第46届国际化学奥林匹克竞赛预备题中文译本

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）。

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。

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）。

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。

高中学生化学奥林匹克竞赛（预赛）模拟试题及答案相对原子质量：H ：1 C ：12 N ：14 O ：16 F ：19 Na ：23 P ：31 S ：32 Cl ：35.5 K ：39 Mn ：55 Fe ：56 Cu ：64 Zn ：65 Ag ：108 I ：1271．本试卷共26题，用时3小时完成，全卷共150分； 2．可使用计算器。

一、选择题（56分）每小题有1—2个选项符合题意；每题4分，共14题。

若有二个选项符合题意，选一个且对得2分；若一对一错，得0分；全选错0分。

1．一些盐的结晶水合物，在温度不太高时就有熔化现象，即熔溶于自身的结晶水中，又同时吸收热量。

它们在塑料袋中经日晒就熔化，又在日后缓慢凝结而释放热量。

故可用于调节室内的温度，或作夏日防暑用枕垫或座垫，这些物质可称之为潜热材料。

现有几种盐的根据上述数据和实用性考虑，实际运用时常采用的应该物质是 （ ） A ．① B ．② C ．③ D ．④ 2．长期以来，人们一直认为氟的含氧酸不可能存在，但是自1971年两位美国科学家斯图查尔和阿佩里曼成功地合成次氟酸后，这种观点强烈地动摇了。

他们在0℃以下将氟从细冰上面通过，得到了毫克量的次氟酸。

已知次氟酸的分子组成与次氯酸相似，且次氟酸与热水剧烈反应，生成既有氧化性又有还原性的物质的溶液。

则下列说法中不正确的是（ ）A ． 次氟酸分子中原子间以共价键相结合B ． 次氟酸分解会产生氟气C ． 次氟酸与热水反应可能有H 2O 2生成D ． 次氟酸的酸性可能比次氯酸强3．现在为婴儿特制成一种新型的尿布—“尿不湿”。

这种尿布表面涂有一种既能吸水又能保留水的物质。

据你的推测，这种特殊物质的结构可能是 （ ）A ．B ．C. D ．—CH 2—CH — ［ O —C —CH 3 ︱ O］ ︱ ︱ n F—CH 2—CH — ［ ︱］ nOH —CH 2—CH — ［ ︱］ n —CCl 2—CCl 2—［ ］ n4．据报道，美国科学家于1998年11月合成了一种名为“N 5”的物质，由于其极强的爆炸性，又称为“盐粒炸弹”。

化学奥赛国际试题及答案一、选择题（每题5分，共20分）1. 下列哪种元素的原子序数是26？A. 铁B. 钴C. 镍D. 锌答案：C2. 哪种化合物在水溶液中会形成氢氧化铝沉淀？A. 硫酸铝B. 氯化铝C. 硝酸铝D. 碳酸铝答案：B3. 下列哪种气体在标准状况下（0°C，1atm）的密度是1.429g/L？A. 氧气B. 氮气C. 二氧化碳D. 氦气答案：C4. 以下哪个反应是氧化还原反应？A. 2H2O → 2H2 + O2B. 2H2 + O2 → 2H2OC. 2H2 + CO2 → CH4 + H2OD. 2H2O + 2CO2 → C + 4H2O + O2答案：B二、填空题（每题5分，共20分）1. 元素周期表中，第VIIA族元素的名称是______。

