中文翻译-第38届ICHO理论试题(中文版)答案
国际化学奥林匹克竞赛-国际化学奥林匹克竞赛-第38届ICHO理论试题(中文版)答案
1-1.The mass of a water droplet:m = V ρ = [(4/3) π r3] ρ = (4/3) π (0.5x10-6 m)3 (1.0 g/cm3)= 5.2x10-16 kg⇒10 marksAverage kinetic energy at 27o C:KE = mv2/2 = (5.2x10-16 kg) (0.51x10-2 m/s)2/2= 6.9x10-21 kg m2/s2= 6.9 x10-21 J ⇒15 marks*.The average kinetic energy of an argon atom is the same as that of a water droplet.KE becomes zero at –273 o C.From the linear relationship in the figure, KE = aT (absolute temperature)where a is the increase in kinetic energy of an argon atom per degree.a = KE/T = 6.9x10-21 J/(27+273K) = 2.3x10-23 J/K⇒25 marksS: specific heat of argon N: number of atoms in 1g of argonS = 0.31 J/g K = a x NN = S/a = (0.31 J/g K) / (2.3x10-23 J/K)= 1.4x1022 ⇒30 marksAvogadro’s number (N A) : Number of argon atoms in 40 g of argonN A = (40)(1.4x1022)= 5.6 x1023⇒20 marks2-1. ⇒ 30 marksmass of a typical star = (4/3)(3.1)(7x108 m)3(1.4 g/10-6 m 3) = 2×1033 g mass of protons of a typical star = (2×1033 g)(3/4 + 1/8) = 1.8×1033 g number of protons of a typical star = (1.8×1033 g)(6×1023/g) = 1×1057number of stellar protons in the universe = (1×1057)(1023) = 1×1080Partial credits on principles:Volume = (4/3)(3.14)radius 3×density; 4 marks 1 mole = 6×1023; 4 marksTotal number of protons in the universe = number of protons in a star ×1023; 2 marks Mass fraction of protons from hydrogen = (3/4)(1/1); 5 marks Mass fraction of protons from helium = (1/4)(2/4); 10 marks2-2. ⇒ 30 marks∆E(2→3) = C(1/4 - 1/9) = 0.1389 C λ(2→3) = 656.3 nm ∆E(1→2) = C(1/1 - 1/4) = 0.75 Cλ(1→2) = (656.3)(0.1389/0.75) = 121.5 nmNo penalty for using Rydberg constant from memory. 15 marks penalty if answered in a different unit (Hz, etc.)2-3.T = (2.9×10-3 m K)/1.215×10-7 m = 2.4×104 K ⇒ 10 marks2-4..⇒ 20 marksλ = 3 × 108 m/1.42 × 109 = 0.21 mT = (2.9 × 10-3 m K)/0.21 m = 0.014 K2-5. ⇒ 10 marks14N + 4He → (17O ) + 1HO-17, O acceptable1783-1.k des = A exp(-E des/R T)= (1x1012 s-1)(5x10-32) = 5x10-20 s-1 at T = 20 K ⇒10 markssurface residence time, τresidence = 1 / k des = 2x1019 s = 6x1011 yr ⇒20 marks(full credit for τhalf-life = ln2 / k des = 1x1019 s = 4x1011 yr)residence time = 2x1019s3-2.The distance to be traveled by a molecule: x = πr = 300 nm.k mig = A exp(-E mig/R T)= (1x1012 s-1)(2x10-16 ) = 2x10-4 s-1 at T = 20 K ⇒ 5 marksaverage time between migratory jumps,τ = 1 / k mig = 5x103 sthe time needed to move 300 nm= (300 nm/0.3 nm) jumps x (5x103 s/jump) = 5x106 s = 50 days ⇒15 marks(Full credit for the calculation using a random-walk model. In this case:t = τ (x/d) 2 = 5 x 109 s = 160 yr. The answer is still (b).)(a) (b)(c) (d) (e)10 marks3-3.k(20 K) / k(300 K) = exp[(E/R) (1/T1 - 1/T2)]= e-112 = ~ 10-49 for the given reaction ).) ⇒15 marks The rate of formaldehyde production at 20 K= ~ 10-49 molecule/site/s = ~ 10-42 molecule/site/ yr⇒10 marks(The reaction will not occur at all during the age of the universe (1x1010 yr).)rate = 10-42molecules/site/yr3-4. circle one(a) (b) (c) (a, b) (a, c) (b,c)(a, b, c)(15 marks, all or nothing)4-1.H PNumber of atoms ( 11.3 ) 1⇒ 10 marksTheoretical wt % ( 3.43 )⇒ 10 marks4-2.adenineN NN NN H H guanineNN N NO N HH HNN O N H H cytosineNN H O O thymine(10 marks on each)4-3. 7 marks each, 20 marks for threeadenineNNNNNHHguanine NN NNON HHH NNH OOthymineNNONHH cytosine NNH OOthymineguanine NN NNON HHHcytosineNNONHHcytosineNNON HHNNHOO thyminethymineNNHOONNH OOthyminethymine NNHOONNONHH cytosineadenineNNNNNHH adenineNNNNNHHadenine NNNNNHHguanineguanine NNNNON HHHNNNNONHHH4-4. 2.5 marks for each bracketadenineN NN N HNH 2guanine N NH N N HO NH 2Uracil N H NH O cytosineN H N NH 2OOHCN ( 5 ) ( 5 ) ( 4 )( 4 )H 2O ( 0 ) ( 1 ) ( 2 ) ( 1 )5-1.(20 marks)1st ionization is complete: H2SO4→ H+ + HSO4-[H2SO4] = 02nd ionization: [H+][SO42-]/[HSO4-] = K2 = 1.2 x 10-2 (1)Mass balance: [H2SO4] + [HSO4-] + [SO42-] = 1.0 x 10-7 (2)Charge balance: [H+] = [HSO4-] + 2[SO42-] + [OH-] (3)Degree of ionization is increased upon dilution.[H2SO4] = 0Assume [H+]H2SO4 = 2 x 10-7From (1), [SO42-]/[HSO4-] = 6 x 104 (2nd ionization is **plete)[HSO4-] = 0From (2), [SO42-] = 1.0 x 10-7 [5 marks]From (3), [H+] = (2 x 10-7) + 10-14/[H+][H+] = 2.4 x 10-7(pH = 6.6) [8 marks][OH-] = 10-14/(2.4 x 10-7) = 4.1 x 10-8[2 marks]From (1), [HSO4-] = [H+][SO42-]/K2= (2.4 x 10-7)(1.0 x 10-7)/(1.2 x 10-2) = 2.0 x 10-12[5 marks]Check charge balance:2.4 x 10-7≈ (2.0 x 10-12) + 2(1.0 x 10-7) + (4.1 x 10-8)Check mass balance:0 + 2.0 x 10-12 + 1.0 x 10-7≈ 1.0 x 10-7Species Concentration** x 10-12HSO4-** x 10-7SO42-** x 10-7H+** x 10-8 OH-5-2. (20 marks)mmol H3PO4 = 0.85 ⨯ 3.48 mL ⨯ 1.69g/mL ⨯ 1 mol/98.00 g ⨯ 1000 = 51.0 [5 marks]The desired pH is above p K2.A 1:1 mixture of H2PO4- and HPO42- would have pH = p K2 = 7.20.If the pH is to be 7.40, there must be more HPO42- than H2PO4-.We need to add NaOH to convert H3PO4to H2PO4-and to convert to the right amount of H2PO4-to HPO42-.H3PO4 + OH-→ H2PO4- + H2OH2PO4- + OH-→ HPO42- + H2OThe volume of 0.80 NaOH needed to react with to to convert H3PO4 to H2PO4- is:51.0 mmol / 0.80M = 63.75 mL [5 marks]To get pH of 7.40 we need:H2PO4- + OH-→ HPO42-Initial mmol 51.0 x 0Final mmol 51.0-x 0 xpH = p K2 + log [HPO42-] / [H2PO4-]7.40 = 7.20 + log {x / (51.0-x)}; x = 31.27 mmol [5 marks]The volume of NaOH needed to convert 31.27 mmol is :31.27 mmol / 0.80 M = 39.09 mLThe total volume of NaOH = 63.75 + 39.09 =102.84 mL , 103 mL [5 marks]Total volume of 0.80 M NaOH (mL) 103 mL5-3. (20 marks)p K = 3.52pH = pK a + log ([A-]/[HA])[A-]/[HA] = 10(pH-pKa) [5 marks]In blood, pH =7.40, [A-]/[HA] = 10(7.40-3.52) = 7586Total ASA = 7586 +1 = 7587 [5 marks]In stomach, pH = 2.00, [A-]/[HA] = 10(2.00-3.52) = 3.02x10-2Total ASA = 1+ 3.02x10-2 = 1.03 [5 marks]Ratio of total aspirin in blood to that in stomach = 7587/1.03 = 7400 [5 marks]** ( 103Ratio of total aspirin in blood to that in stomach6-1. (5 marks)4 H2O + 4 e-→ 2 H2(g) + 4 OH- (or 2 H2O + 2 e-→ H2(g) + 2 OH-)6-2. (5 marks)2 H2O → O2 + 4 H+ + 4 e-(or H2O → 1/2 O2 + 2 H+ + 2 e- )6-3. (5 marks)Cu → Cu2+ + 2e-6-4. (20 marks)Reduction of sodium ion seldom takes place.It has a highly negative reduction potential of –2.710 V.Reduction potential for water to hydrogen is negative (water is very stable).But, it is not as negative as that for sodium ion. It is –0.830 V.Reduction of both copper ion and oxygen takes place readily and the reduction potentials for both are positive.In the present system, the reverse reaction (oxidation) takes place at the positive terminal. Copper is oxidized before water.Reduction potential for hydrogen ion is defined as 0.000 V.6-5. (15 marks)pOH = 14.00 – 4.84 = 9.16[OH-] = 6.92 x 10-10K sp = [Cu2+][OH-]2 = 0.100 x (6.92 x 10-10) = 4.79 x 10-206-6.E = E o Cu2+/Cu + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log (K sp / [OH-]2)= +0.340 + (0.0592/2) log (K sp) - (0.0592/2) log [OH-]2= +0.340 + (0.0592/2) log (K sp) - 0.0592 log [OH-],3 marksBy definition, the standard potential for Cu(OH)2(s) + 2e-→ Cu(s) + 2OH- is the potential where [OH-] = 1.00.E = E o Cu(OH)2/Cu = +0.340 + (0.0592/2) log (K sp)= +0.340 + (0.0592/2) log (4.79 x 10-20)= +0.340 - 0.5722 marks= -0.232 V10 marks-------------------------------------------------------------------------------------------------------------- One may solve this problem as following.Eqn 1: Cu(OH)2(s) + 2e -→ Cu + 2OH-E+o = E o Cu(OH)2/Cu = ?Eqn 2: Cu(OH)2(s) → Cu2+ + 2OH-E o = (0.05916/n) logK sp= (0.05916/2) log(4.79×10-20)= -0.5715 V3 marksEqn 1 – Eqn 2 : Cu2+ + 2e-→ CuE-o = E+o - E o = E o Cu2+/Cu = 0.34 VTherefore, E+o = E-o + E o = + 0.34 + (-0.5715)2 marks= -0.232 V10 marks-0.232 V6-7.Below pH = 4.84, there is no effect of Cu(OH)2 because of no precipitation.Therefore,E = E Cu2+/Cu = +0.340 + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log 0.1003 marks= +0.340 – 0.0296 = +0.310 V7 marks** V6-8.** g graphite = 0.0833 mol carbon6 mol carbon to 1 mol lithium; 1 g graphite can hold 0.0139 mol lithiumTo insert 1 mol lithium, 96487 coulombs are needed.Therefore, 1 g graphite can charge 96487 × 0.0139 = 1340 coulombs. 5 marks1340 coulombs / g = 1340 A sec / g = 1340 x 1000 mA × (1 / 3600) h = 372 mA h / g 5 marks372 mA h / g7-1. (10 marks)n/V = P/RT = (80 x 106 / 1.013 x 105 atm)/[(0.082 atm L/mol/K)(298K)] = 32 mol/L5 marksdensity = mass/volume = d = 32 x 2 g/L = 64 kg/m 3 5 marks64 kg/m 37-2.** or 0.23H 2(g) + 1/2 O 2(g) → H 2O(l); ∆H rexn-1 = ∆H f [H 2O(l)] = -286 kJ/mol = -143 kJ/g 7 marksC(s) + O 2(g) → CO 2(g); ∆H rexn-2 = ∆H f [CO 2(g)] = -394 kJ/mol = -33 kJ/g 7 marks(-∆H rexn-1) / (-∆H rexn-2) = 4.3 or (-∆H rexn-2) / (-∆H rexn-1)= 0.236 marks7-3. (a) (-)1.2 x 105 kJ, (b) (-)6.9 x 104 kJ** x 108 sec or 3.3 x 104 hr or 1.4 x 103 days or 46 month or 3.8 yrI = 0.81 AH 2(g) + 1/2 O 2(g) → H 2O(l)∆H c = -286 kJ/mol = -143 kJ/g = -143 x 103 kJ/kg 5 marksΔG = ΔH – T ΔSΔS c= 70 – 131 – 205/2 = -163.5 J/K/mol5 marksΔG c = -286 kJ/mol + 298K x 163.5 J/K/mol = -237 kJ/mol = -1.2 x 105 kJ/kg 5 marks(a) electric motor W max = ΔG c ⨯ 1 kg = - 1.2 x 105 kJ 5 marks (b) heat engine W max = efficiency x ∆H c 5 marks= (1 – 298/573) x (-143 x 103 kJ) = -6.9 x 104 kJ 5 marks119 x 103 kJ = 1 W x t(sec)t = 1.2 x 108 sec = 3.3 x 104 hr = 1.4 x 103 days = 46 month = 3.8 yr 5 marksΔG = -nFE n = # of electrons involved in the reaction F = 96.5 kC/molH 2(g) + 1/2 O 2(g) → H 2O(l) n = 2 5 marksE = - ΔG/nF = 237 kJ/mol / 2 / 96.5 kC/mol = 1.23 V5 marksI = W/E = 0.81 A5 marks8-1-1. (5 marks on each)①C②C③CO8-1-2.③ Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) 5marks① C(s) + O2(g) → CO2(g) ΔH①◦ = -393.51 kJ = ΔH f◦(CO2(g))② CO2(g) + C(s) → 2CO(g) ΔH②◦ = 172.46 kJFrom ① and ②,ΔH f◦(CO(g)) = (1/2){172.46 + (-393.51)} = -110.525 kJΔH f◦(Fe2O3) = -824.2 kJΔH③◦ = 3ⅹΔH f◦(CO2(g)) - ΔH f◦(Fe2O3) - 3ⅹΔH f◦(CO(g))= 3ⅹ(-393.51) – (-824.2) - 3ⅹ(-110.525) = -24.8 kJ 7 marks ΔS③°=2ⅹ27.28+3ⅹ213.74-87.4-3ⅹ197.674=15.36 J/K 3 marks ΔG③°=ΔH°-TΔS°=-24.8kJ-15.36J/Kⅹ1kJ/1000Jⅹ1473.15K=-47.43 kJ5 marksK = e(-ΔG°/RT)= e(47430J/(8.314J/Kⅹ1473.15K)) = 48 5 marksBalanced equation of ③:K = 48Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)8-2-1. (20 marks)One AB2O4 unit has available 4 (= 1 + (1/4)ⅹ12) octahedral sites.48-2-2. (20 marks)Since one face-centered cube in AB2O4 represents one Fe3O4 unit in this case, it has 8 available tetrahedral sites. In one Fe3O4 unit, 1 tetrahedral site should be occupied by either one Fe2+ (normal-spinel) or one Fe3+ (inverse-spinel). Therefore, in both cases, the calculation gives (1/8) ⅹ100% = 12.5% occupancy in available tetrahedral sites.**%8-2-3. (10 marks for d-orbital splitting, 10 marks for elec. distribution)9-1-1. 1 answer for 8 marks, two for 15 marksH 3CN NNH 3CNNN :::+_+::_:9-1-2. ( 10 marks)H 3CN::9-1-3.H 3CNCH 2CH 2:H 3CN HH CCH 2:(10 marks) (10marks )9-2-1. 5 marks eachHONN +_::ONN:H+:HH_O NN:H+:H_::::::9-2-2.( 10 marks)CH 2CO ::9-3-1.(40 marks)CH 3H 3CH 3C+BC H 2CCH 3CH 3CO 2DEOOO_9-3-2.(10 marks)O OH O n+F10-1. 10 marks eachNMLCH 2OHCH 2OHMeOOMeH HH HOMeMeO CHOCHOCH 2OHCH 2OHHHH H OHOMeMeO OH10-2. 8 marks each for correct structuresNumber of possible structures24 marks12OH(OH)OH(H)HH HHOMeOMeOH COOMeOH(OH)OH(H)HH HHOMeOMeOHCOOMe34OH(OH)OH(H)OH(OH)OHe(H)10-3. 10 marks eachGICH 2OHCH 2OHHHHHMeOOMeOHOMeCH 2OHCH 2OHHHHOMeOMeOMe10-4. 10 marksNumber of the correct structure for C from 10-2110-5.BOH(OH)OH(H)HHHH OHCOOHOHOH10 marks eachDJOH(OH)OH(H)HHHHOMeOMeCOOMeOMeOH(OMe)OMe(H)HHHHOMeOMeOMeCOOMe10-6. 20 marksHOOCOHHH OOOHOOH COOHOOHOHOH COOH11-1. 10 marks311-2. 30 marksCOOHHOOCOOH11-3. 2.5 marks eacha, c, d11-4 30 marksOOCOCOOOHTransition State11-5.For the enzyme-catalyzed reaction, Arrehnius equation could be applied.k cat/k uncat = A exp (-E a, cat/ RT) / A exp (-E a, uncat / RT)= exp [-∆E a, cat-uncat/ RT]= exp [-∆E a, cat-uncat(J/mol) / (2,480 J/mol)] = 106Therefore, -∆E a, cat-uncat = 34,300 J/mol 15 marksk uncat, T/k uncat, 298 = exp (-∆H≠ uncat/ RT) / exp (-∆H≠uncat / 298R)= exp [(-∆H≠ uncat/R)(1/T-1/298)]ln(k uncat, T/k uncat, 298 )= 13.8 = [(-86900/8.32)(1/T-1/298)]Therefore, T = 491 K, or 218o C 15 marks-E a, cat-uncat = 34,300 J/molT = 491 K, or 218o C。
IChO2016国际化学奥林匹克竞赛英文真题理论部分(包括答案)
