第46届国际化学奥林匹克竞赛预备题中文译本
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。
中国化学奥林匹克竞赛预赛试题及参考答案
中国化学奥林匹克竞赛浙江省预赛试题一、选择题（本题包括10小题，每小题4分，共40分。
每小题只有一个选项符合题意）1．中国药学家屠呦呦因发现青蒿素及其抗疟疗效，荣获2015年诺贝尔生理学或医学奖，成为在我国本土首位获得科学类诺贝尔奖的女科学家。
青蒿素结构式如下图所示，下列有关青蒿素研究的说法不正确的是A．提取过程中为防止破坏青蒿素结构，应避免高温，故采用低沸点溶剂乙醚进行萃取B．青蒿素是脂溶性的，既可看作是醚类也可看作是酯类，既有氧化性又有还原性C．可使用红外光谱仪测出分子中可能的官能团，也可通过核磁共振技术检测分子中C、H原子所处的化学环境D．可使用质谱仪测出这个分子的相对分子质量，也可用紫外光谱确定这个分子的环状结构2．2004年，英国曼彻斯特大学物理学家安德烈·盖姆和康斯坦丁·诺沃肖洛夫，成功从石墨中分离出石墨烯，两人因此共同获得2010年诺贝尔物理学奖。
石墨烯被称为“黑金”，是“新材料之王”，科学家甚至预言石墨烯将“彻底改变21世纪”，极有可能掀起一场席卷全球的颠覆性新技术产业革命。
石墨烯结构如右图所示，下列说法不正确的是A．石墨烯是碳原子以sp2杂化构成的六边形蜂巢形结构、只有一层原子厚度的二维晶体，碳原子排列与石墨的单原子层相同，是平面多环芳烃，而碳纳米管就是石墨烯卷成了筒状B．石墨烯中如果有五边形和七边形存在，则会构成缺陷。
含12个正五边形的石墨烯则转化形成了富勒烯C60，它是足球分子C．石墨烯中碳碳键的键能大于金刚石中碳碳键的键能，是目前世界上最薄也是最强韧的纳米材料D．石墨烯是导电性最好的新型纳米材料，在微电子领域有巨大的应用潜力，有可能成为硅的替代品，用来生产未来的超级计算机3．好莱坞影片《终结者》中始终不会被击败的液态金属机器人，在被散弹枪和手榴弹击中后，既能像液体一样重新聚集在一起，恢复之前的结构，又能改变外形，呈现出各种造型。
2015年3月26日清华大学宣布，该校刘静教授等人联合中国科学院理化技术研究所研发出世界首个自主运动的可变液态金属机器（AdvancedMaterials,2014,26(34) 6:025）。
国际化学奥林匹克竞赛-40thIchotheoreticalanswers
Yes, but only in quite dilute solutions can this happen. 1 pt for ticking yes
c = [HA] + [A–] = [H+]
(1 pt)
[H+] = [A–] + [OH–]
(1 pt)
This means that [HA] = [OH–]
(1 pt for reasonable guess – between 6 and 7)
A good approximation is: [H+ ] = 3 (KKw )
The full equation can be solved through iteration: [H+ ] = 3 (K + [H+ ])Kw
In equilibrium constant calculations all concentrations are referenced to a standard concentration of 1 mol/dm3. Consider all gases ideal throughout the exam.
4
Name:
Code: XXX-
c)
Could it be possible that the solution contained acetic acid?
Acetic acid: pKa = 4.76
Yes No
If yes, calculate the pH (or at least try to estimate it) and show your work.
Official English version
第 届国际化学奥林匹克竞赛中文版理论试卷
2018年7月19 – 29日布拉迪斯拉发斯洛伐克布拉格捷克理论试卷第50届 IChO 2018国际化学奥林匹亚斯洛伐克和捷克回到一切开始的地方BACK TO WHERE IT ALL BEGANCHN-1国际化学奥林匹亚/斯洛伐克和捷克 2018目录规则说明 (1)物理常数和公式 (2)第1题 DNA (4)第 2题中世纪遗骸的归国 (10)第3题新兴的电动汽车 (18)第4题放射性铜的柱层析 (25)第5题波希米亚石榴石 (30)第6题让我们一起去采蘑菇 (36)第7题西多福韦 (41)第8题石竹烯 (48)规则说明∙本理论考卷共55页。
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物理常数和公式Avogadro's constant（阿佛加德罗常数）: N A = 6.022 × 1023 mol−1Universal gas constant（普适气体常数）:R = 8.314 J K−1 mol−1 Speed of light（光速）: c = 2.998 × 108 m s−1 Planck's constant（普朗克常数）: h = 6.626 × 10−34 J s Faraday constant（法拉第常数）: F = 9.6485 × 104 C mol−1 Standard pressure（标准压力）: p = 1 bar = 105 Pa Normal (atmospheric) pressure（正常大气压）:p atm = 1.01325 × 105 Pa Zero of the Celsius scale（零摄氏度）: 273.15 KMass of electron（电子质量）: m e= 9.109 × 10−31 kg Unified atomic mass unit（原子质量单位）:u = 1.6605 × 10−27 kgÅngström（埃）: 1 Å = 10−10 m Electronvolt（电子伏特）: 1 eV = 1.602 × 10−19 J Watt（瓦特）: 1 W = 1 J s−1Ideal gas equation（理想气体方程）: pV = nRTThe first law of thermodynamics（热力学第一定律）:ΔU = q + WPower input for electrical device（电子设备的输入功率）: P = U Iwhere U is voltage and I electric current（U表示电压，I表示电流）Enthalpy（焓）: H = U + pVGibbs free energy（吉布斯自由能）: G = H – TSΔG o = – RT ln K= – zFE celloΔG = ΔG o + RT ln QReaction quotient Q（反应商）for a reaction a A+ b B⇌c C+ d D:Q =[C]c[D]d[A]a[B]bEntropy change （熵变）:ΔS =q rev Twhere q rev is heat for the reversible process （q rev 指可逆过程的热量）Heat changefor temperature-independent c m （不随温度变化的热量变化）:Δq = nc m ΔTwhere c m is molar heat capacity （c m 是摩尔热容）Van ’t Hoff equation （范特霍夫方程）: d ln K d T = Δr H m RT 2⇒ln (K 2K 1) = –Δr H m R (1T 2 –1T 1) Henderson –Hasselbalch equation （亨德森-哈塞尔巴赫方程）:pH = p K a + log[A –][]Nernst –Peterson equation （能斯特-彼得森方程）:E = E o –RTzFln Q Energy of a photon （光子能量）: E =hc λRelation between E in eV and in J （E 用 eV 和 J 做单位的换算关系）: E eV ⁄ = E J ⁄q e C ⁄Lambert –Beer law （比尔-朗伯定律）:A = logI 0I= εlc Wavenumber （波数）:ν̃ = νc = 12πc √k μReduced mass µ for a molecule AX （分子AX 的折合质量µ）: μ =m A m X m A + m XEnergy of harmonic oscillator （谐振子的能量）:E n = hν (n +12)Arrhenius equation （阿伦尼乌斯方程）: k = A e − E aRTRate laws in integrated form （速率的积分表达式）:Zero order （零级反应）: [A] = [A]0 – kt First order （一级反应）: ln[A] = ln[A]0 – kt Second order （二级反应）:1[A] = 1[A]0+ kt第1题 DNA回环（Palindromic ）序列是DNA 的一种有趣特征。
第46届国际化学奥林匹克竞赛准备题中文译本
第 25 题：头孢噻吩 第 26 题：乙酰丙酸 第 27 题：莲 第 28 题：NMR 谱 第 29 题：IR 谱 第 2 部分：实验试题 第 30 题：香兰素和苄胺的缩合反应 第 31 题：丁香酚氧基乙酸的合成 第 32 题：水溶液中铁、铝、镁的络合滴定 第 33 题：ZnO 粉末中 Zn 和 Pb 含量的测定 第 34 题：乙酰丙酮合铜(II)的制备 第 35 题：阿司匹林水解的动力学分析 第 36 题：水杨酸与 Fe3+配合物的形成
50 51 52 54 55 57 57 60 63 66 70 73 78
进阶知识
理论部分
