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。

最近十年全国化学竞赛决赛理论试题的分析及给我们的启示

最近十年全国化学竞赛决赛理论试题的分析及给我们的启示1 问题的提出自1984年中国化学会组织全国化学竞赛至今，已经开展了二十多年。

回顾这些年的历程可知：全国化学竞赛的发展经历了三个阶段[1]，且第三个阶段是从1995年肇始，延续至今。

第三阶段的具体举措是：在中国化学会的指导下，把从全国化学竞赛（冬令营）到组织参加国际化学奥赛（简称：IChO，下同！）等全部工作，由承办的省、市、自治区化学会具体组织实施。

因此，分析和研究这一时期在这一级别的赛题，无论是对了解各地竞赛水平和风格，还是总结和归纳近年这一级别竞赛试题考查的特点和热点，以及对竞赛培训方略和竞赛发展方向的导向作用等都有重要义。

本文，笔者正是基于这些方面的考虑，提出一些粗浅的意见，虽难免会挂一漏万，但旨在抛砖引玉，供大家研究，讨论。

2 近十年决赛理论试题的简析2.1 近十年决赛所在地的回顾要分析试题，首先对命题者进行分析是有必要的，因为不同的省市有着各自的特色，竞赛水平也各不相同，所以他们命制的试题在内容、形式、风格、试题质量等方面也是各不相同。

表1 十年全国高中化学竞赛决赛所在地一览表年份 1996 1997 1998 1999 2000 2001 举办地黑龙江大庆市甘肃兰州市河南郑州市福建厦门市浙江杭州市湖南长沙市2002 2003 2004 2005 2006山东济南市湖北武汉市广东广州市上海市上海市[注：感谢《化学之约》站长李德文（化学竞赛高级教练员）老师提供以上信息]2.2 近十年决赛试题的特点分析2.2.1 近十年试题整体评价参加这一级别的选手都是各省的优胜者，为了使全国决赛既符合国内中学化学教学的实际情况和选手水平，又要能初步做到与IChO初步接轨,所以近年竞赛试题的知识水平都是以《全国高中学生化学竞赛基本要求》(以当时最新的为准,因为从以前的《竞赛大纲》到现在的《基本要求》是几经修订的！)为主,以《IChO大纲》(同样是以当时最新的为准!)为辅来作为命题的依据。