48th InternationalChemistry Olympiad Theoretical Problems28 July 2016Tbilisi, GeorgiaInstructions∙Begin only when the START command is given. You have 5 hours to work on the problems.∙Use only the pen and calculator provided.∙The problem booklet has 23 pages, the answer sheet is 28 pages.∙Make sure that your code is on every page of the answer sheet.∙Questions are identical in the problem text and on the answer sheets.∙All results must be written in the appropriate boxes on the answer sheets.Anything written elsewhere will not be graded. Use the reverse of the problempages if you need scratch paper.∙Write relevant calculations in the appropriate boxes when necessary. If you provide only correct end results for complicated questions, you will receive noscore.∙Raise your hand if you need a restroom break.∙When you have finished the examination, put your answer sheets into the envelope provided. Do not seal the envelope.∙You can keep the problem booklet.∙You must stop your work immediately when the STOP command is given. A delay in doing this may lead to cancellation of your exam.∙Do not leave your seat until permitted by the supervisors.∙The official English version of this examination is available on request only for clarification.Constants and formulaePeriodic table with relative atomic massesNitrogen trifluoride is a surprisingly stable compound that was first prepared in the melt electrolysis of a mixture of ammonium fluoride and hydrogen fluoride.1.1.On which electrode does nitrogen trifluoride form? Write a balanced chemicalequation for the electrode half reaction for the formation of NF3.Interestingly the related fluoroamine (NH2F) and difluoroamine (NHF2) are very unstable materials; decomposition of either pure substance can even be explosive. This is dangerous as they are formed in the electrolysis as side products.1.2.Which of NF3, NHF2 or NH2F compound is expected to condense at the lowesttemperature?The N-F bond lengths in these molecules were determined to be 136, 140 and 142 pm. The change in the bond lengths can be explained with a simple electrostatic model taking into account the partial charges on the atoms.1.3.Assign the N-F bond lengths (136, 140, 142 pm) to the molecules.When NHF2 is bubbled through a solution of KF in HF, a binary nitrogen – fluorine compound can be obtained as a mixture of two geometric isomers.1.4.Write a balanced chemical equation for the formation of the binary nitrogen-fluorine compound.Tetrafluoroammonium ion (NF4+) and its corresponding salt can form from NF3 with elementary fluorine in the presence of an appropriate reagent.1.5.Propose a suitable reagent and write a balanced chemical equation for thereaction.NF4+ ions form stable salts with a number of anions. These are very sensitive to humidity, because NF4+ ion hydrolyzes forming NF3 and O2. Interestingly nitrogen trifluoride always forms quantitatively, while the quantity of oxygen is often less than expected due to side reactions.1.6.Write a balanced chemical equation for the hydrolysis of NF4+. Write a balancedchemical equation for a possible side reaction that can decrease thetheoretically expected O2:NF3 mole ratio.Tetrafluoroammonium salts were investigated for use as solid rocket fuels, because NF3 and F2 are released from them on heating. One of them has a fluorine content of65.6 m/m%, all of which is converted into NF3 and F2 upon decomposition. During the decomposition 2.5 times as many moles of F2 are formed as of NF3.1.7.Determine the formula of the salt in question.One of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.ABThe two figures above depict the cubic unit cell of the Cu2O crystal. The lattice constant of the structure is 427.0 pm.2.1.1.Which of the atoms (A or B) is copper?Which basic structure (primitive cubic, face centered cubic, body centered cubic, diamond) is formed by the A atoms and which structure is formed by the Batoms?What are the coordination numbers of the atoms?2.1.2.Calculate the smallest O-O, Cu-O and Cu-Cu distances in the structure.2.1.3.What is the density of pure copper(I) oxide?A common defect in this crystal is some copper atoms missing with the oxygen lattice unchanged. The composition of one such crystal sample was studied, and 0.2% of all copper atoms were found to be in oxidation state +2.2.2.What percentage of normal copper sites are empty in the crystal sample? Whatis x in the empirical formula Cu2-x O of the crystal?Copper(I) oxide is insoluble in water. It is stable in dry air, but humidity in the air catalyzes a transformation (Reaction 1).When copper(I) oxide is dissolved in dilute sulfuric acid, a blue solution containing a precipitate is formed without evolution of a gas (Reaction 2). When hot, concentrated sulfuric acid is used, no precipitate remains, but an odorous gas forms (Reaction 3). The same gas forms when the precipitate from reaction 2 is dissolved in hot concentrated sulfuric acid.2.3.Write balanced chemical equations for reactions (1-3).Copper(I) oxide can be produced in a number of ways. Heating copper in air is a common method in the synthesis of semiconductor Cu2O. In a pure oxygen atmosphere, the three species containing copper (Cu(s), Cu2O(s) or CuO(s)) can potentially interconvert.Suppose that the Δf H o and S o data given for 105 Pa are independent of temperature:2.4.Determine the temperature ranges, if any, of thermodynamic stability of copperand its oxides between 500 and 1500 K in a 105 Pa oxygen atmosphere. Important data are given for 298 K. Use this temperature in the following calculations: K sp(Cu(OH)2)=2∙10−19Cu2O(s) + H2O(l) + 2 e−⟶2 C u(s) + 2 O H−(aq) E o= −0.360 VCu2+(aq) + e−⟶ C u+(aq) E o= +0.159 VCu2+(aq) + 2 e−⟶ C u(s)E o= +0.337 VOne possibility for producing Cu2O is the anodic oxidation of copper. Electrolysis of an aqueous basic solution (e.g. NaOH) with a copper anode and platinum cathode can lead to formation of copper(I) oxide on the anode.2.5.Write the half reaction equations for the electrode processes during the anodicproduction of Cu2O in NaOH solution with a platinum cathode and copperanode.Electrolytic reduction of copper(II) ions in solution is another possibility.2.6.1.Write the half reaction equation of the cathode process giving Cu2O in acidicmedium.Let us use 0.100 mol dm−3 Cu2+ solution and carry out electrolysis with platinum electrodes.2.6.2.What is the maximum pH at which the concentration of copper(II) can bemaintained at 0.100 mol dm−3?If the pH is too low, reduction to metallic copper is preferred to the formation of copper(I) oxide.2.6.3.What is the minimum pH at which the cathodic production of Cu2O in a 0.100mol dm−3 Cu2+ solution is still possible?Problem 3 9% of the total Iodine deficiency is of special concern in Georgia because it occupies a region where iodine is scarce in soil and water. Iodine deficiency can be effectively and inexpensively prevented if salt for human consumption is fortified with small amounts of iodine. Methods for analyzing salt for iodine content are thus important. Current regulations in Georgia are that iodized salt must contain between 25-55 ppm iodine (1 ppm =1 mg iodine/kg salt).Most salt is iodized by fortification with potassium iodate (KIO3). Iodate content can be determined in salt samples using iodometric titration. In a typical procedure, 10.000 g of an iodized salt sample is dissolved in 100 cm3 of 1.0 mol/dm3 aqueous HCl to which1.0 g KI has been added. The solution is then titrated with 0.00235 mol/dm3 aqueous sodium thiosulfate solution to a starch endpoint; this requires 7.50 cm3 of titrant.3.1.1.Write a balanced net ionic equation for the reaction when iodate reacts withexcess iodide in acidic solution.3.1.2.Write a balanced net ionic equation for the reaction taking place during thetitration with thiosulfate.3.1.3.Calculate the iodization level, in ppm, of this salt sample.A less common agent for iodizing salt is potassium iodide, which cannot be easily measured by iodometric titration.One possible method for analyzing iodide in the presence of chloride is potentiometric titration. However, this method is not very precise in the presence of large amounts of chloride.In this method, a silver wire is immersed in the solution (containing iodide and chloride) to be analyzed and silver ion is gradually added to the solution. The potential of the silver wire is measured relative to a reference electrode consisting of a silver wire in a 1.000 mol/dm3 solution of AgNO3. The measured potentials are negative and the absolute values of these potentials are reported. The solution to be analyzed has a volume of 1.000 dm3 (which you may assume does not change as silver ion is added), and T= 25.0°C.The results of this experiment are governed by three equilibria: the solubility of AgI(s) [K spI] and AgCl(s) [K spCl] and the formation of AgCl2−(aq) [K f]. (Iodide also forms complex ions with silver but this may be neglected at the very low concentrations of iodide present in this experiment).AgI(s) ⇌ Ag+(aq) + I−(aq) K spIAgCl(s) ⇌ Ag+(aq) + Cl−(aq) K spClAg+(aq) + 2 Cl−(aq) ⇌ AgCl2−(aq) K fBelow are shown the results of two experiments measuring the observed potential as a function of added number of moles of silver ion. Experiment A (solid circles) was carried out with 1.000 dm 3 of solution containing 1.00∙10−5 mol/dm 3 iodide and no chloride ion. Experiment B (open circles) was done using 1.000 dm 3 of solution containing 1.00∙10−5 mol/dm 3 iodide and 1.00∙10−1 mol/dm 3 chloride.μmol Ag + added3.2.1. Select an appropriate data point from the experiments and use it to calculate thesolubility product of AgI (K spI ). 3.2.2. Select an appropriate data point from the experiments and use it to calculate thesolubility product of AgCl (K spCl ). 3.2.3. Select an appropriate data point from the experiments and use it to calculate K f .You may need to use values of K spI or K spCl to do this calculation. If you wereunable to carry out the calculations in 3.2.1. or 3.2.2., you may use the arbitrary values of K spI = 1.00∙10−15 and K spCl = 1.00∙10−9 without penalty.An analytical method that is more practical, because it is not sensitive to the presence of chloride, uses the Sandell-Kolthoff reaction. This is the reaction of H3AsO3 with Ce(IV) to give Ce(III) in acidic solution, which is strongly catalyzed by iodide ion.3.3.1.Write balanced net ionic equations for the reaction of cerium(IV) with H3AsO3 inacidic solution, as well as reactions of cerium(IV) with a species containing theelement iodine and H3AsO3 with a species containing the element iodine, thatcould reasonably account for the catalysis of the net reaction by iodide.The reaction of Ce(IV) with H3AsO3 can be monitored by measuring the absorbance at 405 nm, as Ce(IV) is orange and absorbs significantly at 405 nm, while the other reactants and products are colorless and do not absorb appreciably. Three runs were carried out, all in 0.50 mol/dm3 H2SO4at 25.0°C using the following initial concentrations:An analyst initiated the reactions by mixing the reagents in a cuvette. After a short variable delay absorbance measurements were started, with the first measurement recorded at t=0 s. The data obtained are shown below:Under these conditions (0.5 mol/dm3 H2SO4, 25.0°C), the rate law for the reaction can be written asRate = k[H3AsO3]m[Ce(IV)]n[I−]p where m, n, and p are integers.3.3.2.Determine the values of m, n, and p and calculate the value of k (be sure tospecify its units).A 1.000 g sample of iodized salt is dissolved in water to give 10.00 cm 3 of solution. A 0.0500 cm 3 aliquot of this solution is added to a mixture of 1.000 cm 3 0.025 mol/dm 3 H 3AsO 3 in 0.5 mol/dm 3 H 2SO 4 and 0.800 cm 3 0.5 mol/dm 3 H 2SO 4. To this mixture is added 0.200 cm 3 0.0120 mol/dm 3 Ce(NH 4)2(NO 3)6 in 0.5 mol/dm 3 H 2SO 4 and the absorbance at 405 nm is measured as a function of time at 25.0°C:3.3.3.Calculate the iodization level, in ppm, of this salt sample.Problem 4 8% of the total Application of kinetic studies in water treatmentIndustrial waste is a major cause of water pollution and kinetic studies are carried out in a laboratory to design effluent treatment. 