B Cl : 0.199×0.7577 = 0.151 11 35 B Cl : 0.801×0.7577 = 0.607 m/z = 46 10 37 m/z = 47 B Cl : 0.199×0.2423 = 0.048 11 37 m/z = 48 B Cl : 0.801×0.2423 = 0.194 Hence, the base peak has nominal mass M = 46 and the relative intensities are: M?1 = 45 (0.151/0.607)×100 = 24.9% M = 46 = 100% M+1 = 47 (0.048/0.607)×100 = 7.9% M+2 = 48 (0.194/0.607)×100 = 32.0% Therefore, the correct answer is C. 3 For the ion N2+ we have: 14 14 M: N N = (0.99634)2 = 0.9927 M+1: 14N 15N + 15N 14N = 2×(0.99634×0.00366) = 0.007293 hence, (M+1)/M = 0.007293/0.9927 = 0.00735 or 0.735% For the ion CO+ we have: 12 16 M: C O = 0.989×0.99762 = 0.9866 12 17 M+1: C O + 13C 16O = (0.989×0.00038) + (0.011×0.99762) = 0.01135 hence, (M+1)/M = 0.01135/0.9866 = 0.0115 or 1.15% For the ion CH2N+ we have: 12 1 M: C H2 14N = 0.989×(0.99985)2×0.99634 = 0.9851 13 1 M+1: C H2 14N + 12C 1H 2H 14N + 12C 2H 1H 14N + 12C 1H2 15N = 0.011×(0.99985)2×0.99634+2×0.989×0.99985×0.00015×0.99634+ +0.989×(0.99985)2×0.00366 = 0.01487 hence, (M+1)/M = 0.01487/0.9851 = 0.0151 or 1.51% For the ion C2H4+ we have: 12 C2 1H4 = (0.989)2×(0.99985)4 = 0.9775 M: M+1: 13C 12C 1H4 + 12C 13C 1H4 + 12C2 2H 1H3 + 12C2 1H 2H 1H2 + + 12C2 1H2 2H 1H +12C2 1H3 2H = 2×0.011×0.989×(0.99985)4 + + 4×0.989×0.00015×(0.99985)3 = 0.02234 hence, (M+1)/M = 0.02234/0.9775 = 0.0229 or 2,29% Therefore the correct answer is (b) CO+

59

Problem 26
H CH3CH2CH2

Chemical Structure and Absolute Stereochemistry of Coniine
1. CH3I (Excess) H H CH3CH2CH2 OH CH3 CH3 + N
-

?

N H

2. Ag2O, H2O

H CH3CH2CH2

N(CH3)2 A

1. CH3I A 2. Ag2O, H2O

H + Me CH3CH2CH2 N OHMe Me

?

H H + CHCH2CH3 H

CH2CH2CH3 1,4 octadiene

CHCH2CH3 1,5-octadiene

H CH3CH2CH2 N H

C6H5CH2OCOCl (Z-Cl) NaOH, 0 C
0

KMnO4 N Z ?

H CH3CH2CH2

ZHN CH3CH2CH2

COOH H

H2, Pd/C

H2N CH3CH2CH2

COOH H

(S)- 5-aminooctanoic acid

The KMnO4 oxidation reaction step is based on A. M. Castano, J.M. Cuerva, A. M. Echavarren, Tetrahedron Letters, 35, 7435-7438 (1994)

60

Problem 27: The chemistry and identification of flavonoids 1.

OCH3 H3CO
5 8 1 3''

O
4

1' 6'

4''

H3CO

O B OCOCH3

H3COCO
5

8

1 3''

O
4

1' 6'

4''

H3COCO

O C

2. a) down field. The 1H-NMR resonance of phenolic proton involvement in hydrogen bonding will be observed at very low magnetic field (~ 12ppm). 3.
OH

HO

OH

OH

HO E

4. 13C-NMR would be expected to show three characteristic peaks of the three different carbonyl groups.