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。

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）。

38届国际化学竞赛试题

问题1：宇宙生命的“简史”化学是生命的语言。

生命是以原子、分子和涉及原子分子间的复杂化学反应为基础的。

一个很自然的问题就是各种原子是从何而来。

按照一种被广泛接受的模型，宇宙起始于大约150亿年前的一次大爆炸。

宇宙的历史总体上讲可以被认为是当宇宙冷却时从简单到复杂的粒子的一系列的缩合。

就现在所知，生命是地球上在一定适中温度下发生的特殊现象。

轻元素，主要是氢元素和氦元素，形成于大爆炸的最初几分钟的迅速扩张，因而迅速地冷却为早期宇宙。

恒星是宇宙空间的特殊物体，因为在恒星形成的过程中温度不是下降而是升高的。

在化学上恒星是很重要的，因为重元素，尤其是构成生命的重元素都是在恒星内部超过数百万度的高温条件下形成的。

膨胀宇宙的温度可以用以下公式简单估计出来：T = 1010/t1/2。

其中T是宇宙的平均温度（K），t是时间（宇宙的年龄），以秒为单位。

回答问题1－1～问题1－6。

保留一位有效数字。

1－1 当宇宙年龄为1秒时，质子和中子因温度太高导致熔融，不能聚变成氦原子核。

估算此时的宇宙温度。

1－2 当宇宙年龄为3分钟时，氦原子核的合成近乎完全，估算此时的宇宙温度。

1－3 当宇宙的温度为3,000K时，由氢和氦的原子核与电子组合成第一个中性原子。

估算此时的宇宙年龄。

1－4 仅当膨胀的宇宙温度低到允许分子中原子保持键连状态时（约1,000K），宇宙中第一个稳定的分子才可能形成。

估算温度为1,000K时的宇宙年龄。

1－5 估算当宇宙年龄约为3亿年且第一个恒星和银河系诞生时的宇宙平均温度。

1－6 估算当前的宇宙温度，注意这就象宇宙的本底微波测量一样（3K）只是粗略的估计。

1－7 将以下缩合反应进行逻辑排序，保持其与在膨胀宇宙中超过99％的原子都是氢或氦原子的事实相一致。

a －( ) －( ) －( ) －( ) －( ) －( ) －( ) －( ) －( )a、夸克质子、中子b、1014个细胞人类c、H、C、N、O H2、CH4、NH3、H2O （在星际空间中）d、质子、氦原子核＋电子中性的H、He原子e、蛋白质、核酸、膜第一个细胞f、质子、中子氦原子核g、H2、He、CH4、NH3、H2O、粉尘太阳系h、H、He原子去离子作用、第一代恒星和银河系i、质子、氦原子核（轻元素）重元素如C、N、O、P、S、Fe、U；超新星爆炸j、H2、CH4、NH3、H2O等地球上的氨基酸、糖、核苷酸、磷脂问题2：星际空间中的氢元素氢元素是宇宙中最丰富的元素，宇宙中大约有75％以上的元素是氢元素，其余的主要是氦元素和少量的其他元素。

31届iCho

235 92 U
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≠..。

第49届(2017)年IChO试题及答案

*
G T* S H + + lower + higher + lower higher + + + lower + + higher + + lower + higher None of the above is correct
Relative to the initial temperature at the same partial pressure.
Part A. Propene can be synthesized via a direct dehydrogenation of propane in the presence of a heterogeneous catalyst. However, such a reaction is not economically feasible due to the nature of the reaction itself. Provide a concise explanation to each of the questions below. Additional information: Hbond(C=C) = 1.77Hbond(C-C), Hbond(H-H) = 1.05Hbond(C-H), and Hbond(C-H) = 1.19Hbond(C-C), where Hbond refers to average bond enthalpy of the indicated chemical bond. 1-A1) What is the enthalpy change of the direct dehydrogenation of propane? Show your calculation and express your answer in terms of Hbond(C-C). Calculation: ΔHrxn = -{Hbond(C=C)+ Hbond(C-C)+ 6Hbond(C-H)+ Hbond(H-H)} +{2Hbond(C-C)+8Hbond(C-H)} (2 points) = -{1.77Hbond(C-C)+ Hbond(C-C)+ 6(1.19Hbond(C-C)+ 1.05(1.19Hbond(C-C)} +{2Hbond(C-C)+8(1.19Hbond(C-C)} = +0.360Hbond(C-C) (2 points)