1,4-dioxane, more commonly known as dioxane (C4H8O2), an industrial solvent and by-product, is a significant water contaminant. It can be oxidised to hazard free chemicals using oxidants such as peroxodisulfate, ozone or hydrogen peroxide.The data obtained in the kinetic study of oxidation of dioxane with potassium peroxodisulfate (K2S2O8) as oxidant and AgNO3 as catalyst at T = 303.15 K are given below. The reaction was monitored by the estimation of unreacted peroxodisulfate. The concentration of AgNO3 used in this study was 1.00∙10−3 m mol∙dm−3.In many countries the accepted maximum level of dioxane in drinking water is specified as 0.35 μg dm−3.A water sample contains an initial dioxane concentration of 40.00 μg dm−3. Assume that1 mol dioxane requires 1 mol of peroxodisulfate for oxidation. The concentration of AgNO3 used in this study was 1.00∙10−3 m mol∙dm−3.4.1.1.Calculate the time in minutes the oxidation process has to continue in order toreach the accepted level of dioxane at 303.15 K if the initial concentration ofK2S2O8is 5.0∙10−6 mol dm−3. Assume that the rate law obtained from the dataabove is valid under these conditions.Various mechanisms have been proposed for the peroxodisulfate oxidation of dioxane. Misra and Ghosh (1963) proposed the following mechanism:k1S2O82− + Ag+⇄Ag3+ + 2SO42−k2k3Ag3+ + D (dioxane) ⟶D’ (dioxane oxidised) + 2H+ + Ag+4.1.2.Assuming Ag(III) to be in steady state, deduce the rate equation for theoxidation of dioxane.4.1.3.Which of the following is/are correct?A) The rate equation based on the mechanism given in 4.1.2, at very highconcentrations of dioxane, is consistent with the experimental data in 4.1.1.B) The rate equation based on the mechanism given in 4.1.2, at very lowconcentrations of dioxane, is consistent with the experimental data in 4.1.1.C) The units of the observed rate constant are dm3∙mol−1∙s−1 at very highconcentrations of dioxane.D) The units of the observed rate constant are dm3∙mol−1∙s−1 at very lowconcentrations of dioxane.Degradation of pharmaceutical products – a kinetic overviewKinetic studies are important in deciding the shelf life of a pharmaceutical product. Several chemical reactions can affect the shelf life of pharmaceutical products and the rates of these reactions depend on conditions such as pH, temperature, humidity. Lysine acetylsalicylate (LAS) is prescribed as a pain killer and anti-inflammatory drug under the brand name Aspegic. LAS on hydrolysis forms lysine salicylate and acetic acid.Hydrolysis of LAS can proceed via three different pathways (a) acid catalysed,(b) uncatalysed and (c) base catalysed.If [LAS] denotes the concentration of LAS at time ‘t’, the overall rate of the hydrolysis reaction can be written as−d[LAS]dt=k H[LAS][H+]+k0[LAS]+k OH[LAS][OH−]where k H, k0 and k OH are the rate constants of the acid catalysed, uncatalysed and base catalysed pathways of hydrolysis, respectively. The observed rate constant is defined by:−d[LAS]dt=k obs[LAS]4.2.1. Write an expression for k obs in terms of k H, k0, k OH and [H+].Hydrolysis of LAS was carried out at 298.15 K at various pH values (0.50 to 13.0). A very low initial concentration of LAS ensured that the pH did not change during the course of the reaction.The following graph shows the pH dependence of the hydrolysis of LAS.4.2.2.Which of the following is/are correct?A) k obs≌k0 at pH = 12B) k obs≌k0 at pH = 5.0C) The rate of the reaction increases when the pH is changed from 0.50 to 1.0.D) The rate of the reaction increases when the pH is changed from 10 to 12.ing the diagram and the data given below, calculate k H, k0 and k OH. Make sureto specify the units.Acetylsalicylic acid, more commonly known as aspirin, is a medicine often used for reducing fever, pain and inflammation. Like LAS, the hydrolysis of aspirin can also take different pathways depending on the pH. The pH rate profile of aspirin hydrolysis at 333.15 K is given below:The following are possible reactions for the hydrolysis of aspirin. Depending on the pH, one or more of these reactions will predominate.I. CH3COOC6H4COOH + H3O+⟶ HOC6H4COOH + CH3COOH + H+II. CH3COOC6H4COOH + H2O ⟶ HOC6H4COOH + CH3COOHIII. CH3COOC6H4COOH + OH−⟶ HOC6H4COO− + CH3COOHIV. CH3COOC6H4COO− + H3O+⟶ HOC6H4COOH + CH3COOHV. CH3COOC6H4COO− + H2O ⟶ HOC6H4COO− + CH3COOHVI. CH3COOC6H4COO− + OH−⟶ HOC6H4COO− + CH3COO−ing the pH-rate profile diagram and the reactions given above, state which of the following statements is/are correct. (pK a of aspirin = 3.57 at 333.15 K)a) In the region C-D, reaction IV is predominantb) In the region C-D, reaction V is predominantc) In the region D-E reaction VI is predominantd) In the region A-B, reaction II is predominantA separate plot of k obs vs pH for the hydrolysis of aspirin has been confirmed to show a minimum at a particular pH. At 290.15 K the following rate constants for reactions I, II and III were determined:The ionic product of water at 290.15 K can be taken as 1.0∙10−14.4.3.2.Assuming that only reactions I, II and III occur, calculate the value of the pH atthe minimum of k obs.Problem 5 8% of the total 5500 years ago in ancient Egypt people learned for the first time how to synthesize a blue pigment. Now we know this pigment as Egyptian blue. About 2000 years later in ancient China another pigment was widely used, which is now referred to as Chinese blue. The two pigments are similar in structure, but have different elemental compositions.Ushabti figurines from Egyptian pharaoh tomb covered with Egyptian blue anda Chinese blue soap dispenser sold at AlibabaThe ancient method of preparation for these pigments can be easily reproduced in a modern laboratory.When considering the amounts, assume that all of the compounds in this task are pure, and the yields are quantitative.To make Egyptian blue, one should heat 10.0 g of mineral A with 21.7 g of SiO2 and9.05 g of mineral B at 800–900°C for a prolonged time. 16.7 dm3 of a mixture of two gaseous products are released (the volume is measured at 850°C and 1.013∙105 Pa (1.013 bar) pressure. In result, 34.0 g of the pigment was obtained. No other products are formed. As the gas mixture is cooled, one component of the mixture condenses. As the remaining gas is further cooled to 0°C, the gaseous volume reduces to 3.04 dm3.5.1.1.Find the mass of the gaseous mixture formed upon heating of A with B and SiO2.5.1.2.Determine the quantitative composition of this gas mixture.When 10.0 g of mineral A is heated with 21.7 g of SiO2 in the absence of B, it forms8.34 dm3of gaseous products (measured at 850°C and 1.013∙105 Pa = 1.013 bar pressure). Mineral A contains only one metal.5.1.3. Calculate the molar mass and determine the formula of mineral B. Hint: it is anionic solid insoluble in water and containing no water of crystallization.In order to obtain Chinese blue, one should take 17.8 g of mineral C instead of mineral B (keeping the amounts of mineral A and SiO2 same as for Egyptian blue), and run the reaction at higher temperatures. Besides the pigment, the same gaseous products in the same quantities are formed as in the preparation of Egyptian blue.5.1.4.Determine the formula of mineral C.5.1.5.Determine the formulae of Egyptian blue and Chinese blue.5.1.6. Determine the formula of mineral A.Elemental analysis of some samples of Chinese blue shows traces of sulfur. This led to a conclusion that those were synthesized using another common mineral instead of C. 5.2.1.Suggest a formula for the mineral used in place of C.5.2.2. Could the temperature of synthesis of Chinese blue be decreased if this mineralis used instead of C?If during the synthesis of Chinese blue we take a smaller amount of silica than in the process above, we will obtain a purple pigment: Chinese violet. It was used, in particular, for coloring the famous Terracotta army soldiers.Terracotta army from Xian, China and reconstruction of its original coloring 5.3. Write down the formula of a binary compound that forms under the conditionsrequired for Chinese violet and is responsible for the change of the color.Problem 6 7% of the total Although there is currently no known cure for Alzheimer’s disease, there are medications available to manage the neurodegenerative disorder. Among these are acetylcholinesterase inhibitors, of which galantamine 1 is an example. This molecule can be isolated from the Caucasian snowdrop, a plant native to Georgia; however, the large amounts needed for therapy require a synthetic route. Shown below is the route used to prepare galantamine industrially.resolutionNotes about the synthesis:∙1H NMR of A indicates 2 aromatic protons in a para arrangement.∙C is labile in aqueous conditions, so it is not isolated, but rather reacted immediately with NaBH4 to convert it to D.6.1.1.Suggest structures for A, B, C, D, F, and G. None of the reactions except for thefinal transformation with L-selectride are stereoselective. Therefore,stereochemistry does not need to be indicated in your answers.6.1.2. Give the formula for a possible reagent, X , to convert compound D to E .The optical rotation of the material obtained by resolution was –400° cm 2 g -1, while that of the enantiomerically pure compound is –415° cm 2 g -1 when measured under the same conditions. You may assume that the only optical impurity is the other enantiomer. One way of describing optical purity is enantiomeric excess (ee ). It is defined as the difference in the percentages of the enantiomers in a mixture. For example in a mixture of 70% R and 30% S , the ee is 40%.6.2.1. What is the enantiomeric excess of the resolved compound as prepared by theindustrial route?L-selectride is a commercial reagent that performs the final reaction stereoselectively.6.2.2. Assign the labelled stereocentres ( ) in (–)-1as R or S .6.2.3. Give the formula for a reagent that carries out thesame reaction as L-selectride, converting H to 1.You need not worry about stereoselectivity.An alternative route to galantamine occurs with theseven-membered ring being the last ring to form.6.3.1. Give the formula for compound Y to carry out the first step of the route.6.3.2. Suggest structures for J and K. (4 equivalents)pH = 5Problem 78% of the totalThis question looks at the synthesis of dolasetronmesylate , Z (shown right), a drug sold under thetradename Anzemet and used to treat post-operative nausea and vomiting.The synthesis begins as shown below.First cyclic compound A is made, which contains C, H, and O only. Compound G is achiral and can be prepared directly from D using ozone under reductive conditions, or via stereoisomers E1 and E2 using OsO 4, or via stereoisomers F1 and F2 using the peracid shown.7.1.Determine the empirical formula of G from the percentage masses given. 7.2. Give the structures of A , B , C , D , E1, E2, F1, F2 and G .Compound G is used in the next stage of the synthesis, under buffered conditions, to form H (as a mixture of two achiral diastereoisomers). Reduction of H with NaBH4 gives alcohol I (as a mixture of four achiral diastereoisomers). I reacts with acidified dihydropyran to form J (as a mixture of even more diastereoisomers). J is then treated first with t-butoxide base, then refluxed with acid before finally extracting under weakly basic conditions to form K as a mix of two diastereomers, K1 (major product) and K2 (minor product). These could be separated, and K1 was used in the final stages of the synthesis.7.3.1.Give the structures of H, I, and J. There is no need to show the differentdiastereoisomers formed.7.3.2. Give the structures of diastereoisomers K1, and K2.In the final stage of the synthesis, L and M react to form intermediate N. N then reacts with K1 to form, after extraction, the neutral amine which gives the target compound upon protonation with CH3SO3H.7.4.Give the structure of N.Problem 8 7% of the totalAn exotic, but biologically relevant sugar analogue can be prepared from D -glucose in the following manner. Heating a mixture of D -glucose and acetone with a few drops ofconcentrated acid results in the formation of a diacetonide A . Then A can be hydrolyzed selectively to B .Acetone/H +XD -glucose A B8.1.1. Which of the following sentences is true?A is an α isomer. A is neither α nor β.A is a β isomer.A is a mixture of α and β isomers.8.1.2. Which of the following sentences is true?We can get product A only if we use α-D glucose as starting material. We can get product A only if we use β-D glucose as starting material.We can get product A either from α- or from β-D glucose as starting material.8.1.3. Which one of these reagents can be utilized as X for the selective hydrolysis ofA ?50% acetic acid concentrated H 2SO 4 6 M HCl in water1 M NaOH in water6 M HCl in acetic acid8.1.4. Which is the stereochemically correct structure for compound B ?Neither of theseB is treated with sodium metaperiodate to get C.C is then reacted with an aqueous solution of NaCN, then heated with 10% NaOH solution to get a mixture of two diastereomeric compounds D1 and D2. These compounds can be separated by column chromatography.IO4−1. NaCN2. NaOH/H2OB C188.2 g/molD1 + D2Reaction of D1 with LiAlH4 followed by heating with 1M HCl solution gives sugar F that is the hydrolysis product of the most abundant natural polysaccharide.LiAlH41M HCl/H2OD1 E F8.2.1.Draw the structures of C, D1, D2, E and F including stereochemical information.Show F as the more stable 6-membered ring containing isomer using the ringskeleton. Indicate with a wavy line if absolute chirality around a carbon is notknown.8.2.2.The reaction sequence from glucose to F does not seem to be useful. In somecases, however, this is the most economical way to produce F. In which case?13C labelling at carbon 6 of F13C labelling at carbon 5 of F13C labelling at carbon 1 of F15O labelling at glycosidic OH of Fsynthesis of an uncommon isomer of FNeutralization of D2 with HCl followed by heating in toluene results in dehydration and formation of G, which has a tricyclic structure in water-free solvents. Boiling G in 1M HCl solution gives H (C6H10O7), which is a natural sugar derivative containing a 6 membered ring. H is a building block of heparin, an anticoagulant polysaccharide produced by our bodies.1. Equimolar HCl2. Heat, toluene 1M HCl/H2OD2G H−H2O8.3.1.Draw the structure of G including the stereochemistry.Draw H as the more stable 6-membered ring containing isomer using the ringskeleton. Indicate with a wavy line if absolute chirality around a carbon is notknown.。
第51 届国际化学奥林匹克试题（理论部分）
126
Vol. 35
大 学 化 学
总则
本理论考卷共 58 页。
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:醛
:羰基 α 位
苄基 CHn—C6H5:
芳环上的:
11.0
10.0
9.0
8.0
7.0
6.0
50
. All
Rights
Reserved.