答案：卤素2. 摩尔质量的单位是______。

答案：g/mol3. 标准大气压下，水的沸点是______°C。

答案：1004. 根据路易斯酸碱理论，能够接受电子对的物质是______。

答案：路易斯酸三、简答题（每题10分，共30分）1. 描述如何通过实验确定一个未知溶液是否含有硫酸根离子。

答案：首先，取少量未知溶液于试管中，加入稀盐酸，观察是否有气泡产生。

若有气泡，则说明溶液中可能含有碳酸根离子，需要排除其干扰。

接着，加入钡离子试剂，如氯化钡溶液，若产生白色沉淀，则说明溶液中含有硫酸根离子。

2. 简述什么是同位素，并给出一个例子。

答案：同位素是指具有相同原子序数但不同质量数的原子，即它们属于同一元素的不同核素。

例如，氢元素的三种同位素是氕（^1H）、氘（^2H）和氚（^3H）。

3. 解释什么是化学平衡，并给出一个常见的化学平衡方程式。

答案：化学平衡是指在一个封闭系统中，正向反应和反向反应的速率相等，导致系统中各组分的浓度不再发生变化的状态。

一个常见的化学平衡方程式是氮气和氢气合成氨的反应：N2(g) + 3H2(g) ⇌2NH3(g)。

46th International Chemistry OlympiadJuly 25, 2014Hanoi, VietnamTHEORETICAL EXAMINATION WITH ANSWER SHEETS GRADINGCountry:Name as in passport:Student Code:Language:GENERAL INTRODUCTIONYou have additional 15 minutes to read the whole set.This booklet is composed of 9 problems. You have 5 hours to fulfill the problems. Failure to stop after the STOP command may result in zero points for the current task.Write down answers and calculations within the designated boxes. Give your work where required.Use only the pen and calculator provided.The draft papers are provided. If you need more draft paper, use the back side of the paper. Answers on the back side and the draft papers will NOT be marked.There are 52 pages in the booklet including the answer boxes, Cover Sheet and Periodic Table.The official English version is available on demand for clarification only.Need to go to the restroom – raise your hand. You will be guided there.After the STOP signal put your booklet in the envelope (do not seal), leave at your table. Do not leave the room without permission.Physical Constants, Units, Formulas and EquationsAvogadro's constant N A = 6.0221 × 1023 mol–1Universal gas constant R = 8.3145 J·K–1·mol–1Speed of light c = 2.9979 × 108 m·s–1Planck's constant h= 6.6261 × 10–34 J·sStandard pressure p° = 1 bar = 105 PaAtmospheric pressure 1 atm = 1.01325 × 105 Pa = 760 mmHg Zero of the Celsius scale 273.15 KMass of electron m e = 9.1094 × 10–31kg1 nanometer (nm) = 10–9 m ; 1 angstrom (Å) = 10–10 m1 electron volt (eV) = 1.6022 × 10–19 J = 96485 J·mol–1Problem 1. Particles in a box: polyenesIn quantum mechanics, the movement of π electrons along a neutral chain ofconjugated carbon atoms may be modeled using the ‘particle in a box’ method. The energy of the π electrons is given by the following equation:2228mLh n E n = where n is the quantum number (n = 1, 2, 3, …), h is Planck’s constant, m is the mass of electron, and L is the length of the box which may be approximated by L = (k + 2)×1.40 Å (k being the number of conjugated double bonds along the carbon chain in the molecule). A photon with the appropriate wavelength λ may promote a π electron from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). An approximate semi-empirical formula based on this model which relates the wavelength λ, to the number of double bonds k and constant B is as follows:λ (nm) = B )12()2(2++×k k Equation 11. Using this semi-empirical formula with B = 65.01 nm calculate the value of the wavelength λ (nm) for octatetraene (CH 2 = CH – CH = CH – CH = CH – CH = CH 2).Code:Question 1 2 3 4 5 Total Examiner Mark 3 7 6 4 7 27 Theoretical Problem 1 5.0 % of thetotalGrade2.Derive Equation 1 (an expression for the wavelength λ (nm) corresponding to the transfer of an electron from the HOMO to the LUMO) in terms of k and the fundamental constants, and hence calculate