动力学：一级反应与二级反应的速率方程积分式；运用稳态近似法分析简单的复杂反应； Arrhenius 方程的应用。 热力学：化学电池；平衡常数、电动势、Gibbs 自由能之间的关系；平衡常数与温度的定量 关系。 量子力学：箱中粒子相关计算。 光谱学：红外光谱的解读；简单的 1H，13C，27Al NMR 谱的解读；NMR 相关概念：化学位 移，偶合与裂分，偶合常数及积分曲线。 高等无机化学：反位效应；运用简单晶体场理论解释并推测四面体配合物与八面体配合物中 电子排布；使用唯自旋方程（ n n 2 ）计算分子磁矩。
ql
偶极矩是一矢量，由正电荷中心指向负电荷中心。
偶极矩常用单位是 Debye (D)。在 SI 单位制中 Debye (D)与库仑米(C•m) 的转换关系为：
1 D 3.33 1030 C m 。
1. 分子偶极矩与分子几何结构紧密相关。将一个多原子分子中各个键的偶极矩进行一定的
能力） 4. 气态时水分子和甲醚分子的净偶极矩分别为 1.84 D 和 1.29 D。水分子中两个 O-H 键的键
第44届国际化学奥林匹克实验试题中文版0
一般步驟
量取你所選擇之體積的鹽酸、蒸餾水和三碘化鉀(標示為 “I2”) 溶液置於反應瓶中。反應 混合物之各試劑的初濃度須在以下之範圍內 (你不一定要涵蓋整個給定的範圍，但你的數值 不可以明顯超出這些範圍)： [H+]：0.2～1.0 M 之間 [I3−]：0.0005～0.002 M 之間 [acetone]：0.5～1.5 M 之間
你只能在提供給你的範圍內工作。
只能用大會提供給你的筆作答 (不可以用鉛筆作答)。
只能用大會提供給你的計算機。
必須在規定的範圍內作答，超出規定範圍的部份將不予給分！ 若有需要，可以使用紙張的背面當計算紙用。
請將廢棄的樣品瓶丟入標示 “Broken Glass Disposal” 的容器中。
你只能在限定的位置或範圍內工作。 限用大會提供給你的筆作答(不可以用鉛筆作答)。
限用大會提供給你的計算機。
必須在規定的範圍內作答，超出規定範圍的部分將不予給分！ 但你可以使用紙張背面空白的部分來計算。
請將裝有反應溶液之密封反應瓶丟入標示 “Used Vials” 的容器中。 請將廢液倒入標示 “Liquid Waste” 的容器中。 請將廢棄的棕色瓶各部位丟入標示 “Broken Glass Disposal” 的容器中。
器材 − Kit #1
玻璃瓶一個，已裝蒸餾水供實驗用 20 mL 具旋轉瓶蓋之玻璃血清瓶，15 個 1-mL 具有 0.25 mL 刻度之塑膠(PE) 吸管，10 個 3-mL 具有 0.50 mL 刻度之塑膠(PE) 吸管，10 個 碼錶一個
1.00 mL 0.75 mL 0.50 mL 0.25 mL
1
2
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）。
2014年第46届国际化学奥林匹克竞赛实验题及解答和评分标准(英文)
46th International Chemistry OlympiadJuly 23, 2014Hanoi, Vietnam PRACTICAL EXAMINATIONCountry:Name as in passport:Student Code:Language:This booklet contains 25523 charactersGENERAL INTRODUCTIONSafety•Safety is the most important issue in the laboratory. You are expected to follow the safety rules given in the IChO regulations. Safety glasses and lab coats must be worn in laboratory ALL TIMES.•If you behave in an unsafe manner, you will receive one warning before you are asked to leave the laboratory. If required to leave due to a second warning, you will receive a score of zero for the rest practical examination.•Eating, drinking, or smoking in the laboratory or tasting a chemical is strictly forbidden.•Pipetting by mouth is strictly forbidden.•Use the labeled waste containers near you for disposal of liquids and solids. A waste container (plastic can) is also available on each bench for organic and inorganic waste.Discard used glass capillaries into a solid trash.•In case of emergency, follow the instructions given by the lab assistants. Examination Procedures•This practical examination has 28 pages for 3 practical problems. Periodic Table of Elements is at the end of this booklet. Do not attempt to separate the sheets.•You have 5 hours to complete practical problems 1, 2, and 3. You have 30 min to read through the problems before the START command is given.•DO NOT begin working on the tasks until the START command is given.•When the STOP command is given, you must stop your work on the tasks immediately.A delay in doing so may lead to your disqualification from the examination.•After the STOP command has been given, wait in your lab space. A supervisor will check your lab space. The following items should be left behind:o The practical examination booklet (this booklet),o Your chosen TLC plates in Petri dish with your student code (Problem 2).•Do not leave the laboratory until you are instructed to do so by the lab assistants. •You may need to reuse some glassware during the examination. If this is the case, clean it carefully in the sink closest to you.•Replacement of chemicals and laboratory ware will be provided if necessary. Other than the first, for which you will be pardoned, each such incident will result in the loss of1 point from your 40 practical points. Refilling of wash-bottle water is permitted with noloss of points.Notes•Use only the pen provided for filling in the answer boxes. You may also use the calculator and the ruler provided. Do not use the mechanical pencil for filling in the answer boxes.•All results must be written in the appropriate areas with the working shown. Results written elsewhere will not be graded. If you need to do rough calculations, etc., use thedraft papers or the back of the sheets. All answers on the draft papers or the back of the sheets will NOT be graded .• You should take care to report answers to an appropriate number of significant figures and give the appropriate unit.