H―H
耦合常数(Hz)
氢的种类
|Jab| (Hz)
R2CHaHb
4–20
R2HaC―CR2Hb
2–12
如果 σ 键可以自由转动：6–8
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国际化学奥林匹克竞赛试题汇编-第38届ICHO理论试题(中文版)答案
1-1.The mass of a water droplet:m = V ρ = [(4/3) π r3] ρ = (4/3) π (0.5x10-6 m)3 (1.0 g/cm3)= 5.2x10-16 kg⇒10 marksAverage kinetic energy at 27o C:KE = mv2/2 = (5.2x10-16 kg) (0.51x10-2 m/s)2/2= 6.9x10-21 kg m2/s2= 6.9 x10-21 J ⇒15 marks*.The average kinetic energy of an argon atom is the same as that of a water droplet.KE becomes zero at –273 o C.From the linear relationship in the figure, KE = aT (absolute temperature)where a is the increase in kinetic energy of an argon atom per degree.a = KE/T = 6.9x10-21 J/(27+273K) = 2.3x10-23 J/K⇒25 marksS: specific heat of argon N: number of atoms in 1g of argonS = 0.31 J/g K = a x NN = S/a = (0.31 J/g K) / (2.3x10-23 J/K)= 1.4x1022 ⇒30 marksAvogadro’s number (N A) : Number of argon atoms in 40 g of argonN A = (40)(1.4x1022)= 5.6 x1023⇒20 marks2-1. ⇒ 30 marksmass of a typical star = (4/3)(3.1)(7x108 m)3(1.4 g/10-6 m 3) = 2×1033 g mass of protons of a typical star = (2×1033 g)(3/4 + 1/8) = 1.8×1033 g number of protons of a typical star = (1.8×1033 g)(6×1023/g) = 1×1057number of stellar protons in the universe = (1×1057)(1023) = 1×1080Partial credits on principles:Volume = (4/3)(3.14)radius 3×density; 4 marks 1 mole = 6×1023; 4 marksTotal number of protons in the universe = number of protons in a star ×1023; 2 marks Mass fraction of protons from hydrogen = (3/4)(1/1); 5 marks Mass fraction of protons from helium = (1/4)(2/4); 10 marks2-2. ⇒ 30 marks∆E(2→3) = C(1/4 - 1/9) = 0.1389 C λ(2→3) = 656.3 nm ∆E(1→2) = C(1/1 - 1/4) = 0.75 Cλ(1→2) = (656.3)(0.1389/0.75) = 121.5 nmNo penalty for using Rydberg constant from memory. 15 marks penalty if answered in a different unit (Hz, etc.)2-3.T = (2.9×10-3 m K)/1.215×10-7 m = 2.4×104 K ⇒ 10 marks2-4..⇒ 20 marksλ = 3 × 108 m/1.42 × 109 = 0.21 mT = (2.9 × 10-3 m K)/0.21 m = 0.014 K2-5. ⇒ 10 marks14N + 4He → (17O ) + 1HO-17, O acceptable1783-1.k des = A exp(-E des/R T)= (1x1012 s-1)(5x10-32) = 5x10-20 s-1 at T = 20 K ⇒10 markssurface residence time, τresidence = 1 / k des = 2x1019 s = 6x1011 yr ⇒20 marks(full credit for τhalf-life = ln2 / k des = 1x1019 s = 4x1011 yr)residence time = 2x1019s3-2.The distance to be traveled by a molecule: x = πr = 300 nm.k mig = A exp(-E mig/R T)= (1x1012 s-1)(2x10-16 ) = 2x10-4 s-1 at T = 20 K ⇒ 5 marksaverage time between migratory jumps,τ = 1 / k mig = 5x103 sthe time needed to move 300 nm= (300 nm/0.3 nm) jumps x (5x103 s/jump) = 5x106 s = 50 days ⇒15 marks(Full credit for the calculation using a random-walk model. In this case:t = τ (x/d) 2 = 5 x 109 s = 160 yr. The answer is still (b).)(a) (b)(c) (d) (e)10 marks3-3.k(20 K) / k(300 K) = exp[(E/R) (1/T1 - 1/T2)]= e-112 = ~ 10-49 for the given reaction ).) ⇒15 marks The rate of formaldehyde production at 20 K= ~ 10-49 molecule/site/s = ~ 10-42 molecule/site/ yr⇒10 marks(The reaction will not occur at all during the age of the universe (1x1010 yr).)rate = 10-42molecules/site/yr3-4. circle one(a) (b) (c) (a, b) (a, c) (b,c)(a, b, c)(15 marks, all or nothing)4-1.H PNumber of atoms ( 11.3 ) 1⇒ 10 marksTheoretical wt % ( 3.43 )⇒ 10 marks4-2.adenineN NN NN H H guanineNN N NO N HH HNN O N H H cytosineNN H O O thymine(10 marks on each)4-3. 7 marks each, 20 marks for threeadenineNNNNNHHguanine NN NNON HHH NNH OOthymineNNONHH cytosine NNH OOthymineguanine NN NNON HHHcytosineNNONHHcytosineNNON HHNNHOO thyminethymineNNHOONNH OOthyminethymine NNHOONNONHH cytosineadenineNNNNNHH adenineNNNNNHHadenine NNNNNHHguanineguanine NNNNON HHHNNNNONHHH4-4. 2.5 marks for each bracketadenineN NN N HNH 2guanine N NH N N HO NH 2Uracil N H NH O cytosineN H N NH 2OOHCN ( 5 ) ( 5 ) ( 4 )( 4 )H 2O ( 0 ) ( 1 ) ( 2 ) ( 1 )5-1.(20 marks)1st ionization is complete: H2SO4→ H+ + HSO4-[H2SO4] = 02nd ionization: [H+][SO42-]/[HSO4-] = K2 = 1.2 x 10-2 (1)Mass balance: [H2SO4] + [HSO4-] + [SO42-] = 1.0 x 10-7 (2)Charge balance: [H+] = [HSO4-] + 2[SO42-] + [OH-] (3)Degree of ionization is increased upon dilution.[H2SO4] = 0Assume [H+]H2SO4 = 2 x 10-7From (1), [SO42-]/[HSO4-] = 6 x 104 (2nd ionization is **plete)[HSO4-] = 0From (2), [SO42-] = 1.0 x 10-7 [5 marks]From (3), [H+] = (2 x 10-7) + 10-14/[H+][H+] = 2.4 x 10-7(pH = 6.6) [8 marks][OH-] = 10-14/(2.4 x 10-7) = 4.1 x 10-8[2 marks]From (1), [HSO4-] = [H+][SO42-]/K2= (2.4 x 10-7)(1.0 x 10-7)/(1.2 x 10-2) = 2.0 x 10-12[5 marks]Check charge balance:2.4 x 10-7≈ (2.0 x 10-12) + 2(1.0 x 10-7) + (4.1 x 10-8)Check mass balance:0 + 2.0 x 10-12 + 1.0 x 10-7≈ 1.0 x 10-7Species Concentration** x 10-12HSO4-** x 10-7SO42-** x 10-7H+** x 10-8 OH-5-2. (20 marks)mmol H3PO4 = 0.85 ⨯ 3.48 mL ⨯ 1.69g/mL ⨯ 1 mol/98.00 g ⨯ 1000 = 51.0 [5 marks]The desired pH is above p K2.A 1:1 mixture of H2PO4- and HPO42- would have pH = p K2 = 7.20.If the pH is to be 7.40, there must be more HPO42- than H2PO4-.We need to add NaOH to convert H3PO4to H2PO4-and to convert to the right amount of H2PO4-to HPO42-.H3PO4 + OH-→ H2PO4- + H2OH2PO4- + OH-→ HPO42- + H2OThe volume of 0.80 NaOH needed to react with to to convert H3PO4 to H2PO4- is:51.0 mmol / 0.80M = 63.75 mL [5 marks]To get pH of 7.40 we need:H2PO4- + OH-→ HPO42-Initial mmol 51.0 x 0Final mmol 51.0-x 0 xpH = p K2 + log [HPO42-] / [H2PO4-]7.40 = 7.20 + log {x / (51.0-x)}; x = 31.27 mmol [5 marks]The volume of NaOH needed to convert 31.27 mmol is :31.27 mmol / 0.80 M = 39.09 mLThe total volume of NaOH = 63.75 + 39.09 =102.84 mL , 103 mL [5 marks]Total volume of 0.80 M NaOH (mL) 103 mL5-3. (20 marks)p K = 3.52pH = pK a + log ([A-]/[HA])[A-]/[HA] = 10(pH-pKa) [5 marks]In blood, pH =7.40, [A-]/[HA] = 10(7.40-3.52) = 7586Total ASA = 7586 +1 = 7587 [5 marks]In stomach, pH = 2.00, [A-]/[HA] = 10(2.00-3.52) = 3.02x10-2Total ASA = 1+ 3.02x10-2 = 1.03 [5 marks]Ratio of total aspirin in blood to that in stomach = 7587/1.03 = 7400 [5 marks]** ( 103Ratio of total aspirin in blood to that in stomach6-1. (5 marks)4 H2O + 4 e-→ 2 H2(g) + 4 OH- (or 2 H2O + 2 e-→ H2(g) + 2 OH-)6-2. (5 marks)2 H2O → O2 + 4 H+ + 4 e-(or H2O → 1/2 O2 + 2 H+ + 2 e- )6-3. (5 marks)Cu → Cu2+ + 2e-6-4. (20 marks)Reduction of sodium ion seldom takes place.It has a highly negative reduction potential of –2.710 V.Reduction potential for water to hydrogen is negative (water is very stable).But, it is not as negative as that for sodium ion. It is –0.830 V.Reduction of both copper ion and oxygen takes place readily and the reduction potentials for both are positive.In the present system, the reverse reaction (oxidation) takes place at the positive terminal. Copper is oxidized before water.Reduction potential for hydrogen ion is defined as 0.000 V.6-5. (15 marks)pOH = 14.00 – 4.84 = 9.16[OH-] = 6.92 x 10-10K sp = [Cu2+][OH-]2 = 0.100 x (6.92 x 10-10) = 4.79 x 10-206-6.E = E o Cu2+/Cu + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log (K sp / [OH-]2)= +0.340 + (0.0592/2) log (K sp) - (0.0592/2) log [OH-]2= +0.340 + (0.0592/2) log (K sp) - 0.0592 log [OH-],3 marksBy definition, the standard potential for Cu(OH)2(s) + 2e-→ Cu(s) + 2OH- is the potential where [OH-] = 1.00.E = E o Cu(OH)2/Cu = +0.340 + (0.0592/2) log (K sp)= +0.340 + (0.0592/2) log (4.79 x 10-20)= +0.340 - 0.5722 marks= -0.232 V10 marks-------------------------------------------------------------------------------------------------------------- One may solve this problem as following.Eqn 1: Cu(OH)2(s) + 2e -→ Cu + 2OH-E+o = E o Cu(OH)2/Cu = ?Eqn 2: Cu(OH)2(s) → Cu2+ + 2OH-E o = (0.05916/n) logK sp= (0.05916/2) log(4.79×10-20)= -0.5715 V3 marksEqn 1 – Eqn 2 : Cu2+ + 2e-→ CuE-o = E+o - E o = E o Cu2+/Cu = 0.34 VTherefore, E+o = E-o + E o = + 0.34 + (-0.5715)2 marks= -0.232 V10 marks-0.232 V6-7.Below pH = 4.84, there is no effect of Cu(OH)2 because of no precipitation.Therefore,E = E Cu2+/Cu = +0.340 + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log 0.1003 marks= +0.340 – 0.0296 = +0.310 V7 marks** V6-8.** g graphite = 0.0833 mol carbon6 mol carbon to 1 mol lithium; 1 g graphite can hold 0.0139 mol lithiumTo insert 1 mol lithium, 96487 coulombs are needed.Therefore, 1 g graphite can charge 96487 × 0.0139 = 1340 coulombs. 5 marks1340 coulombs / g = 1340 A sec / g = 1340 x 1000 mA × (1 / 3600) h = 372 mA h / g 5 marks372 mA h / g7-1. (10 marks)n/V = P/RT = (80 x 106 / 1.013 x 105 atm)/[(0.082 atm L/mol/K)(298K)] = 32 mol/L5 marksdensity = mass/volume = d = 32 x 2 g/L = 64 kg/m 3 5 marks64 kg/m 37-2.** or 0.23H 2(g) + 1/2 O 2(g) → H 2O(l); ∆H rexn-1 = ∆H f [H 2O(l)] = -286 kJ/mol = -143 kJ/g 7 marksC(s) + O 2(g) → CO 2(g); ∆H rexn-2 = ∆H f [CO 2(g)] = -394 kJ/mol = -33 kJ/g 7 marks(-∆H rexn-1) / (-∆H rexn-2) = 4.3 or (-∆H rexn-2) / (-∆H rexn-1)= 0.236 marks7-3. (a) (-)1.2 x 105 kJ, (b) (-)6.9 x 104 kJ** x 108 sec or 3.3 x 104 hr or 1.4 x 103 days or 46 month or 3.8 yrI = 0.81 AH 2(g) + 1/2 O 2(g) → H 2O(l)∆H c = -286 kJ/mol = -143 kJ/g = -143 x 103 kJ/kg 5 marksΔG = ΔH – T ΔSΔS c= 70 – 131 – 205/2 = -163.5 J/K/mol5 marksΔG c = -286 kJ/mol + 298K x 163.5 J/K/mol = -237 kJ/mol = -1.2 x 105 kJ/kg 5 marks(a) electric motor W max = ΔG c ⨯ 1 kg = - 1.2 x 105 kJ 5 marks (b) heat engine W max = efficiency x ∆H c 5 marks= (1 – 298/573) x (-143 x 103 kJ) = -6.9 x 104 kJ 5 marks119 x 103 kJ = 1 W x t(sec)t = 1.2 x 108 sec = 3.3 x 104 hr = 1.4 x 103 days = 46 month = 3.8 yr 5 marksΔG = -nFE n = # of electrons involved in the reaction F = 96.5 kC/molH 2(g) + 1/2 O 2(g) → H 2O(l) n = 2 5 marksE = - ΔG/nF = 237 kJ/mol / 2 / 96.5 kC/mol = 1.23 V5 marksI = W/E = 0.81 A5 marks8-1-1. (5 marks on each)①C②C③CO8-1-2.③ Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) 5marks① C(s) + O2(g) → CO2(g) ΔH①◦ = -393.51 kJ = ΔH f◦(CO2(g))② CO2(g) + C(s) → 2CO(g) ΔH②◦ = 172.46 kJFrom ① and ②,ΔH f◦(CO(g)) = (1/2){172.46 + (-393.51)} = -110.525 kJΔH f◦(Fe2O3) = -824.2 kJΔH③◦ = 3ⅹΔH f◦(CO2(g)) - ΔH f◦(Fe2O3) - 3ⅹΔH f◦(CO(g))= 3ⅹ(-393.51) – (-824.2) - 3ⅹ(-110.525) = -24.8 kJ 7 marks ΔS③°=2ⅹ27.28+3ⅹ213.74-87.4-3ⅹ197.674=15.36 J/K 3 marks ΔG③°=ΔH°-TΔS°=-24.8kJ-15.36J/Kⅹ1kJ/1000Jⅹ1473.15K=-47.43 kJ5 marksK = e(-ΔG°/RT)= e(47430J/(8.314J/Kⅹ1473.15K)) = 48 5 marksBalanced equation of ③:K = 48Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)8-2-1. (20 marks)One AB2O4 unit has available 4 (= 1 + (1/4)ⅹ12) octahedral sites.48-2-2. (20 marks)Since one face-centered cube in AB2O4 represents one Fe3O4 unit in this case, it has 8 available tetrahedral sites. In one Fe3O4 unit, 1 tetrahedral site should be occupied by either one Fe2+ (normal-spinel) or one Fe3+ (inverse-spinel). Therefore, in both cases, the calculation gives (1/8) ⅹ100% = 12.5% occupancy in available tetrahedral sites.