theoretical value of the constant B calc..3. We wish to synthesize a linear polyene for which the excitation of a π electron from the HOMO to the LUMO requires an absorption wavelength of close to 600 nm. Using your expression from part 2, determine the number of conjugated double bonds (k) in this polyene and give its structure. [If you did not solve Part 2, use the semi-empirical Equation 1 with B = 65.01 nm to complete Part 3.]Thus, k = 15.So, the formula of polyene is:CH 2 = CH – (CH = CH)13 – CH = CH 22 points4. For the polyene molecule found in Part 3, calculate the difference in energy between the HOMO and the LUMO, ΔE , (kJ·mol –1).In case Part 3 was not solved, take k = 5 to solve this problem.5. The model for a particle in a one-dimensional box can be extended to a three dimensional rectangular box of dimensions L x , L y and L z , yielding the following expression for the allowed energy levels:⎟⎟⎠⎞⎜⎜⎝⎛++=2222222,,8z z y y x x n n n L n L n L n m h E zy xThe three quantum numbers n x , n y , and n z must be integer values and are independentof each other.5.1 Give the expressions for the three different lowest energies, assuming that the box is cubic with a length of L .Levels with the same energy are said to be degenerate. Draw a sketch showing all the energy levels, including any degenerate levels, that correspond to quantum numbers having values of 1 or 2 for a cubic box.Problem 2. Dissociating Gas CycleDininitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide:N 2O 4(g) ⇌ 2NO 2(g)1.00 mole of N 2O 4 was put into an empty vessel with a fixed volume of 24.44 dm 3.The equilibrium gas pressure at 298 K was found to be 1.190 bar. When heated to 348 K, the gas pressure increased to its equilibrium value of 1.886 bar. 1a. Calculate ∆G 0 of the reaction at 298K, assuming the gases are ideal.1b. Calculate ∆H 0 and ∆S 0 of the reaction, assuming that they do not change significantly with temperature.Code: Question 1a 1b 2 3 TotalExaminerMark12 8 3 10 33Theoretical Problem 2 5.0 % of thetotalGrade∆S 0∆G 0348 = - 4.07 kJ = ∆H – 348∆S (1) ∆G 0298 = 4.72 kJ = ∆H – 298∆S (2) (2) - (1) → ∆S = 0.176 kJ·mol –1·K –1 ∆H 0∆H 0 = 4.720 + 298 × 0.176 = 57.2 (kJ·mol –1)4pts 4ptsIf you cannot calculate ∆H 0, use ∆H 0 = 30.0 kJ·mol –1 for further calculations.The tendency of N 2O 4 to dissociate reversibly into NO 2 enables its potential use in advanced power generation systems. A simplified scheme for one such system is shown below in Figure (a). Initially, "cool" N 2O 4 is compressed (1→2) in a compressor (X ), and heated (2→3). Some N 2O 4 dissociates into NO 2. The hot mixture is expanded (3→4) through a turbine (Y ), resulting in a decrease in both temperatureand pressure. The mixture is then cooled further (4→1) in a heat sink (Z ), to promote the reformation of N 2O 4. This recombination reduces the pressure, thus facilitates the compression of N 2O 4 to start a new cycle. All these processes are assumed to take place reversibly.out(a)To understand the benefits of using reversible dissociating gases such as N 2O 4, we will focus on step 3 → 4 and consider an ideal gas turbine working with 1 mol of air (which we assume to be an inert, non-dissociating gas). During the reversible adiabatic expansion in the turbine, no heat is exchanged .2. Give the equation to calculate the work done by the system w(air) during the reversible adiabatic expansion for 1 mol of air during stage 3 → 4. Assume that C v,m(air) (the isochoric molar heat capacity of air) is constant, and the temperature changes from T3 to T4.