第38届国际化学奥林匹克实验试题

第３８届国际化学奥林匹克实验试题实验1反相色谱分离和分光光度分析用色谱法将物质分离后再进行分光光度分析是一种化学实验室中广泛使用的分析技术。

例如，复杂的有机混合物经常用反相液相色谱结合分光光度测定进行分析。

反相色谱是利用固定相材料（通常是C18）与被分析物中非极性部分之间的相互作用实现分离的。

色谱可以被简化，并且对有兴趣的化合物可通过选择适当的检测波长来选择性地测定。

在本实验中，将对经过分离和未经分离的染料进行分光光度分析。

1—1混合溶液中R和B的分光光度分析a)测定溶液R(3.02×10-5M)和溶液B(1.25×10-5M)的吸光度，将测定结果填入答卷的表格中。

用红色笔画出红染料的吸收光谱，用蓝色笔画出蓝染料的吸收光谱（图1—1-略）。

b) 溶液MD是溶液R和B按一定比例的混合溶液。

重复上面的操作，测量溶液MD的吸光度，并将所得的吸收光谱用黑色笔画到答卷的图中。

c)应用表中数据根据Beer-Lambert定理确定溶液MD中2种染料各自的摩尔浓度。

不能用1减去一种染料所占的分数来得到另一种染料所占的分数。

1—2色谱分离后进行分光光度分析a)使用10 mL的注射器吸取约10 mL溶液E淋洗柱子。

b)将1.00 mL溶液MD装载到柱子上。

c)使用1 mL的注射器吸取溶液E淋洗柱子。

通过外接的1个10 mL容量瓶收集洗出的溶液。

重复进行操作直到红色化合物完全被洗出并被收集。

d)将溶液E加入容量瓶中到10 mL刻度为止并摇匀。

标记该溶液为F。

e)如实验1-1那样测定溶液F的吸收光谱。

由于溶液在洗出过程中被稀释，因而在画溶液F的吸收光谱图时，用10乘以所测得的吸光度。

并且使用红色笔在答卷的图上用虚线画出它的吸收光谱。

f)为了分析溶液F中的红色染料R，按需要稀释溶液R，并以你所选择的波长做一条工作曲线。

在答卷上画出该工作曲线（在该图中，X轴表示浓度，Y轴表示吸光度）。

指出所选取的波长。

第38届中国化学奥林匹克初赛试题及答案

第38届中国化学奥林匹克(初赛)试题参考答案和评分标准(2024年9月1日9:00~12:00)提示：1) 试卷共8页。

2) 凡题目中要求书写反应方程式，须配平且系数为最简整数比。

3) 只有题1-3和题2-1-1的计算结果要求修约有效数字。

4) 每个解释题的文字不得超过20个。

5) 可能用到的常数：法拉第常数F = 9.6485×104 C mol -1;气体常数R = 8.3145 J K -1 mol -1阿伏加德罗常数N A = 6.0221×1023 mol -1;玻尔兹曼常数k B = R /N A缩写：Ac ：乙酰基；Ar ：芳基；Et ：乙基；DCM ：二氯甲烷；Me ：甲基；rt ：室温；TFAA ：三氟乙酸酐； tol ：对甲基苯基。

仅供参考！一切版权问题归中国化学会所有！第1题炼丹与化学 (22分)十六世纪一位托名为Basil V alentine 的炼金术士系统研究了制备“红龙血”的方法，并在他的著作中进行了详细记载。

后来，英国化学家Robert Boyle 验证了他的实验。

在Basil V alentine 的记载中，某种天然矿物因其颜色而被称为“灰狼”,加热熔融的“灰狼”可以“吞噬”金属铜，得到另一种灰白色金属和漂浮在熔融金属上的“矿渣”。