**%8-2-3. (10 marks for d-orbital splitting, 10 marks for elec. distribution)9-1-1. 1 answer for 8 marks, two for 15 marksH 3CN NNH 3CNNN :::+_+::_:9-1-2. ( 10 marks)H 3CN::9-1-3.H 3CNCH 2CH 2:H 3CN HH CCH 2:(10 marks) (10marks )9-2-1. 5 marks eachHONN +_::ONN:H+:HH_O NN:H+:H_::::::9-2-2.( 10 marks)CH 2CO ::9-3-1.(40 marks)CH 3H 3CH 3C+BC H 2CCH 3CH 3CO 2DEOOO_9-3-2.(10 marks)O OH O n+F10-1. 10 marks eachNMLCH 2OHCH 2OHMeOOMeH HH HOMeMeO CHOCHOCH 2OHCH 2OHHHH H OHOMeMeO OH10-2. 8 marks each for correct structuresNumber of possible structures24 marks12OH(OH)OH(H)HH HHOMeOMeOH COOMeOH(OH)OH(H)HH HHOMeOMeOHCOOMe34OH(OH)OH(H)OH(OH)OHe(H)10-3. 10 marks eachGICH 2OHCH 2OHHHHHMeOOMeOHOMeCH 2OHCH 2OHHHHOMeOMeOMe10-4. 10 marksNumber of the correct structure for C from 10-2110-5.BOH(OH)OH(H)HHHH OHCOOHOHOH10 marks eachDJOH(OH)OH(H)HHHHOMeOMeCOOMeOMeOH(OMe)OMe(H)HHHHOMeOMeOMeCOOMe10-6. 20 marksHOOCOHHH OOOHOOH COOHOOHOHOH COOH11-1. 10 marks311-2. 30 marksCOOHHOOCOOH11-3. 2.5 marks eacha, c, d11-4 30 marksOOCOCOOOHTransition State11-5.For the enzyme-catalyzed reaction, Arrehnius equation could be applied.k cat/k uncat = A exp (-E a, cat/ RT) / A exp (-E a, uncat / RT)= exp [-∆E a, cat-uncat/ RT]= exp [-∆E a, cat-uncat(J/mol) / (2,480 J/mol)] = 106Therefore, -∆E a, cat-uncat = 34,300 J/mol 15 marksk uncat, T/k uncat, 298 = exp (-∆H≠ uncat/ RT) / exp (-∆H≠uncat / 298R)= exp [(-∆H≠ uncat/R)(1/T-1/298)]ln(k uncat, T/k uncat, 298 )= 13.8 = [(-86900/8.32)(1/T-1/298)]Therefore, T = 491 K, or 218o C 15 marks-E a, cat-uncat = 34,300 J/molT = 491 K, or 218o C。
2009年第41届ICHO预备试题及答案中文
质谱：分子离子以及常见的碎片离子。
3
预备试题
理论试题
第一题 月岩年代推算
通过分析样品中不同矿物内87Rb/86Sr以及87Sr/86Sr的含量比，人们已经算出了由阿波罗16 号飞船采集的月球岩石标本的年代
矿物 A (斜长石) B (ະ矿ੴ)
c) 在横截面积为A，深度为d的空间内充满海水，写出海水质量的表达式。 d) 推导出由(c)中的海水所产生的压力的表达式，并由此进一步写出，潜水员在d米
深处比在水面所多感受到的压强的表达式。 e) 潜水员如果想要能顺利地用呼吸管进行呼吸，计算她最多只能下潜多少米。
第三题 理想气体与实际气体
气体施加在器壁上的压力，会随着气体分子与器壁碰撞的激烈程度加剧而增大。在单 次碰撞中，力给表面的冲量等于动量改变量，即mΔv。然后这个力的大小，就可以通 过把冲量乘以气体分子与表面的碰撞频率算得。 由于气体分子的运动取向是随机的，所以在恒温下，单位时间内发生的碰撞次数是一 定值。 温度反映了气体分子速度的分布情况。对于一种给定的气体而言，分子的平均速率是 随着温度的升高而增大的。
CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g)
ΔrH° = –802.7 kJ mol–1
7
预备试题
第五题 氢的工业生产
在工业上，氢气可以通过加热碳氢化合物（比如甲烷）和水蒸气的混合物来制备：
CH4 (g) + H2O (g)
3H2 (g) + CO (g)
A
a) 通过下面给出的热力学数据，计算反应A在298K下的ΔrG°值，然后由它计算该 反应的平衡常数Kp。
【VIP专享】第44届国际化学奥林匹克实验试题中文版0
根據 IChO 的規定，你必須遵守安全守則，實驗時必須配戴大會所提供的護目鏡，或是你 自備且經核可的有度數的護目鏡。操作及使用藥品時可以戴手套。
若你違反實驗安全守則，第一次將給予警告；若仍然再犯，則你將被驅逐出場，且實驗 考試以零分計算！
當你有任何關於實驗安全的問題，或是需要暫時離開實驗室，請立即向監考人員求助。
3.0 mL 2.0 mL 1.0 mL
35
Name:
Task 1
Code: TAI
18% of the total
abcde f
g Task 1 18%
10 2 10 12 16 12 8
70
動力學、同位素效應和丙酮碘化反應之機制
揭開化學反應之機制可促進催化及合成之進展。探測反應機制最強而有力的工具之一為
I C R2
+ R+ + 2I-
R = H or D
本反應的速率定律式為
Rate＝k [acetone]m [I3-]n [H+]p
式中速率常數 k 和整數的反應級數 m、n、p 將由你來決定。
你也須比較 (CH3)2CO 和 Acetone-d6 之反應速率以決定同位素效應 (kH/kD)。Acetone-d6 係將 (CH3)2CO 之 6 個氕(1H)置換成氘(2H, D)。根據這些數據，你將可推測本反應之機制。
在進行實驗前，請詳讀本試題之全部描述並規畫你的實驗方法及步驟。
36
Name:
Code: TAI
步驟
反應速率與溫度有關。請記錄實驗室溫度 (問實驗室助教)。 oC
碼表使用說明： (1) 按壓 [MODE] 鈕直到螢幕顯示 COUNT UP。 (2) 按壓 [START/STOP] 鈕以開始計時。 (3) 再次按壓 [START/STOP] 鈕以停止計時。 (4) 按壓 [CLEAR] 鈕以歸零。
第32届IChO预备题中译本(简译本)
科目化学年级高三文件aosai003.doc考试类型化学竞赛考试时间2000关键词国际奥林匹克竞赛标题第32届IChO预备题中译本（简译本）内容第1题酸雨纯水pH为7.0。
天然雨水因溶解大气二氧化碳而呈弱酸性。
但许多地区的雨水酸性更强，其原因有的是天然的，有的则是人为的。
大气中的二氧化硫和一氧化氮会被氧化为三氧化硫和二氧化氮，并分别与水反应生成硫酸和硝酸。
所谓“酸雨”的平均pH为4.5，最低可达1.7。
二氧化硫在水溶液中是一个二元酸，在25o C时酸式电离常数如下：SO2(aq) + H2O(l) ⇌ HSO3-(aq) + H+(aq)K a1= 10-1.92 M HSO3-(aq) ⇌ SO32-(aq) + H+(aq)K a2= 10-7.18 M注：“M”是原文用于代替国际符号mol·dm-3的欧洲国家中学教科书通用符号，请同时熟悉这两种符号。
本译文未将此符号改为国际符号。
下同。
请注意平衡常数的指数表达式和以SO2而非H2SO3为反应物。
a.在二氧化硫的分压为1bar时它在每升水中的溶解度为33.9升(25o C, 全题同)。
i）计算被二氧化硫饱和的水中的二氧化硫总浓度（忽略因溶解SO2引起的水的体积变化）。
ii）计算亚硫酸氢根离子的百分含量。
iii）计算溶液的pH。
b．计算含0.0100 M亚硫酸钠的水溶液的氢离子浓度。
c．在亚硫酸钠水溶液中存在的主要平衡为：2HSO3-(aq) ⇌ SO2(aq) + SO32-(aq) + H2O(l)i)计算它的平衡常数。
ii)若只考虑此平衡，计算0.0100 M亚硫酸钠水溶液中的二氧化硫浓度。
d．亚硫酸钡在水中的溶解度为0.016g/100ml。
i）计算饱和溶液中的钡离子浓度。
ii）计算饱和溶液中的压硫酸根离子浓度。
iii）计算亚硫酸钡的溶度积。
e．亚硫酸银的溶度积为10-13.87 M3。
计算亚硫酸银饱和水溶液中的银离子浓度（忽略亚硫酸根离子的碱性）。
第38届国际化学奥林匹克理论试题
２００６年第９期化学教育・５５・第３８届国际化学奥林匹克纪实段连运（北京大学化学与分子工程学院１００８７１）第３８届国际化学奥林匹克于２００６年７月２～１１日在韩国举行。
来自世界５大州６６个国家和地区共２５４名选手参加了本届竞赛。
中国派出了由９人组成的代表队：领队段连运教授（北京大学），副领队高翔教授（复旦大学），科学观察员杨宇翔教授（华东理工大学）以及范康年教授（复旦大学）和徐华龙教授（复旦大学）；４名队员分别是：蔡李超（湖南长沙市第一中学）、曾毅（广东深圳中学）、刘艺斌（江西鹰潭市第一中学）和叶钦达（上海华东师大第二附属中学）。
本届竞赛设奖牌１６５枚，其中金牌２８枚，银牌５６枚，铜牌８１枚。
我国４名选手全部获得金牌。
竞赛通讯一ＣａｔａＩｙｚｅｒ登载的金牌、银牌、铜牌获得者名单次序是按姓氏（ｆａｍｉｌｙｎａｍｅ）英文字母次序排列的，并非竞赛成绩高低次序。
本届竞赛还设立了特殊奖项。
韩国选手ＨｗａｎＢＡＥ总成绩第一，荣获ＰＯＳＣ０奖（ＰｏｓＣｏＡｗＡＲＤ）；我国选手蔡李超赢得理论竞赛成绩第一（ＥｘｃｅｌｌｅｎｃｅｉｎＴｈｅｏｒｅｔｉｃａｌＴｅｓｔ），获得ＬＧ奖（ＬＧＡｗＡＲＤ一功能齐全的笔记本手提电脑）；中国台北选手Ｃｈｅｎｇ—ＹｉＫＡｏ实验成绩最高（ＥｘｃｅｌｌｅｎｃｅｉｎＰｒａｃｔｉｃａｌＴｅｓｔ），获得三星奖（ＳＡＭＳＵＮＧＡＷＡＲＤ）；土耳其女选手ＨａｎｄｅＢＯＹＡＣＩ获得ＳＫ奖（ＳＫＡＷＡＲＤ—ＡｗａｒｄｆｏｒＢｅｓｔＦｅｍａｌｅＳｔｕｄｅｎｔ）。
除我国代表队外，在本届竞赛中成绩较好的代表队有：中国台北队（３金１银），韩国队（３金１银），俄罗斯队（３金１银），越南队（２金２银）。
第一集团的代表队仍然主要来自亚洲。
本次竞赛的主题是“Ｃｈｅｍｉｓｔｒｙｆｏｒｌｉｆｅ”，“Ｃｈｅｍｉｓｔｒｙｆｏｒｂｅｔｔｅｒｌｉｆｅ”。
理论题涉及许多基本的物理、化学概念和原理，但是这些概念和原理是以科学发展的历史及发展前沿的成果为载体的。
31届iCho
226 88 Ra 222
(t1/2 = 1620 years) which,
b)
In the neutron induced binary fission products
are often found. Assuming that these nuclides have come from the original fission process, find (i) (ii) (iii) what elementary particles are released, energy released per fission in MeV and in joules, energy released per 1 gram of 235 U in unit of kW-hour. 235 136 Atomic masses: 92 U = 235.04393 u, 54 Xe = 135.90722 u,
31st International Chemistry Olympiad Preparatory Problems
Problem 2
a) Acetone (denoted as A) and chloroform (denoted as C) are miscible at all proportions. The partial pressure of acetone and chloroform have been measured at 35oC for the following solutions: Xc 0.00 0.20 0.40 0.60 0.80 1.00 0.00 35 82 142 219 293 Pc (torr) PA (torr) 347 270 185 102 37 0.00 where Xc is the mole fraction of chloroform in the solution. (i) (ii) (iii) each component in the solution. Activity (a) may be found from the following equation (taking chloroform as an example):
第38届ICHO理论试题(中文版)
理論競賽2006. 7. 7 Gyeongsan, KoreaChemistry for Life, Chemistry for better Life一般規定-每一頁的答案紙上都必須寫上你的名字和代碼。
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常數與有用的公式氣體常數 R = 8.314 J K -1 mol -1 法拉第常數 F = 96485 C mol -1 標準壓力: p = 1.013∙105 Pa 標準溫度: T = 25°C = 298.15 K 亞佛加厥數 N A = 6.022∙1023 mol -1 普朗克常數 h = 6.626∙10-34 J s 光速（真空） c = 3.00∙108 m s -1∆G = ∆H - T ∆S ∆G = - nFE∆G 0 = - RT∙lnK∆G = ∆G 0 + R T∙lnQ with Q =)()(reactands c of product products c of product∆H(T 1) = ∆H 0 + (T 1 - 298.15 K)∙C p (C p = constant)Arrhenius (阿瑞尼士) 方程式 k = A ∙TR E a e⋅-理想氣體定律 pV = nRT Nernst 方程式E = E 0 +redox c c ln nF RT⋅ Beer- Lambert 定律 A = logPP 0= ε∙c∙dV (圓柱體積) = πr 2h A( 球表面積) = 4πr 2 V (球體積) =34πr 31 J = 1 N m 1 N = 1 kg m s -2 1 Pa = 1 N m -2 1 W = 1 A V = 1 J s -1 1 C = 1 A s1. 亞佛加厥數(5 分)現有相同大小之球形水滴分散於氬氣之中。
第38届ICHO理论试题及答案
Theoretical Test2006. 7. 7 Gyeongsan, KoreaChemistry for Life, Chemistry for better LifeGeneral Directions-Write your name and code number on each page of the answer sheet.-You have 5 hours to finish the task. Failure to stop after the STOP command may result in zero points for the task.-Write answers and calculations within the designated box.-Use only the pen and the calculator provided.-There are 23 pages of Problems and 19 pages of Answer Sheet.-An English-language version is available.-You may go to the restroom with permission.-After finishing the examination, place all sheets including Problems and Answer Sheet in the envelope and seal.-Remain seated until instructed to leave the room.Constants and useful formulasGas constantR = 8.314 J K -1mol -1Faraday constant F = 96485 C mol -1Use as standard pressure:p = 1.013∙105PaUse as standard temperature: T = 25°C = 298.15 K Avogadro ’s number N A = 6.022∙1023mol -1Planck constant h = 6.626∙10-34J sSpeed of lightc = 3.00∙108 m s-1∆G = ∆H - T ∆S ∆G = - nFE∆G 0= - RT ∙lnK∆G = ∆G 0+ RT ∙lnQ with Q =)()(reactands c of product products c of product∆H(T 1) = ∆H 0+ (T 1 - 298.15 K)∙C p (C p = constant)Arrhenius equationk = A ∙TR E a e⋅-Ideal gas law pV = nRTNernst equationE = E 0 +redox c c ln nF RT⋅ Beer- Lambert Law A = logPP 0= ε∙c ∙dV(cylinder) = πr 2h A(sphere) = 4πr 2 V(sphere) = 34πr 31 J = 1 N m1 N = 1 kg m s -21 Pa = 1 N m -21 W = 1 A V = 1 J s -11 C = 1 A s1. Avogadro's number(5 pts)Spherical water droplets are dispersed in argon gas. At 27o C, each droplet is 1.0 micrometer in diameter and undergoes collisions with argon. Assume that inter-droplet collisions do not occur. The root-mean-square speed of these droplets was determined to be 0.50 cm/s at 27o C. The density of a water droplet is 1.0 g/cm3.1-1. Calculate the average kinetic energy (mv2/2) of this droplet at 27o C. The volume of a sphere is given by (4/3) π r3 where r is the radius.If the temperature is changed, then droplet size and speed of the droplet will also change. The average kinetic energy of a droplet between 0o C and 100o C as a function of temperature is found to be linear. Assume that it remains linear below 0o C.At thermal equlibrium, the average kinetic energy is the same irrespective of particle masses (equipartition theorem).The specific heat capacity, at constant volume, of argon (atomic weight, 40) gas is 0.31 J g-1 K-1.1-2. Calculate Avogadro's number without using the ideal gas law, the gas constant, Boltzmann’s constant).2. Detection of Hydrogen (5 pts)Hydrogen is prevalent in the universe. Life in the universe is ultimately based on hydrogen.2-1. There are about 1023 stars in the universe. Assume that they are like our sun (radius, 700,000 km; density, 1.4 g/cm3; 3/4 hydrogen and 1/4 helium by mass).Estimate the number of stellar protons in the universe to one significant figure.In the 1920s, Cecilia Payne discovered, by spectral analysis of starlight, that hydrogen is the most abundant element in most stars.2-2. The electronic energy of a hydrogen atom is given by -C/n2relative to zero energy at infinite separation between the electron and the proton (n is the principle quantum