∆U = q + w; work done by turbine w(air)=-w 1 ptq = 0, thus w(air) = ∆U = C v,m(air)[T3-T4] 2 pts3.Estimate the ratio w(N2O4)/w(air), in which w(N2O4) is the work done by the gas during the reversible adiabatic expansion process 3 → 4 with the cycle working with 1 mol of N2O4, T3 and T4 are the same as in Part 2. Take the conditions at stage 3 to be T3 = 440 K and P3 = 12.156 bar and assume that:(i) the gas is at its equilibrium composition at stage 3;(ii) C v,m for the gas is the same as for air;(iii) the adiabatic expansion in the turbine takes place in a way that the composition of the gas mixture (N2O4 + NO2) is unchanged until the expansion is completed.Oxidation number of Ag1 : ……….+1 Oxidation number of Ag2 : ……… +3 2 pointsCode: Question 1 2 3 4 Total ExaminerMarks 8 14 2 12 36 Theoretical Problem 3 9.0 % of the totalGrade1c.What is the coordination number of O atoms in the lattice of A?The coordination number of O atoms =……… 3 1 point1d.How many Ag I and Ag III bond to one O atom in the lattice of A?Number of Ag I = (1)Number of Ag III = ……. 2 2 points1e.Predict the magnetic behaviour of A. Check the appropriate box below.S2O82-(aq) + 2Ag+(aq) + 2H2O (l) 2SO42-(aq) + Ag I Ag III O2 (s) + 4H+(aq)1 point2. Among the silver oxides which have been crystallographically characterized, the most surprising is probably that compound A is not a Ag II O. Thermochemical cycles are useful to understand this fact. Some standard enthalpy changes (at 298 K) are listed:Atom Standard enthalpyof formation(kJ·mol–1)1st ionization(kJ·mol–1)2nd ionization(kJ·mol–1)3rd ionization(kJ·mol–1)1st electronaffinity(kJ·mol–1)2nd electronaffinity(kJ·mol–1)Cu(g) 337.4 751.7 1964.1 3560.2Ag(g) 284.9 737.2 2080.2 3367.2O(g)249.0 -141.0844.0Compounds ΔH o f (kJ ·mol –1) Ag I Ag III O 2 (s) –24.3 Cu II O (s) –157.3The relationship between the lattice dissociation energy (U lat ) and the lattice dissociation enthalpy (ΔH lat ) for monoatomic ion lattices is: nRT U H lat lat +=Δ, where n is the number of ions in the formula unit.2a. Calculate U lat at 298 K of Ag I Ag III O 2 and Cu II O. Assume that they are ionic compounds. U lat of Ag I Ag III O 2 Calculations:ΔH lat (Ag I Ag III O 2) = 2 ΔH o f (O 2-) + ΔH o f (Ag +) + ΔH o f (Ag 3+) –ΔH o f (Ag I Ag III O 2) = (2×249 – 2 × 141 + 2 × 844) + (284.9 + 737.2) + (284.9 + 737.2 + 2080.2 + 3367.2 ) – (–24.3)= +9419.9 (kJ·mol –1)U lat (Ag I Ag III O 2) = ΔH lat (Ag I Ag III O 2) – 4RT= + 9419.9 – 10.0 = + 9409.9 (kJ·mol –1)3 points(no penalty if negative sign)U lat of Cu II OCalculations for: U lat of Cu II OΔH lat (Cu II O) = ΔH o f (O 2–) + ΔH o f (Cu 2+) – ΔH o f (Cu II O)= (249 – 141 + 844) + (337.4 + 751.7 + 1964.1) – (–157.3)= 4162.5 (kJ ·mol –1)U lat (Cu II O) = ΔH lat (Cu II O) – 2RT = 4162.5 – 5.0 = 4157.5 (kJ ·mol –1)3 points(no penalty if negative sign)If you can not calculate the U lat of Ag I Ag III O 2 and Cu II O, use following values forfurther calculations: U lat of Ag I Ag III O 2 = 8310.0 kJ·mol –1; U lat of Cu II O = 3600.0 kJ·mol –1.The lattice dissociation energies for a range of compounds may be estimated using this simple formula:311C ⎟⎟⎠⎞⎜⎜⎝⎛×=m lat V UWhere: V m (nm 3) is the volume of the formula