现代研究证明，“灰狼”和“矿渣”均为二元化合物，上述反应过程中只有金属的化合价发生了改变。

“矿渣”难溶于水和稀盐酸，其化学式中两种元素的计量比为1。

每得到1.000g 灰白色金属需要“吞噬”0.7826g 铜。

1-1写出“灰狼”A 、灰白色金属B 和“矿渣”C 的化学式。

A Sb 2S 3B SbC CuS 各2分，共6分。

1-2 Basil V alentine 还进行了后续实验：(i) “灰狼”可以提纯一种金黄色的金属“国王”。

在加热条件下，“灰狼”可“吞噬”“国主”，然后除去漂浮在熔融金属上的固体，高温加热剩下的物质；如此重复三次便可得到“经过救赎的国王”。

31ICho理论试题

第31届国际化学奥林匹克竞赛理论试题中文译本译者注：注意事项（略）；竞赛时限：5小时；试题分题面与答卷两种，答卷略，题面发稿时已作压缩处理。

为正确把握国际竞赛试题的水平必须对照竞赛前由命题委员会下发的预备题。

本届国际竞赛试题所有超过竞赛大纲一、二级水平的知识点均在预备题中反映了。

凡具有竞赛大纲一、二级知识水平的选手在学习预备题之前不必预先掌握这些三级知识点，在竞赛前的半年中完全可以按预备题的指示达到这个水平。

如果没有预备题，参加竞赛的选手能够完满地做出这种竞赛试题是难以想象的.也是毫无必要的。

因此，以为这份国际竞赛试题反映的知识水平是我国全国竞赛决赛前各省市自治区选手就应该达到的知识水平，是一种误解。

我们刊发国际竞赛试题决不愿造成这种误解。

第一题化合物Q (摩尔质量 122.0 g mol -1) 含碳、氢、氧。

A 分题CO 2(g) 和 H 2O(l) 在 25.00 o C 的标准生成焓分别为 –393.51和–285.83 kJ ·mol -1。

气体常数R 为 8.314 J ·K -1·mol -1。

(原子量: H = 1.0, C = 12.0, O = 16.0)质量为0.6000g 的固体Q 在氧弹量热计内充分燃烧,在25.000o C 下开始反应, 此时量热计内含水710.0 g 。

反应完成后温度上升至27.250 o C, 反应生成1.5144 g CO 2 (g) 和 0.2656 g H 2O(l)。

问题1-1 确定Q 的分子式，并写出Q 燃烧反应的配平的化学方程式（标明物质状态）。

问题1-2 若水的比热为4.184 J ·g -1·K -1，反应内能变化(∆U o )为–3079kJ ·mol -1，请计算量热计的热容（不包括水）。

问题1-3 计算Q 的标准生成焓（∆H f o ）。

B 分题在6o C 时Q 在苯和水中的分配情况见下表，表中C B 和C w 分别表示Q 在苯和水中的平衡。

38届化学竞赛试卷

38届化学竞赛试卷一、选择题（每题3分，共30分）1. 下列物质在常温下为液态且不溶于水的是（）A. 乙醇B. 苯C. 乙酸D. 葡萄糖。

2. 下列关于化学键的说法正确的是（）A. 离子键就是阴阳离子间的静电引力。

B. 共价键只存在于共价化合物中。

C. 非极性键只能存在于单质分子中。

D. 化学反应的实质是旧化学键的断裂和新化学键的形成。

3. 设N_A为阿伏伽德罗常数的值。

下列说法正确的是（）A. 1mol Fe与足量的稀硝酸反应，转移电子数为2N_AB. 标准状况下，22.4LSO_3所含分子数为N_AC. 常温常压下，18gH_2O中含有的氢原子数为2N_AD. 0.1mol/L的CaCl_2溶液中Cl^-的数目为0.2N_A4. 下列离子方程式书写正确的是（）A. 碳酸钙与盐酸反应：CO_3^2 - + 2H^+=H_2O + CO_2↑B. 铁与氯化铁溶液反应：Fe + Fe^3 + = 2Fe^2 +C. 氢氧化钡溶液与稀硫酸反应：Ba^2 + +OH^-+H^++SO_4^2 - =BaSO_4↓+H_2OD. 铜与稀硝酸反应：3Cu + 8H^++2NO_3^-=3Cu^2 + +2NO↑+4H_2O5. 下列实验操作能达到实验目的的是（）A. 用排水法收集NO气体时，导管口刚有气泡冒出就开始收集。