• Contact a supervisor near you if you need a refreshment/toilet break. • Read the whole description of the problems before you begin•An official English version of this examination is available upon request if you require clarification.Attention:Pipetting by mouth is strictlyforbidden. Student was provided a pipettebulb. Make sure that you properly use thepipette bulb shown in Figure below.Description of three-way pipette bulb.An adapter is provided for larger pipettes. Instructions for using the thermometer 1. Press the [ON/OFF] button to display thetemperature reading in Celsius. 2. Insert the stainless steel probe (at least 5cm) in the solution to be measured. 3. Wait for display to stabilize (display valueis unchanged and stable for 3 seconds) and read the temperature on the display.4. Press the[ON/OFF]button again to turnthe thermometer off, then rinse the stainlesssteel probe with distilled water.List of chemicalsThe concentration indicated on the label is approximate. The exact values are indicated in the table.Chemical/Reagent QuantityPlacedin Labeled Safety Practical Problem 10.100 M KI solution 120 mL Glass bottle 0.1 M KI H320Solution #A1 contains KI,Na2S2O3, and starch indicator in distilled water 40 mL Glass bottle Solution #A1H314, H302,H315, H319Solution #B1 contains Fe(NO3)3, HNO3 in distilled water 40 mL Glass bottle Solution #B1H314, H315,H319, H335Solution #A2-1 contains 5.883×10–4 M Na2S2O3, KNO3, andstarch indicator in distilled water360 mL Glass bottle Solution #A2-1H314 H272 Solution #B2 contains 0.1020 MFe(NO3)3 and HNO3 in distilled water. 100 mL Glass bottle Solution #B2H314, H272,H315, H319Distilled water 1 L Glass bottle H2O (Practical Problem 1)Practical Problem 2Artemisinin 1.000 g Small bottle ArtemisininSodium borohydride, NaBH4 0.53gSmallbottleNaBH4H301-H311CH3OH 20 mL Glass bottle Methanol H225, H301n-Hexane 30mLBottlen-Hexane H225cerium staining reagent for TLC 3-5 mL Bottle Ceri reagentCH3COOH 1 mL 1.5 mL vial Acetic Acid H226, H314Ethyl acetate 5 mL Glass bottle Ethyl acetateBag of NaCl for salt bath 0.5 kg Ice bath NaCl bagCaCl2 in drying tube 5-10 g Tube CaCl2H319 Practical Problem 3~ 30 wt% H2SO4, solution inwater40 mL Bottle ~30 wt% H2SO4 H314 1.00×10–2M KMnO4, aqueoussolution50 mL Bottle ~0.01 M KMnO4,H272, H302,2.00×10-3M EDTA, aqueoussolution40 mL Bottle 2.00×10-3 M EDTA H319pH = 9-10 Buffer aqueous Solution, NH4Cl + NH340 mL Bottle pH = 9-10 BufferSolutionH302 , H319~20 wt% NaOH, aqueoussolution20 mL Plastic bottle ~20 wt% NaOH, H314 ~3 M H3PO4, solution in water 15 mL Bottle ~3 M H3PO4H314 Indicator: ETOO, solid in KCl ca. 0.5 g Plastic bottle ETOO H301List of Glassware and EquipmentsProblem Item on every working placeQuantityHotplate stirrer1 Magnetic stirring bar (seek in Kit #1)1Plastic wash bottle filled with distilled water (refill if necessary from the 1 L glass bottle of distilled water provided) 1 1-L glass beaker for inorganic waste liquid 1 250-mL conical flask for organic waste liquid1 Pipette rack with:1-mL graduated pipette5-mL graduated pipette (One for Problem 1; another labeled ‘MeOH’ for Problem 2) 10-mL graduated pipette 10-mL volumetric pipette 25-mL graduated pipette Pasteur pipette and bulb Glass spatula spoon Cleaning brushLarge glass stirring rod Glass funnel1 12 1 1 1 2 2 1 1 1 Bag of paper towels 1 Goggles1 Digital thermometer1 Three-way pipette bulb with a little rubber adapter for bigger pipettes 1 Ceramic Büchner funnel with fitted rubber bung 1 Büchner flask1 Pair of rubber gloves 1 P r a c t i c a l P r o b l e m s 1-3 One cotton glove1Practical Problem 1 (KIT # 1) Digital stop watch 1Insulating plate for the hotplate stirrer labeled I.P. 1 K I T # 1100-mL glass beaker 6 Practical Problem 2 (KIT # 2) 5-mL graduated measuring cylinder 1 50-mL graduated measuring cylinder 2 100-mL two-neck round bottom flask with plastic stopper (in ice bath) 1 100-mL conical (Erlenmeyer) flask 1 Hair dryer 1 Petri dish with cover containing 1 TLC plate, 2 capillaries in paperholder1Plastic pot for ice bath 1 Stand & clamp 1 K I T # 2TLC developing chamber with glass lid 1Replacement or extrachemicals Lab assistant’s signatureStudent’s signaturePenalty_________________ _________________ _______________________________ ______________ ____________________________ ______________ _____________________ _______ _______Tweezers 2 Metal spatula 1 Very small test tubes for TLC in container 2 Zipper store bag (containing cotton wool, round filter paper, watch glassfor Problem 2 labeled with WHITE student code) 1Empty Petri dish with cover 1 Practical Problem 3 (KIT # 3)50-mL glass