number, and C is a constant). For detection of the n=2→ n=3transition (656.3 nm in the Balmer series), the electron in the ground state of the hydrogen atom needs to be excited first to the n=2 state. Calculate the wavelength (in nm) of the absorption line in the starlight corresponding to the n=1→n=2 transition.2-3. According to Wien's law, the wavelength (λ) corresponding to the maximum light intensity emitted from a blackbody at temperature T is given by λT = 2.9×10-3 m K. Calculate the surface temperature of a star whose blackbody radiation has a peak intensity corresponding to the n = 1 → n = 2 excitation of hydrogen.The ground state of hydrogen is split into two hyperfine levels due to the interaction between the magnetic moment of the proton and that of the electron. In 1951, Purcell discovered a spectral line at 1420 MHz due to the hyperfine transition of hydrogen in interstellar space.2-4. Hydrogen in interstellar space cannot be excited electronically by starlight.However, the cosmic background radiation, equivalent to 2.7K, can cause the hyperfine transition. Calculate the temperature of a blackbody whose peak intensity corresponds to the 1420 MHz transition.2-5. Wien generated hydrogen ions by discharge of hydrogen gas at a very low pressure and determined the e/m(charge/mass) value, which turned out to be the highest among different gases tested. In 1919, Rutherford bombarded nitrogen with alpha-particles and observed emission of a positively charged particle which turned out to be the hydrogen ion observed by Wien. Rutherford named this p article the “proton”. Fill in the blank in the answer sheet.14N + 4He → ( ) + 1H3. Interstellar Chemistry (5 pts)Early interstellar chemistry is thought to have been a prelude to life on Earth. Molecules can be formed in space via heterogeneous reactions at the surface of dust particles, often called the interstellar ice grains (IIGs). Imagine the reaction between H and C atoms on the IIG surface that forms CH. The CH product can either desorb from the surface or further react, through surface migration, with adsorbed H atoms to form CH2, CH3, etc.Depending on how energetically a molecule “jumps” from its anchored site, it either leaves the surface permanently (desorption) or returns to a new position at the surface (migration). The rates of desorption and migratory jump follow the Arrhenius formula, k = A exp(-E/R T), where k is the rate constant for desorption or migratory jump, A the jumping frequency, and E the activation energy for the respective event.3-1. Desorption of CH from the IIG surface follows first-order kinetics.Calculate the average residence time of CH on the surface at20 K. Assume that A = 1 x 1012 s-1 and E des = 12 kJ mol-1.3-2. Consider the shortest time it would take for one CH unit to move from its initial position to the opposite side of an IIG by successive migratory jumps. Assume that the activation energy for migration (E mig) is 6 kJ mol-1, and the IIG is a sphere with a 0.1 μm radius. Each migratory jump laterally advances the molecule by 0.3 nm. Show work and choose your answer from (a)-(e) below.(a) t≤ 1 day (b) 10 day ≤t≤ 102 yr (c) 103 yr ≤t≤ 106 yr(d) 107 yr≤t≤ 1010 yr (e) t≥ 1011 yr3-3. Consider the reaction of CO with H2 to form H2CO. The activation energy on a metal catalyst is 20 kJ mol-1, which produces formaldehyde at a rate of 1 molecule/s per site at 300 K. Esitmate the rate of formaldehyde formation per site if the reaction takes place at 20 K.3-4. Which is a set of all true statements? Circle one.(a) Most CH species desorb from the IIG surface before encountering other reactants by surface migration.(b) IIGs can assist transformation of simple molecules to more complex ones in interstellar space.(c) For a reaction on the IIG to occur at an appreciable speed during the age of the Universe (1 x 1010 yr), the reaction energy barrier must be absent or negligible.(a) (b) (c) (a, b) (a, c) (b, c) (a, b, c)4. The Chemistry of DNA (5 pts)4-1. In 1944 Oswald Avery isolated a genetic material and showed, by elemental analysis, that it was a sodium salt of deoxyribonucleic acid. A segment of DNA with formula mass of 1323.72 is shown.Assuming that equimolar amounts of the four bases are present in DNA, write the number of H atoms per P atom. Calculate, to 3 significant figures, the theoretical weight percentage of H expected upon elemental analysis of DNA.4-2. Chargaff extracted the separated bases and determined their concentrations by measuring UV absorbance. The Beer-Lambert law was used to obtain the molar concentration. Chargaff discovered the following molar ratio for bases in DNA:adenine to guanine = 1.43 thymine to cytosine = 1.43 adenine to thymine = 1.02 guanine to cytosine = 1.02Chargaff’s discovery sugge sted that the bases might exist as pairs in DNA. Watson and Crick mentioned in their celebrated 1953 paper in Nature : "It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material."Draw structures of the specific pairing found in DNA. Indicate hydrogen bonds. Omit the sugar-phosphate backbone.4-3. Mutation can occur through base pairings different from the above. Draw structures of any three alternative base pairs.4-4. The plausibility of the formation of purine and pyrimidine bases in the prebiotic atmosphere of the Earth from HCN, NH 3, and H 2O has been demonstrated in the laboratory. Write the minimum number of HCN and H 2O molecules required for formation of the following compounds.adenineN NN N HNH 2guanine N NH N N HO NH 2Uracil N H NH O cytosineN H N NH 2OO5. Acid-Base Chemistry(5 pts)5-1. Calculate [H +], [OH -], [HSO 4-], and [SO 42-] in a 1.0 x 10-7 M solution of sulfuric acid (K w = 1.0 x 10-14, K 2 = 1.2 x 10-2 at 25o C). In your work you may use mass- and charge-balance equations. Answer with two significant figures.5-2. Calculate the volume of 0.80 M NaOH solution that should be added to a 250 mL aqueous solution containing 3.48 mL of concentrated phosphoric acid in order to prepare a pH 7.4 buffer. Answer with three significant figures. (H 3PO 4 (aq), purity = 85 % wt/wt, density = 1.69 g/mL, FW = 98.00) (p K 1 = 2.15, p K 2 = 7.20, p K 3 = 12.44). 5-3. The efficacy of a drug is greatly dependent on its ability to be absorbed into the blood stream. Acid-base chemistry plays an important role in drug absorption.Assume that the ionic form (A -) ofa weakly acidic drugdoes not penetrate the membrane, whereas the neutralform (HA) freely crosses the membrane. Also assume that equilibrium is established so that the concentration of HA is the same on both sides. Calculate the ratio of the total concentration ([HA] + [A -]) of aspirin (acetylsalicylic acid, p K = 3.52) in the blood to that in the stomach.+ Stomach pH = 2.0BloodpH = 7.4 H + + A-HAHAH + A-Membrane6. Electrochemistry (5 pts)Water is a very stable molecule, abundant on earth and essential for life. As such, water was long thought to be a chemical element. However, soon after the invention of a voltaic cell in 1800, Nicholson and Carlyle decomposed water into hydrogen and oxygen by electrolysis.6-1. Water can be thought of as hydrogen oxidized by oxygen. Thus, hydrogen can be recovered by reduction of water, using an aqueous solution of sodium sulfate, at a platinum electrode connected to the negative terminal of a battery. Thesolution near the electrode becomes basic. Write a balanced half-reaction forthe reduction of water.6-2. Water can also be thought of as oxygen reduced by hydrogen. Thus, oxygen can be recovered by oxidation of water at the Pt electrode connected to the positiveterminal. Write a balanced half-reaction for the oxidation of water.6-3. When copper is used at both electrodes, gas is generated only at one electrode during the initial stage of electrolysis. Write the half-reaction at the electrode thatdoes not generate gas.Another species in solution that can be reduced is sodium ion. The reduction of sodium ion to metallic sodium does not occur in aqueous solution, because water is reduced first. However, as Humphrey Davy discovered in 1807, sodium can be made by electrolysis of fused sodium chloride.6-4. Based on these observations, connect the half-reactions with the standard reduction potential (in volts).Reduction of copper ion (Cu2+) · -------------------- · +0.340Reduction of oxygen ·· -2.710Reduction of water ·· -0.830Reduction of sodium ion (Na+) ·· 0.000Reduction of hydrogen ion ·· +1.230The electrode potential is affected by other reactions taking place around theelectrode. The potential of the Cu2+/Cu electrode in a 0.100 M Cu2+ solution changes as Cu(OH)2 precipitates. Answer with 3 significant figures for the following problems.The temperature is 25o C. Note that K w = 1.00 x 10-14 at 25o C.6-5. Precipitation of Cu(OH)2 begins at pH = 4.84. Determine the solubility product of Cu(OH)2.6-6. Calculate the standard reduction potential for Cu(OH)2(s) + 2e-→ Cu(s) + 2OH-.6-7. Calculate the electrode potential at pH = 1.00.Lithium cobalt oxide and specialty carbon are active ingredients for the positive and negative electrodes, respectively, of a rechargeable lithium battery. During the charge/recharge cycles, the following reversible half-reactions occur.LiCoO2Li1-x CoO2 + x Li+ + x e-C + x Li+ + x e-CLi xThe total amount of energy a battery can store is rated in mAh. A battery rated at 1500 mAh can power a device drawing 100 milliamps for 15 hours.6-8. Graphite has lithium intercalation sites between its layers. Assuming a maximum 6:1 carbon-to-lithium intercalation stoichiometry, calculate the theoretical charge capacity of 1.00 gram of graphite to intercalate lithium. Answer in mAh/g with 3 significant figures.7. Hydrogen Economy (4 pts)Hydrogen is more energy-dense than carbon, by mass. Thus, historically there has been a move toward fuel with higher hydrogen content: coal → oil →natural gas → hydrogen. Cost-effective production and safe storage of hydrogen are two major hurdles to the successful inauguration of a hydrogen economy.7-1. Consider hydrogen in a cylinder of 80 MPa at 25 o C. Using the ideal gas law, estimate the density of hydrogen in the cylinder in kg/m3.7-2. Calculate the ratio between