B. 用酒精萃取碘水中的碘。

C. 用托盘天平称取10.58gNaCl固体。

D. 用pH试纸测定新制氯水的pH6. 在一定温度下，反应A(g)+3B(g)⇌ 2C(g)达到平衡的标志是（）A. 单位时间内生成nmolA的同时生成3nmolBB. A、B、C的浓度不再变化。

C. A、B、C的分子数之比为1:3:2D. v(A):v(B):v(C)=1:3:27. 下列关于元素周期表和元素周期律的说法正确的是（）A. 同周期主族元素从左到右，原子半径逐渐增大。

B. 同主族元素从上到下，金属性逐渐减弱。

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。

“Start (开始)”命令发出后，即可开始答题。

考试时间共 5 小时。

所有结果和答案必须用笔清楚地书写在考卷指定的相应答题框中。写在答题框外的答案不予评
判。

若需打草稿，可以利用试卷的背面。但谨记：写在指定区域之外的解答不予评判。
: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
这次比赛。经过各 5 小时的实验和理论考试，共决出 196 枚奖牌——金牌
37 枚，银牌 64 枚，铜牌 95 枚。中国代表队何流(浙江杭州学军中学)、邓
子杰(河北衡水第一中学)、杨景程(湖南长沙市第一中学)、陈劭炎(山东省
广饶县第一中学) 4 名选手发挥出色，获得三金一银的优异成绩。其中，何
流总分第一、实验(满分)与另两位选手并列第一，邓子杰总分第二、理论成绩第一，获得国际纯粹与应用化学联合
教授观摩比赛并为申办工作提供指导；南开大学朱守非教授、梁广鑫教授和郭东升教授观摩比赛并代表中国化学会
向国际化学奥赛组委会提出申请并进行答辩，为我国获得了 2022 年 IChO 的举办权。
兹将本届国际化学奥林匹克理论试题及实验试题中文版刊出，以飨读者。为最大程度展示试题原貌，编辑保持

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。

第38届中国化学奥林匹克竞赛初赛试题

选择题下列关于原子结构的叙述中，正确的是：A. 原子核由质子和电子组成B. 原子序数等于原子核中的中子数C. 电子在核外分层排布，形成电子云D. 所有原子的最外层电子数都达到8个才稳定（C（正确答案））下列哪一项不是化学反应速率的影响因素？A. 反应物的浓度B. 反应温度C. 催化剂的使用D. 反应物的颜色（D（正确答案））下列关于化学键的叙述，错误的是：A. 共价键是通过共用电子对形成的B. 离子键是阴、阳离子之间的静电作用C. 金属键是金属原子内自由电子与阳离子形成的D. 氢键是一种比共价键更强的化学键（D（正确答案））下列物质中，属于非电解质的是：A. 硫酸B. 氢氧化钠C. 蔗糖（正确答案）D. 醋酸下列关于配位化合物的描述，正确的是：A. 配位化合物中，中心原子只接受来自配体的σ键B. 配体只能是中性分子C. 中心原子与配体形成的化学键是纯粹的共价键D. 配位化合物的稳定性与中心原子的电子构型有关（D（正确答案））下列哪个过程不涉及氧化还原反应？A. 金属的腐蚀B. 燃烧反应C. 酸碱中和反应（C（正确答案））D. 电池的放电过程下列关于分子间作用力的说法，正确的是：A. 分子间作用力是化学键的一种B. 分子间作用力随分子间距离的增大而增强C. 分子间作用力只存在于极性分子之间D. 分子间作用力包括范德华力和氢键（D（正确答案））下列关于电解质溶液导电性的说法，错误的是：A. 电解质溶液的导电性是由自由移动的离子提供的B. 电解质溶液的浓度越大，导电性越强（在一定范围内）C. 电解质溶液的温度升高，导电性一定增强D. 相同条件下，强电解质溶液的导电性一般优于弱电解质溶液（C（正确答案））下列关于化学平衡的描述，正确的是：A. 化学平衡是静态平衡，反应物和生成物的浓度不再变化B. 改变反应条件，化学平衡一定会发生移动C. 化学平衡常数K只与反应本身的性质和温度有关D. 在化学平衡状态下，正反应速率和逆反应速率都为零（C（正确答案））。