beaker (for transferring EDTA and KMnO 4 solutions toburettes)225-mL burette with BLUE graduation marks 1 25-mL burette with BROWN graduation marks 1 250 mL glass beaker 2 250 mL conical flask (Erlenmeyer flask) 2 100 mL volumetric flask with stopper 2 10 mL glass graduated measuring cylinder 1 100 mL glass graduated measuring cylinder 1 Burette stand & clamp 1 Reel of pH paper 1 K I T # 3Zipper store bag (containing a large round filter paper for the glassfunnel)1Items on the tables for the common use:Electronic balance with 0.1-mg resolution (6-8 students/each)PRACTICAL EXAMINATIONCode: Question1 2 3 4 5 6 Total Examiner Mark2 4 50 2 2 10 70 Practical Problem 114 % of thetotalGradePractical Problem 1. The oxidation of iodide by iron(III) ions – a kinetic study based on the thiosulfate clock reactionClock reactions are commonly used as demonstrations by chemical educators owing to their visual appeal. Oxidation of iodide by iron(III) ions in a weakly acidic medium is a reaction that can be transformed into a clock reaction. In the presence of thiosulfate and starch, chemical changes in this clock reaction can be presented by the following equations:Fe3+(aq)+S 2O 32-(aq)[Fe(S 2O 3)]+(aq)(1)fast 2Fe 3+(aq)+3I -(aq)2Fe 2+(aq)+I 3-(aq)(2)slow I 3-(aq)+2S 2O 32-(aq)3I -(aq)+S 4O 62-(aq)(3)fast 2I 3-(aq)+starchstarch -I -5+I -(aq)(4)fastReaction (1) is a fast reversible equilibrium which occurs in the reaction mixture giving a reservoir of iron(III) and thiosulfate ions. After being produced in reaction (2), iodine in the form of triiodide ion (I 3–), is immediately consumed by thiosulfate in reaction (3). Therefore, no iodine accumulates in the presence of thiosulfate. When thiosulfate is totally depleted, the triiodide ion accumulates and it may be detected by use of starch indicator according to reaction (4).The kinetics of reaction (2) is easily investigated using the initial rates method. One has to measure the time elapsed between mixing the two solutions and the sudden color change.For the oxidation of iodide by iron(III) ions (reaction 2), the reaction rate can be defined as:3+Fe ⎡⎤⎣⎦=−d v dt(5)The initial reaction rate can then be approximated by:3+0Fe v t⎡⎤Δ⎣⎦≈−Δ (6) with Δ[Fe 3+] being the change in the concentration of iron(III) ions in the initial period of the reaction. If Δt is the time measured, then Δ[Fe 3+] is the change in iron(III) ion concentration from the moment of mixing to the moment of complete thiosulfate consumption (assume that the reaction rate does not depend on thiosulfate concentration). Therefore, from the reactions' stoichiometry it follows:3+2230Fe S O −⎡⎤⎡⎤−Δ=⎣⎦⎣⎦ (7) and consequently:22300S O v t −⎡⎤⎣⎦≈Δ(8)The initial thiosulfate concentration is constant and significantly lower than that of iron(III) and iodide ions. The above expression enables us to determine the initial reaction rate by measuring the time required for the sudden color change to take place, Δt .The rate of reaction is first order with respect to [Fe 3+], and you will determine the order with respect to [I –]. This means the initial reaction rate of reaction can be expressed as:yk v 0030]I []Fe [−+=(9)where k is the rate constant and y is the order with respect to [I –].We assume that the reaction rate does not depend on the thiosulfate concentration, and that the reaction between Fe 3+ and S 2O 32- is negligible. You have to observe carefully the color changes during the clock reaction and to determine the reaction order with respect to [I –], and the rate constant of clock reaction.Experimental Set-upInstructions for using the digital timer (stopwatch)1.Press the [MODE] button until the 00:00:00 icon is displayed.2.To begin timing, press the [START/STOP] button.3.To stop timing, press the [START/STOP] button again.4.To clear the display, press the [SPLIT/RESET] button.PRECAUTIONS¾To minimize fluctuations in temperature only use the distilled water on your bench (in the wash bottle and in the glass 1 L bottle).¾The heating function of the heating magnetic stirrer must be TURNED OFF (as shown in Figure 1 below) and be sure that the stirrer plate is not hot before starting your experiment. Put the insulating plate (labeled I.P.) on top of the stirrer plate for added insulation.¾Start the stopwatch as soon as the solutions #A and #B are mixed. Stop the stopwatch as soon as the solution suddenly turns dark blue.