heat generated when hydrogen is burned and heat generated when the same weight of carbon is burned. The difference comes to a large extent from the fact that the most abundant isotope of hydrogen has noneutron and hydrogen has no inner electron shell. ∆H f o [H2O(l)] = -286 kJ/mol,∆H f o [CO2(g)] = -394 kJ/mol.7-3. Calculate the theoretical maximum work produced by the combustion of 1 kg hydrogen (a) from the electric motor using hydrogen fuel cell and (b) from the heat engine working between 25 o C and 300 o C. The efficiency (work done/heatabsorbed) of an ideal heat engine working between T cold and T hot is given by [1 –T cold/T hot].S o298[H2(g)] = 131 J/(K mol)S o298[O2(g)] = 205 J/(K mol)S o298[H2O(l)] = 70 J/(K mol).If the fuel cell is working at 1 W and the standard potential difference, how long will the electric motor run at what current?8. Chemistry of Iron Oxides (5 pts)The nucleus of iron is the most stable among all elements and, therefore, iron accumulates at the core of massive red giant stars where nucleosynthesis of many elements essential for life (such as C, N, O, P, S, etc.) takes place. As a result, among heavy elements iron is quite abundant in the universe. Iron is also abundant on Earth.8-1. Development of a technology for reducing iron oxide to iron was a key step in human civilization. Key reactions taking place in the blast furnace are summarized below.C(s) + O2(g) → CO2(g) ΔH◦ = -393.51 kJ(/mol) ----- ①CO2(g) + C(s) → 2CO(g) ΔH◦ = 172.46 kJ(/mol) ----- ②Fe2O3(s) + CO(g) → Fe(s) + CO2(g)ΔH◦ = ? ------------------- ③8-1-1. Indicate the reducing agent in each reaction.8-1-2. Balance reaction ③and calculate the equilibrium constant of reaction ③at 1200 o C. (ΔH f◦(Fe2O3(s)) = -824.2 kJ/mol, S°(J/mol/K): Fe(s) = 27.28, Fe2O3(s) =87.40, C(s) = 5.74, CO(g) = 197.674, CO2(g) = 213.74)8-2. In the manufacture of celadon pottery, Fe2O3 is partially reduced in a charcoal kiln to mixed oxides of Fe3O4and FeO. The amount of thedifferent oxides seems to be related to the“mystic” color of celadon ceramics.Fe3O4 (magnetite) itself is a mixed oxide containing Fe2+ and Fe3+ ions and belongs to a group of compounds with a general formula of AB2O4. The oxide ions form a face-centered cubic array. The figure shows the array of oxygens (gray circles) and representative sites for divalent A and trivalent B cations. The dark circle represents a tetrahedral site and the white circle an octahedral site.8-2-1. How many available octahedral sites for iron ions are there in one AB2O4 unit?Certain sites are shared by neighboring units.AB2O4 can adopt a normal- or an inverse-spinel structure. In normal-spinel structure, two B ions occupy two of the octahedral sites and one A occupies one of the tetrahedral sites. In an inverse-spinel structure, one of the two B ions occupies a tetrahedral site. The other B ion and the one A ion occupy octahedral sites.8-2-2. What percentage of available tetrahedral sites is occupied by either Fe2+or Fe3+ ion in Fe3O4?8-2-3Fe3O4 has an inverse-spinel structure. Draw the crystal field splitting pattern of Fe2+ and fill out the electrons. The electron pairing energy is greater than the octahedral field splitting.9. Photolithographic process (5 pts)Photolithography is a process used in semiconductor device fabrication to transfer a pattern from a photomask to the surface of a substrate. In a typical photolithography process, light is projected, through a mask that defines a particular circuitry, onto a silicon wafer coated with a thin layer of photoresist.9-1. The earliest photoresists were based on the photochemistry that generates areactive intermediates from bis(aryl azide). Patterning becomes possible through the cross-linking reaction of the nitrenes generated from the azides .N 3N 3Bis(aryl azide)reactive intermediate called as nitrene+ 2 N 2SO 3- Na ++Na -O 3S9-1-1. Draw two possible Lewis structures of CH 3-N 3, the simplest compound havingthe same active functional group of bis(aryl azide). Assign formal charges.9-1-2. Draw the Lewis structure of nitrene expected from CH 3-N 3.9-1-3. Draw the structures for two possible products, when this nitrene from CH 3-N 3reacts with ethylene gas (CH 2CH 2).9-2. Photoresists consisting of Novolak polymers, utilizes acid to change theirsolubility. The acid component can be produced photochemically from diazonaphthaquinone. In fact, “Novolaks” have been the representative “positive” photoresists of the modern microelectronic revolution.CH 3OHnNovolakWhen irradiated, diazonaphthaquinone undergoes photochemical decomposition followed by rearrangement eventually producing a carboxylic acid.ON 2S O OOR+ N 2+ H 2ODiazonahpthaquinonederivativeCO 2HS OOORcarbene intermediaterearranged intermediate9-2-1. Draw three Lewis structures of diazoacetaldehyde (see below), the simplestcompound having the same active functional group of diazonaphthaquinone. Indicate formal charges.H-C-CHN 2Odiazoacetaldehyde9-2-2. Draw a Lewis structure of the rearranged intermediate, A (see below),generated from diazoacetaldehyde after losing N 2. A satisfies Lewis’ octet rule and reacts with water to form acetic acid, CH 3CO 2H.HCHN 2O carbene intermediate_N 2ACH 3COOHH 2O9-3. Advanced photoresists were invented in 1982 based on chemical amplification.The most popular chemical amplification for positive-tone involves the acid catalyzed deprotection of poly(p -hydroxystyrene) resin protected by various acid-sensitive protecting groups such as t -butyloxycarbonyl (t -BOC).OnOThe thermal decomposition of carbonate ester itself normally occurs well above 150℃.9-3-1. Two plausible mechanisms have been suggested for this decompositionreaction having relatively high activation energy. Draw expected intermediates and products from this reaction.OOOCH 2H CH 3CH3OOOOH+O+BD++pericyclic trans. stateheterolytic cleavageE+B+H +OH+_CC9-3-2. In the presence of a trace amount of acid, the reaction temperature can bereduced to below 100℃. Draw expected intermediate F from the following chemical amplification process based on using t -BOC.O OOnH ++OHn+ H +OOHOnFDC+B10. Natural Products – Structural Analysis (9 pts)Licorice (Glycyrrhizia. Uralensis) Licorice RootThe flavor extracted from the licorice root is 50 – 150 times sweeter than table sugar. The most important and abundant compound responsible for the sweetness and medicinal effects of licorice is glycyrrhizin (C 42H 62O 16).Glycyrrhizin requires three equivalents of NaOH to effect neutralization.``When glycyrrhizin was subjected to acid hydrolysis, Glycyrrhizinic acid (A (C 30H 46O 4)) and B (C 6H 10O 7) were obtained in a 1:2 molar ratio (figure 1).hydrolysis, hydrolysis produced A’ (methyl glycyrrhizinate), C and D (figure 2). B, C and D exist as mixtures of anomers.Methylation of C and D with MeI produced the same isomeric mixture of compounds, J (figure 3.)C was reduced with LiAlH 4 to give K , and L was produced by the reduction of K . Oxidative cleavage of vicinal diol of L with NaIO 4 produced M and two equivalents of formaldehyde. Reduction of M produced N . The structure and stereochemistry of N was confirmed by the synthesis of N from D-(-)-tartaric acid through methylation followed by reduction (figure 4). A 1H-NMR spectrum of L showed two distinct peaks for methyl groups. (There is no symmetry in L)10-1. Complete structures for L , M, and N in the answer sheet.10-2. How many structures for C are possible? Complete possible structures for C .To determine the correct structure of C , following set of reactions were performed.J was reduced to E, and acid hydrolysis of E produced F . Reduction of F generated G, and G was oxidized with NaIO4 to H with formation of one equivalent of formaldehyde. I was obtained from H through reduction. Among all compounds from A to I , only I was optically inactive (figure 5).10-3. Complete structures for G and I .10-4. Which one is the correct structure for C among ones you have drawn in 10-2?10-5. Complete structures for B, D, and J .10-6. Complete the structure for Glycyrrhizin.11. Enzyme Reaction (7 pts)Shikimic acid biosynthesis is an important pathway for amino acids, alkaloids and heterocyclic natural product production. Nature converts shikimic acid to chorismic acid through a cascade of enzymatic reactions. Then chorismate mutase catalyzes the conversion of chorismic acid to prephenic acid at the branch point for the biosynthesis of aromatic amino acids such as tyrosine and phenylalanine.OCOOHChorismic AcidPrephenic AcidChorismate mutase2COOHOShkimic Acid11-1. During the transformation ofshikimic acid to chorismic acid, dehydrationisoccurring. Choose the hydroxyl group in shikimic acid that is lost through above dehydration among all possible reactions.11-2. Chorismate mutase rearranges chorismic acid into prephenic acid withoutchanging the molecular formula. Chorismic acid becomes prephenic acid through the Claisen rearrangement, a concerted pericyclic process like the Cope rearrangement as shown below:DD DDBased on the following spectral data, propose the structure of prephenic acid.1H-NMR (D 2O, 250 MHz): δ 6.01 (2H, d, J = 10.4 Hz), 5.92 (2H, dd J = 10.4, 3.1 Hz), 4.50 (1H, t, J = 3.1 Hz), 3.12 (2H, s). Note that there are three protons, which have been exchanged by D 2O very fast, and two protons at δ 3.12, which are exchanged slowly in prephenic acid. 13C-NMR (D 2O, 75 MHz): δ 203, 178, 173, 132 (for two identical carbons), 127 (for two identical carbons), 65, 49, 48.δ, chemical shift; H, integrals; d, doublet; dd, doublet of doublet; J, coupling constant; t, triplet; s, singletChorismate mutase is believed to stabilize the transition state of Claisen rearrangement. Thus it is an interesting target for inhibitor design. Inhibitors, called transition state analog (TSA)s that resemble the transition state (TS, e.g., the species in brackets “[ ]” above) of the reaction are designed to occupy the active site. Several inhibitors were designed and synthesized, and among them eight turned out to be potent inhibitors of the enzyme. The lower the IC 50 (inhibitor concentration of 50% of the enzymatic activity) value, the better the inhibitor.OHCO 2HCO 2HOHCO 2HCO 2HOHCO 2HCO 2H1IC 50 = 2.5 mM 2IC 50 = 1.3 mM3IC 50 = 0.78 mMOHO 2COHCO 2HOCO 2H OHCO 2HOHO 2COHCO 2H8IC 50 = 0.00015 mM6IC 50 = 0.017 mM 7IC 50 =0.0059 mM OHCO 2H4IC 50 = 1.1 mMHOCO 2HCO 2H5IC 50 = 5.3 mMHaHa11-3. Choose all correct statements based on the structures and IC 50 values of aboveinhibitors. Increase of factor 5 is considered to be important.(a) Configuration of the hydroxyl group plays an important role in the TS and inhibitor design.(b) The presence of both carboxylic groups is important in the TS and inhibitor design.(c) Transition state of the reaction contains two six-membered rings with one chair and one twist-boat conformation.(d) 7 and 8 can be distinguished on the basis of the 1H-NMR of H a .11-4. Draw the transition state of the transformation of chorismic acid to prephenic acid based on the TSA structures and their IC50 values.11-5. Compared with the uncatalyzed thermal conversion, chorismate mutase accelerates conversion of chorismic acid to prephenic acid 1.0 x 106 fold at 25o C by lowering the activation energy of the reaction. Calculate the decrease in activation energy of chorismate mutase at 25o C.∆H≠uncat is 86,900 J/mol for the thermal conversion of chorismic acid to prephenic acid. At what temperature will the rate of the uncatalyzed thermal conversion be the same as that of the enzyme-catalyzed conversion at 25o C, assuming that E a =∆H≠..。