第38届化学奥林匹克试题

选择题
下列哪项不是化学反应速率的影响因素？
A. 反应物浓度
B. 反应温度
C. 催化剂的使用
D. 反应物的颜色（正确答案）
在实验室中，下列哪项操作是正确的？
A. 用嘴吹灭燃着的酒精灯
B. 将实验剩余药品放回原瓶
C. 用灯帽盖灭燃着的酒精灯（正确答案）
D. 随意丢弃废旧电池
下列关于原子结构的说法中，错误的是？
A. 原子由原子核和核外电子构成
B. 原子核由质子和中子构成
C. 核外电子在核外分层排布
D. 原子中质子数一定等于中子数（正确答案）
下列关于化学平衡的说法中，正确的是？
A. 化学平衡是静态平衡，反应物和生成物的浓度不再变化（正确答案）
B. 化学平衡时，反应物和生成物的浓度一定相等
C. 改变反应条件，化学平衡不会移动
D. 化学平衡时，正反应速率和逆反应速率均为零
下列哪种元素不属于主族元素？
A. 钠
B. 铁（正确答案）
C. 氯
D. 硫
下列关于氧化还原反应的说法中，正确的是？
A. 氧化还原反应中，元素化合价一定发生变化
B. 氧化还原反应中，一定有氧元素参与
C. 氧化还原反应中，氧化剂和还原剂一定是不同物质
D 在氧化还原反应中，得到电子的物质被还原（正确答案）
下列关于化学实验基本操作的说法中，错误的是？
A. 过滤时，液面要低于滤纸边缘
B. 蒸发时，要用玻璃棒不断搅拌防止液体飞溅
C. 分液时，下层液体要从分液漏斗的下口放出
D. 萃取时，萃取剂的选择对萃取效果无影响（正确答案）。

生物化学测试题+答案

生物化学测试题+答案一、判断题（共100题，每题1分，共100分）1.原油中所含的无机盐类大部分溶解在水中，也有少量以微小结晶状态悬浮在原油中。

A、正确B、错误正确答案：A2.换热器实际采用的传热面积和计算得到的传热面积相等就可以。

A、正确B、错误正确答案：B3.Word中插入的文本框通过设置文本框格式当中的版式允许在文字的上方或下方，A、正确B、错误正确答案：A4.采取隔离措施，实现微机控制，不属于防尘防毒的技术措施A、正确B、错误正确答案：B5.铬法测定COD的第四测定步骤是加硫酸银。

A、正确B、错误正确答案：B6.在重铬酸钾法测水中COD的加热过程中，溶液颜色由黄变绿，说明重铬酸钾溶液过量。

A、正确B、错误正确答案：B7.反硝化过程中硝酸盐的转化是通过反硝化细菌的同化作用和异化作用来完成的.A、正确B、错误正确答案：A8.化工设备图的视图布置灵活，俯左)视图可以配置在图面任何地方，但必须注明""俯左)视斟”的字样。

A、正确B、错误正确答案：A9.从结构上看，PLC是固定式的。

A、正确B、错误正确答案：B10.反渗透设备中使用的膜称为半透膜，它只允许溶质通过，不允许水通过。

A、正确B、错误正确答案：B11.反渗透和超滤都是以压力为驱动力的。

A、正确B、错误正确答案：A12.干燥进行的必要条件是物料表面的水气或其他蒸气)的压大于干燥介质中水气或其他蒸气) 的分压。

A、正确B、错误正确答案：A13.氧化塘是以太阳能为初始能量，通过稳定塘中多条食物链的物质迁移、转化和能量的逐级传递、转化，将进入塘中污水的有机污染物进行降解和转化。

A、正确B、错误正确答案：A14.好氧消化为放热反应，池内温度要稍高于入池污泥温度•A、正确B、错误正确答案：A15.用浮筒液面计测量液位，当浮筒脱落时，指示最小A、正确B、错误正确答案：B16.在HSE管理体系中,一般不符合项是指个别的、偶然的、孤立的、性质轻微的问题或者是不影响体系有效性的局部问题。