unit and C (kJ·nm·mol –1) is an empirical constant which has a particular value for each type of lattice with ions of specified charges.The formula unit volumes of some oxides are calculated from crystallographic data as the ratio between the unit cell volume and the number of formula units in the unit cell and listed as below:Oxides V m (nm 3)Cu II O 0.02030 Ag III 2O 3 0.06182 Ag II Ag III 2O 4 0.089852b. Calculate U lat for the hypothetical compound Ag II O. Assume that Ag II O and Cu II O have the same type of lattice, and that V m (Ag II O) = V m (Ag II Ag III 2O 4) – V m (Ag III 2O 3).2c. By constructing an appropriate thermodynamic cycle or otherwise, estimate the enthalpy change for the solid-state transformation from Ag II O to 1 mole of Ag I Ag III O 2. (Use U lat Ag II O = 3180.0 kJ·mol -1 and U lat Ag I Ag III O 2 = 8310.0 kJ·mol -1 if you cannot calculate U lat Ag II O in Part 2b).2Ag IIO (s)Ag I Ag III O 2(s)2Ag 2+(g)+2O 2-(g)Ag +(g)+Ag 3+(g)+2O 2-(g)H rxn2U lat (AgO)+4RT-U lat (Ag I Ag III O)-4RTIE 3(Ag)-IE 2(Ag)Calculations:ΔH rxn = 2U lat (Ag II O) + 4RT + IE 3 – IE 2 – U lat (Ag I Ag III O 2) – 4RT= 2 × 3733.6 + 3367.2 – 2080.2 – 9409.9= – 655.7 (kJ/mol) or - 663.0 kJ/mol using given U lat values 4 pts2d. Indicate which compound is thermodynamically more stable by checking the appropriate box below.3. When Ag I Ag III O 2 is dissolved in aqueous HClO 4 solution, a paramagnetic compound (B ) is first formed then slowly decomposes to form a diamagnetic compound (C ). Given that B and C are the only compounds containing silver formed in these reactions, write down the equations for the formation of B and C .For B :Ag I Ag III O 2 (s) + 4 HClO 4 (aq) 2Ag(ClO 4)2 (aq) + 2 H 2O (l) 1 pointFor C : 4Ag(ClO 4)2 (aq) + 2 H 2O (l)4 AgClO 4 (aq) + 4 HClO 4 (aq) + O 2 (g) 1 point4. Oxidation of Ag+ with powerful oxidizing agents in the presence of appropriate ligands can result in the formation of high-valent silver complexes. A complex Z is synthesized and analyzed by the following procedures:An aqueous solution containing 0.500 g of AgNO3 and 2 mL of pyridine (d = 0.982 g/mL) is added to a stirred, ice-cold aqueous solution of 5.000 g of K2S2O8. The reaction mixture becomes yellow, then an orange solid (Z) is formed which has a mass of 1.719 g when dried.Elemental analysis of Z shows the mass percentages of C, H, N elements are38.96%, 3.28%, 9.09%, respectively.A 0.6164 g Z is added to aqueous NH3. The suspension is boiled to form a clear solution during which stage the complex is destroyed completely. The solution is acidified with excess aqueous HCl and the resulting suspension is filtered, washed and dried (in darkness) to obtain 0.1433 g of white solid (D). The filtrate is collected and treated with excess BaCl2 solution to obtain 0.4668 g (when dry) of white precipitate (E).4a.Determine the empirical formula of Z and calculate the percentage yield in the preparation.4b. Ag (IV) and Ag (V) compounds are extremely unstable and found only in few fluorides. Thus, the formation of their complexes with organic ligands in water can be discounted. To confirm the oxidation number of silver in Z, the effective magnetic moment (µeff ) of Z was determined and found to be 1.78 BM. Use the spin only formula to determine the number of unpaired electrons in Z and the molecular formula of Z. (Z contains a mononuclear