1992年第24届IChO试题与答案
1992年第24届IChO试题---理论题和实验题注意事项（1）当监考人员将“开始（start）”写在黑板上后才开始答题。
（2）请用黑色笔（黑墨水笔或黑圆珠笔）答题。
（3）请在每页答卷的右上角写上你的参赛编号。
（4）请将计算过程和答案写在每题后面的空白处，勿写在纸的反面。
若需另页，可向监考人员索取。
原子量H C N O Na Si P1.008 12.01 14.01 16.00 22.99 28.09 30.97S Cl K Ca Cr Fe Ag32.06 35.45 39.10 40.08 52.00 55.85 107.9物理常数R＝8.314J·mol-1·K-1＝8.314m3Pa mol-1K-1F＝96486C·mol-11.硅藻（一种微生物）是海洋中丰富的食物来源，它们通过CO2和水的光合作用产生碳水化合物（即醣）：6CO2＋6H2O＋太阳能→C6H12O6＋6O2a）蓝鲸靠捕食一种称之为“Krill”的小鱼生长，在其头5年的生长期内体重每天增加75kg。
鲸每天吃掉的小鱼量是它自身体重增加量的10.0倍。
而小鱼（Krill）需吃掉10.0kg硅藻才增加1.0kg体重。
假定鲸在头5年内体重的增加完全是由于消耗硅藻碳水化合物，则为提供给该鲸在头5年内生长中所需要的硅藻碳水化合物，硅藻要消耗多少升CO2（0℃、1.0atm）？b）i）在24℃和1.0atm下，每升海水溶解0.23mlCO2，若硅藻能从水中把CO2全部除去，则必须处理多大体积的水才能产生蓝鲸在5年内所需要的碳水化合物？ii）假如海水的总体积是1.37×1018m3。
1000头蓝鲸在5年生长中所消耗的CO2（这些CO2是硅藻用来产生碳水化合物的）应该由这一总体积的百分之多少的水来溶解？c）成年鲸质量的3％是氮。
质量为9.1×104kg的鲸死亡后所产生的NH4+的最大质量是多少（这些NH4+可为水生物利用）？d）鲸的质量的18％是碳。
2020年338真题及答案
2020 年中国科学院大学《生物化学》考研真题答案一、名词解释1转导:是指细菌的基因通过噬菌体从供体转移到受体细胞的过程，是实现细菌基因转移的机制之一。
转导过程主要有普通性转导和局限性转导两种类型。
2.甘油磷脂:是指由甘油-3-磷酸衍生出的用于组成细胞膜结果的磷脂，也称磷、酸甘油酯。
例如磷脂酰乙醇胺、磷脂酰胆碱、磷脂酰丝氨酸等。
甘油磷脂具有两亲性结构，极性投的-端为亲水端，亲脂性为非极性的脂酰取代基端。
3.酶促反应的米氏方程:是指衡量底物浓度与酶促反应速率定量关系的方程式。
具体公式为V=Vmax[S]/Km+[S]。
4.AB0血型系统:是指一种根据红细胞表面的抗原决定簇命名的血型分类系统。
该系统下可将血型分为A、B、AB和O型四种，其中A型血的红细胞具有凝集原A，B型血具有凝集原B,AB型兼有凝集原A和B，O型血既不具有凝集原A，也不具有凝集原B。
5.X染色体失活:是指性染色体在参与生物的性别决定中存在的剂量补偿方式。
哺乳动物通常以该方式进行剂量补偿，将两个x染色体中一个关闭。
x染色体失活从x失活中心开始，然后拓展使x染色体大部分基因失活。
6.断裂基因:是指真核生物由于居间序列即内含子分隔而成的基因结构。
真核生物在DNA复制过程中保留内含子部分，转录后通过RNA剪接将内含子切除，并将外显子部分连接。
7.糖原：是指动物细胞内贮存的多糖形式，又称为动物淀粉，主要存在于肌肉组织和肝脏组织中。
与植物支链淀粉样，糖原是a-1,4 糖苷键连接的葡萄糖残基多聚物，其分支程度更高，此外还有a-1,6 糖苷键连接的分支。
8.必需脂肪酸: 是指必须由外界特别是植物性食物摄入的脂肪酸类型，主要包括亚麻酸和亚油酸。
二、选择题1、识别帽子结合活性: A.eIF4a B、eIF4b C、eIF4c D、eIF4d2、真核线粒体DNA复制酶3、SV40 猴子病毒A、它是环状单链dnaB、人先发现的，C其rna剪辑很简单D、与宿主的h2a, h2b, h3, h4结合。
第38届ICHO实验试题(中文版)
2006. 7. 5 Gyeongsan, Korea實作競賽Chemistry for Life, Chemistry for better Life一般規定●你有五小時來做完本實驗，請將時間做有效分配。
建議你用1小時於實作一(佔總分10分)，2小時在實作二(佔總分15分)，及2小時在實作三(佔總分15分)。
●每一頁的答案紙上都必須寫上你的名字和代碼。
●本試卷共有13頁的試題(和3頁附圖) 和7頁的答案紙。
●把你的答案和計算過程寫在規定的地方。
●只可以使用大會提供筆、尺和計算機。
●你可以要求英文題目。
●解釋光譜儀、C18 (逆相管柱) 和安全吸取管使用方法的圖形在另3頁紙上。
●若需要額外的藥品、試劑或玻璃器具需要扣分，一樣東西扣總分一分。
蒸餾水可以無限提供。
●若要上廁所，要告訴助理人員。
●做完後，將所有的紙張(題目和答案) 全部放進信封袋裡，並封好信封。
●請留在座位上，直到宣布可以離開。
●你可以將鉛筆盒、筆、尺、計算機和C18管柱帶回家。
安全和廢液●實驗室裡，一定要戴安全眼鏡。
●本實驗中沒有特別危險的藥品，所有的酸鹼都是稀釋的。
但若沾到，仍應立刻用溼的拭淨紙擦掉。
●不要聞任何藥品。
●將用過的藥品倒到貼有”DISPOSABLE”的白色塑膠瓶，將用過的試管或破玻璃丟到”WASTE BASKET”的籃子裡。
器材與藥品實作1, 2 (白色籃子內)實作3 (黑色籃子內)本實作中，有三樣未使用過之儀器，使用手續如下，若有任何疑問，都可請助教示範給你看。
如何使用光譜儀光譜儀分為三部份：光源、偵測器和樣品槽座。
你會發現樣品槽座處的蓋子(銀色部分)是打開的，在實驗中就讓它開著，不要蓋上。
有一塑膠樣品槽放在樣品槽座內，樣品槽有一面貼有標籤，此面應面向光源，在做實驗時，方向固定如此擺。
(塑膠樣品槽只有兩面是光學面，另兩面為較不透明的。
測光譜時一定要用光學面。
) 參考圖A。
光譜儀已全部準備好，可以使用。
用下列方法測光譜。
技能认证制氢高级考试(习题卷8)
技能认证制氢高级考试(习题卷8)说明：答案和解析在试卷最后第1部分：单项选择题，共45题，每题只有一个正确答案,多选或少选均不得分。
1.[单选题]若火苗过长,造成转化炉炉膛局部超温,应( )处理。
A)关小火嘴阀B)开大火嘴阀C)提高负荷D)降低负荷2.[单选题]分子式为C8H10的苯的同系物可能的结构式种类有( )。
A)2种B)3种C)4种D)5种3.[单选题]关于防止泵汽蚀的方法中，错误的一项是( )。
A)减小泵入口的阻力损失B)泵的吸入高度应按规定的数值选择C)选择耐汽蚀的材料制造零件时表面应具有足够的光洁度D)泵的入口阀不能开大4.[单选题]在脱吸塔底通入水蒸汽，与溶液逆流接触，将溶质从塔顶带走，从而分离吸收后的溶液，并得到回收的溶质，这一方法称为( )。
A)闪蒸B)汽提C)脱吸D)蒸发5.[单选题]下列器具中，从防止中毒的角度考虑，( )可适用于任何环境下使用。
A)空气呼吸器B)活性碳滤毒罐C)自然供风长管式防毒面具D)滤毒罐6.[单选题]下列物质与氧化剂发生反应的结果是( )。
A)SO3→H2SO4B)MnO2→MnSO4C)FeCl3→FeCl2D)Na2S→SC)酮类D)醛类8.[单选题]双作用往复压缩机的一个往复工作循环有( )膨胀、吸入、压缩、排出过程。
A)二次B)一次C)三次D)四次9.[单选题]中变催化剂还原是采用( )的热源升温还原的。
A)原料预热炉B)转化炉C)专门开工换热D)水蒸汽10.[单选题]停工过程中，如果转化炉循环气还有大量烃类，就停止进入蒸汽，会发生的现象是( )。
A)转化催化剂发生中毒B)转化催化剂发生结炭C)转化催化剂发生破碎D)锅炉发生超压11.[单选题]氧化锌与硫化氢的反应是一种( )过程。
A)吸收B)解吸C)转化D)变换12.[单选题]当油品的( )相近时，油品的特性因数K值大小的顺序为芳香烃＜环烷烃＜烷烃。
A)混合粘度B)临界温度C)相对分子量D)气化热13.[单选题]转化停止配汽必须在停止配氢后，转化炉入口温度在( )，且催化剂允许单独接触水蒸汽的时间内进行。
第49届ICHO预备题
49th IChO预备试题目录目录 (2)翻译说明 (3)序言 (4)致谢 (4)作者 (5)常用常数及公式 (6)元素周期表 (7)NMR化学位移 (8)红外特征吸收峰表 (9)国际三级大纲 (11)试题目录 (12)第一部分理论试题 (14)第二部分实验试题 (52)附录 (63)翻译说明本译稿采用知识共享署名–非商业性使用–相同方式共享3.0中国大陆许可协议[1]进行许可。
全体译稿作者保留追究此协议及相关法律许可内的一切权利。
总编常泰维*北京大学化学与分子工程学院陈胤霖北京大学化学与分子工程学院理论试题翻译耿景行北京大学化学与分子工程学院01–03何诗晴湖南师范大学附属中学04–06陈一乐湖南师范大学附属中学07–09黄纯熙北京大学化学与分子工程学院10–12吴宇飞西安高新第一中学13–15张宇婷北京大学化学与分子工程学院16–18王泽雨长沙市雅礼中学19–21刘程中香港科技大学化学系22–24常泰维北京大学化学与分子工程学院25–27王泽淳北京大学化学与分子工程学院28–30陈思聪清华大学化学系31–33实验试题翻译杨帆北京大学化学与分子工程学院01–03彭路遥北京大学化学与分子工程学院04–05图片编辑朱凯帝南京大学化学化工学院校对陈胤霖北京大学化学与分子工程学院彭路遥北京大学化学与分子工程学院朱凯帝南京大学化学化工学院1/licenses/by-nc-sa/3.0/cn/* ctw@作为科学委员会的代表，我很高兴为第49届国际化学奥林匹克竞赛提供预备试题。
本次预备试题会涵盖现代化学中一些比较具有挑战性的课题，并且这些课题都可以通过高中所学的基本化学知识和在「竞赛三级大纲」项目中明确列出六个三级大纲知识点来解决。
这些问题将会为参赛者的准备提供便利。
参赛者应该借助问题中三级大纲知识点的考查与应用方式来熟悉这些考点。
本书中列出的试题包括33个理论试题和5个实验试题。
希望您能利用好这些试题来为竞赛考试做准备。
IMDG CODE 38-16培训试卷-答案
IMDG CODE 38-16试卷部门：姓名：一、填空题：（共52分，2分/题）1、锂电池对应的EmS____________2、易燃液体是闭杯试验在________(相当于开杯试验65.6℃)或在60℃以下时放出易燃蒸汽的液体或液体混合物，或含有处于溶液中或悬浮状态的固体或液体（如油漆、清漆、真漆等，但不包括由于其它危险性已另列入其它类别中的物质），上述温度通常指闪点。
但本规则不适用于闪点高于_______且__________________的液体。
3、溶剂分离试验中，溶剂分离层的高度低于总高度的________；以及粘度试验中喷嘴直径为6mm时物质的流出时间等于或大于_______;或40S，如果该粘性物质所含________________.4、2016版《IMDG规则》中第4类物质包括______________________、________________、___________________、_____________________________、____________________________、______________________、_________________________。
5、易于燃烧的固体系指__________、粉末状、颗粒状或______________，如果该物质与燃烧的火柴等火源短暂接触时易于点燃且火焰迅速蔓延，该物质的为危险性不仅来自于_______，还可能来自于_________的燃烧产物。
6、自反应物质是____________________，即使没有_________（空气）参与也易产生强烈的放热分解。
7、自反应物质的分解可因___________、与催化性杂质（如酸、重金属化合物、碱）接触、________________或撞击产生。
8、聚合性物质是指在不添加___________的情况下，在正常运输条件下，已于发生强烈的放热反应，形成大分子或者____________的物质。
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(3)
Degree of ionization is increased upon dilution.
[H2SO4] = 0 Assume [H+]H2SO4 = 2 x 10-7
From (1), [SO42-]/[HSO4-] = 6 x 104 (2nd ionization is almost complete)
4-1.
Number of atoms
10 marks
Theoretical wt %
10 marks
H
P
( 11.3 )
1
( 3.43 )
4-2.
H N O HN
N
NH
N
NH
H guanine
N N
O
cytosine
HH N
N
N
NN
O
HN N
O
adenine
thymine
(10 marks on each)
4-3. 7 marks each, 20 marks for three
H
NH
O
N N
O
HN N
O
cytosine
thymine
H
N
O
HN
N
N
NH
N
NH
H
guanine
N
N
N
adenine
O
HN
N
O
N
N
O NH
N
NH
H
guanine
thymine
H NH
O N
N
N N
O
HN H
cytosine
cytosine
From (1), [HSO4-] = [H+][SO42-]/K2 = (2.4 x 10-7)(1.0 x 10-7)/(1.2 x 10-2) = 2.0 x 10-12 [5 marks]
Check charge balance: 2.4 x 10-7 (2.0 x 10-12) + 2(1.0 x 10-7) + (4.1 x 10-8)
30 marks
Avogadro’s number (NA) : Number of argon atoms in 40 g of argon
NA = (40)(1.4x1022) = 5.6 x1023
20 marks
1
T-2 Name: 1 ______________________
5 pts marks
[HSO4-] = 0
From (2), [SO42-] = 1.0 x 10-7
[5 marks]
From (3), [H+] = (2 x 10-7) + 10-14/[H+]
[H+] = 2.4 x 10-7
(pH = 6.6) [8 marks]
[OH-] = 10-14/(2.4 x 10-7) = 4.1 x 10-8 [2 marks]
Check mass balance: 0 + 2.0 x 10-12 + 1.0 x 10-7 1.0 x 10-7
Species HSO4SO42H+ OH-
Concentration 2.0 x 10-12 1.0 x 10-7 2.4 x 10-7 4.1 x 10-8
7
T-52
= (300 nm/0.3 nm) jumps x (5x103 s/jump) = 5x106 s = 50 days
marks
5 marks
15
(Full credit for the calculation using a random-walk model. In this case: t = τ (x/d) 2 = 5 x 109 s = 160 yr. The answer is still (b).)
30 30 25 15
Student code: ____________
3-1. kdes = A exp(-Edes/RT) = (1x1012 s-1)(5x10-32) = 5x10-20 s-1 at T = 20 K
10 marks
surface residence time, τresidence = 1 / kdes = 2x1019 s = 6x1011 yr
[H2SO4] = 0
2nd ionization: [H+][SO42-]/[HSO4-] = K2 = 1.2 x 10-2
(1)
Mass balance: [H2SO4] + [HSO4-] + [SO42-] = 1.0 x 10-7 (2)
Charge balance: [H+] = [HSO4-] + 2[SO42-] + [OH-]
a = KE/T = 6.9x10-21 J/(27+273K) = 2.3x10-23 J/K 25 marks
S: specific heat of argon
N: number of atoms in 1g of argon
S = 0.31 J/g K = a x N
N = S/a = (0.31 J/g K) / (2.3x10-23 J/K) = 1.4x1022
Student ___________________
5 pts marks
code: 20 20 20 20
4-4. 2.5 marks for each bracket
NH2
N
N
O
N
NH
O NH
NH2 N
N H
N
N H
N
NH2
NO H
NO H
adenine
guanine
Uracil
cytosine
H NH
N
N
N
N N
O
HN
N
H
cytosine
adenine
O
N
NH
O
O
HN
N
O
thymine
O
thymine
NH
O
N
O
HN
N
O
thymine
thymine
N
N
O
N
N
H
N
H
H
H
H
N
H N
N
N O
N
N
N
N
H
HN
N
N
NH
H
N
N
N
guanine
guanine
adenine
adenine
5
T-4 Name: 1 ________________________
10 marks
Average kinetic energy at 27oC:
KE = mv2/2 = (5.2x10-16 kg) (0.51x10-2 m/s)2/2
=
6.9x10-21
kg
m2/s2
=
6.9
x10-21
J
15 marks
1.21-2.
The average kinetic energy of an argon atom is the same as that of a water droplet.
number of protons of a typical star = (1.8×1033 g)(6×1023/g) = 1×1057
number of stellar protons in the universe = (1×1057)(1023) = 1×1080
Partial credits on principles: Volume = (4/3)(3.14)radius3×density; 4 marks 1 mole = 6×1023; 4 marks Total number of protons in the universe = number of protons in a star ×1023; 2 marks Mass fraction of protons from hydrogen = (3/4)(1/1); 5 marks Mass fraction of protons from helium = (1/4)(2/4); 10 marks
(full credit for τhalf-life = ln2 / kdes = 1x1019 s = 4x1011 yr)
20 marks
residence time =
2x1019 s
3-2.
The distance to be traveled by a molecule: x = πr = 300 nm.
HCN
( 5)
(5)
(4)
(4)
(2)
H2O (1)
(
0)
(
1
)
6
T-5 Name: 1 ______________________ Student code: ____________
5 pts marks 20 20 20
5-1. (20 marks)
1st ionization is complete: H2SO4 → H+ + HSO4-
(a)
(b)
(c)
10 marks
3-3.
k(20 K) / k(300 K) = exp[(E/R) (1/T1 - 1/T2)]
= e-112 = ~ 10-49 for the given reaction ).)