¾Magnetic stirrer bar (take it with the provided tweezers) and beakers should be washed and rinsed with distilled water and wiped dry with paper towel to reuse. General ProcedureSolution # A (containing Na2S2O3, KI, KNO3 and starch) is first placed in the beaker and is stirred using the magnetic bar. The rate of stirring is set at level 8 as indicated in Figure 1. Solution #B (containing Fe(NO3)3 and HNO3) is quickly added into solution #A and the stopwatch is simultaneously started.The time is recorded at the moment the solution suddenly turns dark blue. The temperature of the solution is recorded using the digital thermometer.Figure 1. The apparatus employed for kinetic study of the clock reaction.1.Practice run to observe the color changes-There is no need to accurately measure the volumes used in this part – just use the marks on the beaker as a guide.-Pour ca. 20 mL of solution # A1 (containing KI, Na2S2O3, and starch in water) to a 100-mL graduated beaker containing a magnetic stirrer bar. Place the beaker on top of the insulating plate on the magnetic stirrer.-Pour ca. 20 mL of solution # B1 (containing Fe(NO3)3 and HNO3 in water) in another 100 mL graduated beaker.-Quickly pour the solution # B1 into solution # A1 and start stopwatch simultaneously. Stop stopwatch when the color of the mixture changes. There is no need to record this time. Answer the following questions.Task 1.1: Write down the molecular formula of the limiting reactant for the given clock reaction.Task 1.2:What are the ions or compounds responsible for the colors observed in this experiment? Tick the appropriate box.Color CompoundPurple Fe3+[Fe(S2O3)]+ Fe2+starch-I5- I3-Dark blue Fe3+[Fe(S2O3)]+ Fe2+starch-I5- I3-2. Determination of the order with respect to [I –] (y), and the rate constant (k) In this section, Δt is determined for different initial concentrations of KI according to the table below. The experiment is repeated as necessary for each concentration of KI.Hint: Use 25 mL graduated pipette for solution #A2-1, 10 mL graduated pipette for KI, 5 mL graduated pipette for solution #B2, and one of the burettes for water (you will need to refill the burette from the wash bottle for each measurement).- Prepare 55 mL of solution # A2 in a 100 mL beaker containing a magnetic stirrer bar and place it on top of the insulating plate on the stirrer. Solution #A2 contains solution #A2-1, KI, and distilled water (see the table below for the volume of each component).- Add 5 mL of solution # B2 in another 100 mL beaker.Quickly pour prepared solution #B2 into solution #A2. Determine the time (Δt ) necessary for the color change by a stopwatch. The temperature of the solution is recorded.Task 1.3: Record the time (Δt) for each run in the table below. (You DO NOT need to fill all three columns for the runs.) For each concentration of KI, record your accepted reaction time (Δt accepted ) and temperature. You will be only graded on your values of Δt accepted and T accepted .55 mL of solution #A2 Run 1 Run 2 Run 3N o#A2-1 (mL) H 2O(mL)0.100MKI(mL)Δt (s) T (ºC) Δt (s) T (ºC) Δt (s) T (ºC) Δt accepted (s)T accepted (ºC)1 20.4 31.6 3.02 20.4 30.1 4.53 20.4 28.6 6.0 4 20.4 27.4 7.25 20.4 25.6 9.0When you are satisfied you have all the necessary data for Problem 1, before continuing further with the analysis, it is strongly recommended that you start the practical procedure for Problem 2 since there is a reaction time of one hour in that Problem.Task 1.4: Fill in the table below and plot the results in the graph.Hint: Make sure your data is graphed as large as possible in the provided space.No. 1 2 3 4 5ln([I-]0 / M) - 5.30 - 4.89 - 4.61 - 4.42 - 4.20Δt accepted (s)ln(Δt accepted / s)Task 1.5: Draw the best fit line on your graph and use this to determine the order with respect to [I–] (y).y = ………………………………Task 1.6: Complete the table below and calculate k for each of the concentrations of iodide. Report your accepted value for the rate constant, giving the appropriate unit. Remember that the order with respect to [Fe3+] is equal to one.No Δt accepted(s)[Fe3+]0(×10-3 M)[I-]0(×10-3 M)[S2O32-]0(×10-3 M)k1 5.02 7.53 10.04 12.05 15.0k accepted = ………………….Code:Task 1 2 3 45TotalExaminer Mark 35 15204 276Practical Problem 2 13 % of thetotalGradePractical Problem 2. Synthesis of a derivative of ArtemisininArtemisinin (also known as Quinghaosu) is an antimalarial drug isolated fromthe yellow flower herbArtemisia annua L., in Vietnam. This drug is highly efficaciousagainst the chloroquine-resistant Plasmodium falciparum. However, artemisinin has apoor solubility in both oil and water so that one needs to prepare its new derivatives toimprove the applicability of this drug. The reduction of artemisinin is an attractive method to synthesize new derivatives of artemisinin as shown in Scheme 1.Scheme 1In this practical exam you are going to reduce artemisinin to product P and check its purity using Thin-Layer Chromatography (TLC).Experimental Set-up- The experimental set-up is shown in Figure 2.1.