complex with only one species of Ag and only one type of ligand in the ligand sphere.)4c. Write down all chemical equations for the preparation of Z, and its analysis.Formation of Z:2Ag+(aq) + 8Py (l) + 3S2O82–(aq) 2[Ag II(Py)4](S2O8) (s) + 2SO42–(aq) 2 ptsDestruction of Z with NH3:[Ag II(Py)4](S2O8) (s) + 6NH3(l) [Ag(NH3)2]+(aq) + ½ N2(g) + 2SO42-(aq)+3NH4+ + 4Py (l) 2 pts(aq)(All reasonable N –containing products and O2 are acceptable)Formation of D:[Ag(NH3)2]+(aq) + 2H+(aq) + Cl– (aq) AgCl (s) + 2NH4+(aq) 1 ptFormation of E:Ba2+(aq) + SO42– (aq) BaSO4(s)1ptProblem 4. Zeise’s Salt1. Zeise's salt, K[PtCl 3C 2H 4], was one of the first organometallic compounds to bereported. W. C. Zeise, a professor at the University of Copenhagen, prepared this compound in 1827 by reacting PtCl 4 with boiling ethanol and then adding potassium chloride (Method 1). This compound may also be prepared by refluxing a mixture of K 2[PtCl 6] and ethanol (Method 2). The commercially available Zeise's salt is commonly prepared from K 2[PtCl 4] and ethylene (Method 3).1a. Write balanced equations for each of the above mentioned preparations of Zeise's salt, given that in methods 1 and 2 the formation of 1 mole of Zeise’s salt consumes 2 moles of ethanol.PtCl 4 + 2 C 2H 5OH → H[PtCl 3C 2H 4] + CH 3CH=O + HCl + H 2O H[PtCl 3C 2H 4] + KCl → K[PtCl 3C 2H 4] + HClK 2[PtCl 6] + 2 C 2H 5OH → K[PtCl 3C 2H 4] + CH 3CH=O + KCl + 2 HCl + H 2O K 2[PtCl 4] + C 2H 4 → K[PtCl 3C 2H 4] + KCl1pt for each (2 pts if the first two reactions combined), total of 4 pts1b. Mass spectrometry of the anion [PtCl 3C 2H 4]– shows one set of peaks with mass numbers 325-337 au and various intensities.Calculate the mass number of the anion which consists of the largest natural abundance isotopes (using given below data).Code: Question 1a 1b 2a 3a 3b 3c Total ExaminerMark 4 1 10 2 6 4 27Theoretical Problem 4 4.0 % of the totalGradeIsotopePt 19278Pt 19478Pt 19578Pt 19678Pt 19878C 126C136Natural abundance,% 0.8 32.9 33.8 25.3 7.2 75.8 24.2 98.9 1.1 99.99Calculations:195 + 3×35 + 2×12 + 4×1 = 328 1 pt2. Some early structures proposed for Zeise’s salt anion were:In structure Z1, Z2, and Z5 both carbons are in the same plane as dashed square. [You should assume that these structures do not undergo any fluxional process byinterchanging two or more sites.]2a. NMR spectroscopy allowed the structure for Zeise’s salt to be determined as structure Z4. For each structure Z1-Z5, indicate in the table below how many hydrogen atoms are in different environments, and how many different environments of hydrogen atoms there are, and how many different environments of carbon atoms there are?StructureNumber of differentenvironments of hydrogen Number of differentenvironments of carbonZ121pt 21 ptZ22 1pt 21 ptZ321pt 21 ptZ41 1pt 11 ptZ52 1pt 11 pt3. For substitution reactions of square platinum(II) complexes, ligands may be arranged in order of their tendency to facilitate substitution in the position trans to themselves (the trans effect). The ordering of ligands is:CO , CN- , C2H4 > PR3 , H- > CH3- , C6H5- , I- , SCN- > Br- > Cl- > Py > NH3 > OH- , H2OIn above series a left ligand has stronger trans effect than a right ligand.Some reactions of Zeise’s salt and the complex [Pt2Cl4(C2H4)2] are given below.3a.Draw the structure of A, given that the molecule of this complex has a centre of symmetry, no