D

OCOCH3 H3COCO
5 8 1 3''

O
4

1' 6'

4''

H3COCO

O C

61

Problem 28: Synthesis of peptides 1
C6H5 H N H2N O Dipeptide I CO2H H2N O Dipeptide II H N CO2H C6H5

H2N O

H N

CO2H H2N C6H5

C6H5 H N O Dipeptide IV

CO2H

Dipeptide III C6H5 O NH C6H5 DP-I DP-II DP-III HN O O NH

The cyclic dipeptides (diketo piperazines) must also be considered:
C6H5 HN O O NH HN O

2. Best answers are 5 and 2. 3.
O N

O N

4. Benzyl chloroformate, reagent No 4, would react easily with an amine in the following way:
base C6H5CH2OCOCl + H2NR C6H5CH2OCONHR + HCl

5. If we assume the intermediate formation of a carbonium ion, the ease of formation of such ion would parallel its stability. Electron delocalization is most extensive in case D:
H + H H

CH3O

And least effective in case A:
O N O CH2

CH3O+

MeO

O N O

CH2

In the same way the cation from B is better stabilized than the cation from C. Therefore, the order of increasing lability is: A<C<B<D.

62

Problem 29: Oleuropein hydrolysis
1.
O R O O O HO OH OH OH OH HO OH O COOCH3 R O O O OH O O COOCH3

(a) 2.
OH HO OH

(b) 3.
OH HO He Hc Hd CH2
a b

CH2 OH

The correct structure is C

63

Problem 30: Stereochemistry of the Addition Reactions to Alkenes
CH3 H a) Br H
(R)

CH3 Br H Br Br H H CH3 Yes CH3 Br Br Br H Br Br CH3 Yes Br H H H Br Br

H C CH3 C CH3

H Br H CH3 No CH3 H Br H H H CH3 No

C

H 3C b) H H3C C C

H
(S)

C

CH3 Br H3C c) H H C C Br
(R) C

CH3 H Br H H Br Br H Br CH3 Yes CH3 H H Br H

CH3 H

Br H Br

(S)

C H

Br CH3 Yes CH3

H H3C d) H

Br C (R) C C H3C H C
(S)

H H Br H H CH3 No

Br Br Br Br

H Br Br H CH3 No

Br H H Br

H

64

Problem 31: Identification of Organic Compounds
CH3 C C H H2/Pd/C CH3

CH2CH2CH3

CH3

(A)

H2/Lindlar CH3 CH CH H O3 KMnO4 CH3 O (B)

COOH H COOH CH3 I2/NaOH -CHI3

COOH H COOH OH O (C)

Problem 32: Lipases
a)
O C CH3 O CH O C CH2Cl O C CCl3 O C CF3 O C CF2CF3

(6)
b)

(5)

(4)

(3)
O

(2)

(1)

HN

H

O O O O

c)

CH3CO2CH3 (1)

CH3CO2CH2CH3 (2)

CH3CO2CH2 (3)

CH3CO2C(CH3)3 (4)

65

Problem 33: Polymers 1. The volume of the Larnax is 40.9 cm x 34.1 cm x 17.0 cm = 23.7 dm3. Consequently, the Larnax will be filled with m=V d = 23.2 kg This quantity corresponds to (23200/106) x NA Vergina Star Copolymers, or 0.0232 x NA Vergina Star Copolymers, where NA is the Avogadro number. 2. The following reaction scheme should be followed in order to synthesize the Vergina Star Copolymer:
s-BuLi + n CH2 CH CH2 CH Li

PS 8 CH2 CH Li

+ 8 LiCl

(A)

Due to the steric hindrance of the styrillithium anion, only one polymeric chain can react with each –SiCl2 group. CH3 (A) + 8 CH2 C CH CH2 Li

66

+ 8 LiCl

Due to the lower steric hindrance of the polyisoprenyllithium living ends, the reaction goes to completion.

67



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