- By moving the finger clamp, you can adjust the position of the two-neck round-bottom flask.41: Digital thermometer; 2: Plastic Stopper; 3: CaCl2 drying tube; 4: Ice BathFigure 2.1. Reaction system for Problem 2ProcedureStep 1. Synthesis of a Derivative of Artemisinin1.Prepare an ice bath with a temperature between –20 and –15 o C by mixing ice andsodium chloride in the plastic pot (approximate ratio of NaCl : crushed ice = 1 scoop : 3 scoops). Use the digital thermometer to monitor the temperature. Place the bath on the magnetic stirrer. Put a layer of three tissues between the bath and the stirrer.2.Connect the CaCl2 drying tube to the small neck of the round-bottom flask andclose the other neck with the plastic stopper.3.Place a magnetic stirring bar into the dry round-bottom flask and set up the reactionsystem onto the clamp-stand so that the system is immersed in the ice bath. Monitor the temperature using the digital thermometer.4.Setting aside a tiny amount (ca. 2 mg) of artemisinin for TLC analysis, open thestopper and add the 1 gram of artemisinin through the bigger neck.e the glass funnel to add 15 mL of methanol (measured using the 50-mLgraduated cylinder). Close the stopper and turn on the magnetic stirrer. (Set the magnetic stirrer to level 4). Start the stopwatch to keep track of the time.6.After ca. 5 min stirring, open the stopper and add carefully 0.53 g of NaBH4 insmall portions over 15 min using a spatula. Close the stopper in between addition.(Caution: Adding NaBH4 rapidly causes side-reactions and overflowing). Keep stirring for 50 min. Maintain the temperature of the ice bath below –5 o C; removesome of the liquid and add more NaCl-crushed ice mixture if necessary.Cool the vial containing the 1 mL of acetic acid in the ice bath.During this waiting time, you are advised to finish calculations from Problem 1, answer the questions below, and prepare further experimental steps.7.Prepare 50 mL of ice-cold distilled water (cooled in the ice bath) in the 100 mL-conical flask. Measure ca. 20-22 mL n-hexane in the 50 mL measuring cylinder and cool it in the ice bath. After the reaction is complete, keep the reaction flask in the ice bath below 0 o C. Remove the CaCl2 tube, open the stopper, and add gradually ca. 0.5 mL of the cold acetic acid from the vial into the reaction flask until the pH is between 6 and 7. (Use the glass rod to spot the reaction mixture on to the pH paper.) With stirring, slowly add the 50 mL of ice cold water over 2 min.A white solid precipitates in the reaction flask.8.Assemble the vacuum filtration apparatus. Put a filter paper onto the Büchnerfunnel, wet the filter paper with distilled water and open the vacuum valve.Transfer the reaction mixture on to the filter, and remove the stirring bar from the reaction flask using the spatula. Wash the product three times with portions of 10 mL ice-cold water (cooled in the ice bath). Wash the product two times with portions of 10 mL ice-cold n-hexane (cooled in the ice bath). Continue to use the pump to dry the solid on the filter. After ca. 5 min, carefully transfer the dried powder on to the watch glass labeled with your code and put into the labeled Petri dish. Turn off the vacuum valve when you do not use it!Note: Your sample will be collected, dried and weighed later by the lab assistant.Task 2.1 – the recording of your yield –will be performed after the exam by the lab assistants.Step 2. TLC Analysis of the product1.Check your TLC plate before use. Unused damaged plates will be replaced uponrequest without penalty. Use the pencil to draw the start front line, and the line where the solvent front will be run to exactly as shown in Figure 2.2. Write your student code on the top of the TLC plate in