Pt-Pt bond, and no bridging alkene.Structure of A2 pt3b.Draw the structures of B, C, D, E, F and G.B1 ptCPtCl NH2C6H5Cl1 ptD1 ptE1 ptF1 ptG1 pt3c.Suggest the driving force(s) for the formation of D and F by choosing one or more of the following statements (for example, i and ii):i) Formation of gasii) Formation of liquidiii) Trans effectiv) Chelate effectStructure D FDriving force(s) i iii and iv2 pts 2 ptsProblem 5. Acid-base Equilibria in WaterA solution (X) contains two weak monoprotic acids (those having one acidicproton); HA with the acid dissociation constant of K HA = 1.74 × 10–7, and HB with the acid dissociation constant of K HB = 1.34 × 10–7. The solution X has a pH of 3.75.1. Titration of 100 mL solution X requires 100 mL of 0.220 M NaOH solution for completion.Calculate the initial (total) concentration (mol·L –1) of each acid in the solution X . Use reasonable approximations where appropriate. [K W = 1.00 × 10–14 at 298 K.]HAH HBH OH Code:Question 1 2 3 4 TotalExaminer Mark 6 4 4 6 20 Theoretical Problem 5 6.5 % of thetotalGrade2. Calculate the pH of the solution Y which initially contains 6.00×10-2 M of NaA and 4.00×10-2 M of NaB.Solution:Solution Y contains NaA 0.06 M and NaB 0.04 M. The solution is basic, OH– was produced from the reactions:NaA + H 2O HA + OH–K b,A = K w/K HA = 5.75 ×10-8NaB + H 2O HB + OH– K b,B = K w/K HB = 7.46 ×10-8H 2O H+ + OH–K w = 1.00 10-14and we have:3. Adding large amounts of distilled water to solution X gives a very (infinitely) dilute solution where the total concentrations of the acids are close to zero. Calculate the percentage of dissociation of each acid in this dilute solution.Solving the equation gives: α = 0.573- The percentage of dissociation of HA = 65.5 %- The percentage of dissociation of HB = 57.3 % 2 points4. A buffer solution is added to solution Y to maintain a pH of10.0. Assume no change in volume of the resulting solution Z.Calculate the solubility (in mol·L–1) of a subtancce M(OH)2 in Z, given that the anions A– and B– can form complexes with M2+:M(OH)2 M2+ + 2OH–K sp = 3.10 ×10-12M2+ + A– [MA]+K 1= 2.1 × 103[MA]+ + A– [MA 2] K2 = 5.0 × 102M2+ + B– [MB]+K’1 = 6.2 × 103[MB]+ + B– [MB 2] K’2 = 3.3 × 102MO H[MB][MBSolve this equation: [A -] = 8.42× 10 –3 M Substitute this value into Eq. 3 and Eq. 4:[MA +] = 0.651 × [A –] = 5.48 × 10 –3 M [MA 2] = 325.5 × [A –]2 = 2.31 × 10 –2 MSimilarly, [B –]total = 0.04 M][92.1][1010.3102.6]][[][432'1−−−−++×=××××==B B B M K MB Eq. 6222'2'12][3.634]][[][−−+×==B B M K K MB Eq.7[B –]total = [B -] + [MB +] + 2 × [MB 2] = 0.04 M Eq. 8 2ptsSubstitute Eq. 6 and Eq. 7 into Eq. 8: [B –] + 1.92 × [B –] + 2 × 634.3 × [B –]2 = 0.04 Solve this equation: [B –] = 4.58 × 10–3 M Substitute this value into Eq. 6 and Eq. 7: [MB +] = 1.92 ×[B –] = 8.79 × 10 –3 M [MB 2] = 634.3 ×[B –]2 = 1.33 × 10–2 MThus, solubility of M(OH)2 in Z is s’s’ = 3.10×10 – 4 + 5.48×10 – 3 + 2.31×10 – 2 + 8.79 × 10 – 3+ 1.33 ×10 – 2 = 5.10×10 – 2 M Answer: Solubility of M(OH)2 in Z = 5.10×10 – 2 M. 2 pointsProblem 6. Chemical KineticsThe transition-metal-catalyzed amination of aryl halides has become one of the mostpowerful methods to synthesize arylamines. The overall reaction for the nickel-catalyzed amination of aryl chloride in basic conditions is:in which NiLL’ is the nickel complex catalyst. The reaction goes through several steps in which the catalyst, reactants, and