pencil.Figure 2.2. Instruction of TLC plate preparation2. Dissolve ca. 1 mg of artemisinin (a spatula tip) in ca. 0.5 mL of methanol in the labeled very small test tube (use the labeled 5 mL graduated pipette). Dissolve ca. 1 mg of the product in ca. 1 mL of methanol in the labeled test tube.3. Spot the artemisinin solution and the product solution on the TLC plate using two different glass capillary spotters so the finished plate is as shown in Figure 2.2.4. Prepare the TLC developing chamber. Use the 5 mL graduated cylinder to make 5 mL of a mixture of n -hexane/ethyl acetate (7/3, v/v) as the solvent system. Pour the mixture of n -hexane/ethyl acetate into the chamber (Note: The solvent level should not reach the spots on the plate if prepared as shown). Cover and swirl the chamber and allow it to stand for 2 min.Aab R f (A )=a bCalculate SolventWatch glassDeveloping chamberTLC plateFigure 2.3. A TLC plate placed in the TLC developing chamber and instruction for R fcalculation of compound A5. Insert the TLC plate upright into the TLC developing chamber. Wait until thesolvent system reaches the pre-drawn solvent front line. (Note: You are advised to work on some question below while you wait for the TLC to run.)6. When the solvent front reaches the line, remove the TLC plate using the tweezersand then dry the solvent using the hair dryer set at level 1.7. Dip the piece of cotton wool into the cerium staining reagent, taking care not to letthe tweezers come into contact with the solution since the metal stains the plate.Carefully apply the stain to the whole TLC plate.8. Heat the TLC plate using the hair dryer set at level 2 (Attention: Do NOT set thehair dryer to COLD) until the blue spots of artemisinin and the product appear on the TLC plate.9. Ask the lab assistant to take a photo of your final TLC plate together with yourstudent code.10. Circle all the visualized spots and calculate the R f values of both artemisinin andthe product (See instruction in Fig. 2.3). Store your TLC plate in the Petri dish.Task 2.2: Fill the values of R f in Table below.R f, Artemisinin R f, Product R f Artemisinin/R f Product ---------------------- -------------------------- -------------------------- Task 2.3: Check the total number of developed spots on the TLC plate:Step 3. Identifying the reaction product PThe reduction of artemisinin leads to the formation of two stereoisomers (P). Comparing the 1H-NMR spectrum (in CDCl3) of one of these isomers with the spectrum of artemisinin shows an extra signal at δH = 5.29 ppm as a doublet, and also an extra signal as a broad singlet at δH= 2.82 ppm.Task 2.4: Suggest structure for P.(You do not need to draw the stereochemistry of the compounds).PTask 2.5:P is mixture of two stereoisomers. What is their stereochemical relationship? Check the appropriate box below.Isomers Enantiomers Diastereomers Constitutional IsomersCode: Task 1 2 3 4 567 8 9 10Total ExaminerMark 0 2522534 3 2 5 271Practical Problem 313 % of thetotalGradePractical Problem 3. Analysis of a hydrated zinc iron(II) oxalate double saltZinc iron(II) oxalate double salt is a common precursor in the synthesis of zincferrite which is widely used in many types of electronic devices due to its interesting magnetic properties. However, such double salts may exist with different compositions and different amount of water depending on how the sample was synthesized.You will analyze a pure sample of hydrated zinc iron(II) oxalate double salt (Z ) in order to determine its empirical formula.ProcedureThe concentration of the standard KMnO 4 is posted on the lab walls.Bring a clean 250 mL beaker to the lab assistant who will be waiting by the balance. You will receive a pure sample of Z for analysis. Accurately weigh between 0.7-0.8 g of the pure sample Z onto the weighing paper (m , grams). This should then be immediately quantitatively transferred into your 250 mL beaker for analysis, and its mass recorded in table below.Task 3.1: Record the mass of the sample of pure Z taken.Mass of sample, m (gram) Lab assistant’s signature---------------------------------------Analysis of Z- Using the 100 mL graduated measuring cylinder, measure ca. 30 mL of 30 wt% H 2SO 4 solution and add it into the 250-mL beaker containing your accurately weighed pure sample of Z . To speed up the dissolving of your sample you may usethe hotplate stirrer to warm up the mixture, but be careful not to boil it. You should not use the digital thermometer as the acid may damage it. After the solid has dissolved, remove the beaker from the hotplate stirrer and cool it to close to room temperature. After the solution has cooled, quantitatively transfer it into the 100 mL volumetric flask. Add distilled water up to the 100 mL–mark. We will now call this solution C.- Use an appropriately labeled beaker to transfer the standardized KMnO4 solution into the burette graduated with brown marks.- Use another appropriately labeled beaker to transfer the standardize EDTA solution into the burette graduated with blue marks.Titration with KMnO4a) Using the 5 mL graduated pipette add 5.00 mL of the solution C into a 250 mLconical flask.b) To this conical flask add about 2 mL of 30 wt% H2SO4 solution, about 3 mL of3.0 M H3PO4 solution, and about10 mL of distilled water. Heat the mixture on thehot plate stirrer until hot, but be careful not to boil it.c) Titrate the hot solution with the standardized KMnO4 solution, recording yourburette readings in the table below. At the end point of the titration, the pink color of the solution appears. Repeat the titration as desired and report your accepted volume of KMnO4 solution consumed (V1 mL) in the table.Task 3.2: Record volumes of standardized KMnO4 solution consumed(You DO NOT need to fill in the entire table)Titration No1 2 3 4Initial reading of the burette of KMnO4, mLFinal reading of the burette of KMnO4, mLConsumed volume of KMnO4,mLAccepted volume, V1 = ________ mL。
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准备题中文译本，胡一征译
6 7 8 8 10 12 14 16 17 19 21 24 26 27 28 30 31 33 35 37 39 40 42 44 46 47 49
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第 46 届国际化学奥林匹克竞赛 越南河内 - 2014
第 25 题：头孢噻吩 第 26 题：乙酰丙酸 第 27 题：莲 第 28 题：NMR 谱 第 29 题：IR 谱 第 2 部分：实验试题 第 30 题：香兰素和苄胺的缩合反应 第 31 题：丁香酚氧基乙酸的合成 第 32 题：水溶液中铁、铝、镁的络合滴定 第 33 题：ZnO 粉末中 Zn 和 Pb 含量的测定 第 34 题：乙酰丙酮合铜(II)的制备 第 35 题：阿司匹林水解的动力学分析 第 36 题：水杨酸与 Fe3+配合物的形成
前言
准备题中文译本，胡一征译
我们很高兴为你提供第 46 届国际化学奥林匹克竞赛准备题。这些题目是按照高中 化学课程的知识以及一些在“进阶知识”中列出的知识来命制的，“进阶知识”的应用在 这些题目中有体现。准备题答案将于 2014 年 3 月 31 日在 .vn 上 公布。在命题过程中，我们已尽力让其尽量完美，但仍难以避免出现一些错误。如果 读者在题目中发现任何错误或对题目本身有任何疑问或建议，欢迎向 icho2014prep@.vn 提出。
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2014 年 1 月
第 46 届国际化学奥林匹克竞赛 越南河内 - 2014
准备题中文译本，胡一征译
特约作者
Le Minh Cam, Hanoi National University of Education Vu Viet Cuong, Hanoi University of Science, VNU-Hanoi Pham The Chinh, Institute of Chemistry, VAST Nguyen Huu Dinh, Hanoi National University of Education Tran Thi Da, Hanoi National University of Education Nguyen Van Dau, Hanoi University of Science, VNU-Hanoi Dao Phuong Diep, Hanoi National University of Education Pham Huu Dien, Hanoi National University of Education Nguyen Hien, Hanoi National University of Education Hoang Van Hung, Hanoi National University of Education Nguyen Hung Huy, Hanoi University of Science, VNU-Hanoi Tu Vong Nghi, Hanoi University of Science, VNU-Hanoi Trieu Thi Nguyet, Hanoi University of Science, VNU-Hanoi Do Quy Son, Vietnam Atomic Energy Institute Ta Thi Thao, Hanoi University of Science, VNU-Hanoi Nguyen Tien Thao, Hanoi University of Science, VNU-Hanoi Lam Ngoc Thiem, Hanoi University of Science, VNU-Hanoi Ngo Thi Thuan, Hanoi University of Science, VNU-Hanoi Vu Quoc Trung, Hanoi National University of Education Dao Huu Vinh, Hanoi University of Science, VNU-Hanoi
准备题中文译本
第 46 届国际化学奥林匹克竞赛(IChO -2014)
编委会 Nguyen Tien Thao, Editor in Chief Nguyen Minh Hai Nguyen Van Noi Truong Thanh Tu
河内科技大学，越南国立大学
电话: 0084 435406151;
希望这些题目能够激励学生参加化学竞 赛，同时也是让世界各地学生了解越南文化的时间。
我们期待与你在河内科技大学与越南国立大学中见面。
河内，2014 年 1 月 31 日 主编
Nguyen Tien Thao
第 46 届 IChO 准备题中文译本，越南河内，2014.7
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第 46 届国际化学奥林匹克竞赛 越南河内 - 2014
目录
进阶知识 物理常数及其符号、单位转换关系 第 1 部分：理论试题 第 1 题：极性分子与非极性分子 第 2 题：离子化合物晶格能的计算 第 3 题：井底之蛙 第 4 题：二维势箱与三维势箱中的粒子 第 5 题：拔河 第 6 题：铀化学 第 7 题：热力学的应用 第 8 题：滴定分析 第 9 题：铅化合物 第 10 题：电化学的应用 第 11 题：磷酸 第 12 题：化学动力学 第 13 题：过氧化氢分解动力学 第 14 题：过度金属配合物的磁性 第 15 题：铝的杂多离子的结构与合成 第 16 题：黄樟素 第 17 题：咪唑 第 18 题：含有 CF3 基团的小杂环 第 19 题：维生素 H 第 20 题：香水中的茉莉 第 21 题：越南肉桂 第 22 题：肉桂酸 第 23 题：三（三甲硅基）硅烷与偶氮二异丁腈 第 24 题：由(+)-δ-3-蒈烯合成(–)-薄荷醇
鸣谢
全体编委会成员在此对为我们提出宝贵意见和建议的国际指导委员会以及与我们鼎力 合作的 Vu Viet Cuong 博士，Nguyen Hung Huy 博士，和 Pham Van Phong 博士表示由 衷的感谢。
此致 编委会
化学：生活的滋味
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第 46 届国际化学奥林匹克竞赛 越南河内 - 2014