solvent may be involved in elementary steps.6a. To determine the reaction order with respect to each reactant, the dependence of the initial rate of the reaction on the concentrations of each reagent was carried out with all other reagents present in large excess. Some kinetic data at 298 K are shown in the tables below. (Use the grids if you like)Code: Question 6a6b 6c 6d 6e Total ExaminerMarks 6 8 4 12 2 32 Theoretical Problem 6 7.0 % of thetotalGradeDetermine the order with respect to the reagents assuming they are integers. -Order with respect to [ArCl] = = 1-Order with respect to [NiLL’] = = 1-Order with respect to [L’] = = -1 6pts6b. To study the mechanism for this reaction, 1H, 31P, 19F, and 13C NMR spectroscopy have been used to identify the major transition metal complexes in solution, and the initial rates were measured using reaction calorimetry. An intermediate, NiL(Ar)Cl, may be isolated at room temperature. The first two steps of the overall reaction involve the dissociation of a ligand from NiLL’ (step 1) at 50 o C, followed by the oxidation addition (step 2) of aryl chloride to the NiL at room temperature (rt):Using the steady state approximation, derive an expression for the rate equation for the formation of [NiL(Ar)Cl].(4 pts for rate calculation)The next steps in the overall reaction involve the amine (RNH2) and t BuONa. To determine the order with respect to RNH2 and t BuONa, the dependence of the initial rates of the reaction on the concentrations of these two reagents was carried with the other reagents present in large excess. Some results are shown in the tables below.6c . Determine the order with each of these reagents, assuming each is an integer. (Use the grids if you like)- Order with respect to [NaO t Bu] = 0 2 pts- Order with respect to [RNH 2] = 02 ptsDuring a catalytic cycle, a number of different structures may be involved which include the catalyst. One step in the cycle will be rate-determining.A proposed cycle for the nickel-catalyzed coupling of aryl halides with amines is as follows:6d. Use the steady-state approximation and material balance equation to derive the rate law for d[ArNHR]/dt for the above mechanism in terms of the initial concentration of the catalyst [NiLL’]0 and concentrations of [ArCl], [NH 2R], [NaO t Bu], and [L’].NiLLNiLLApply the steady-state approximation to the concentrations for the intermediates:[NiL][L’] + k [NiL(Ar)HNR] (Equation 1) 1pt(Equation 2) 1pt6e.Give the simplified form of the rate equation in 6d assuming that k1 is very small. d[ArNHR]/dt = - d[ArCl]/dt =k2[ArCl] [NiL] = k1k2 [ArCl][NiLL’]0 / k-1[L’] (i.e. consistent with all the orders of reaction as found in the beginning) 2 ptsProblem 7. Synthesis of Artemisinin(+)-Artemisinin, isolated from Artemisia annua L.(Qinghao, Compositae ) is a potent antimalarial effective against resistant strains of Plasmodium . A simple route for the synthesis of Artemisinin is outlined below.First, pyrolysis of (+)-2-Carene broke the cyclopropane ring forming, among other products, (1R )-(+)-trans -isolimonene A (C 10H 16), which then was subjected to regioselective hydroboration using dicyclohexylborane to give the required alcohol B in 82% yield as a mixture of diastereoisomers. In the next step, B was converted to the corresponding γ,δ-unsaturated acid C in 80% yield by Jones’ oxidation.7a. Draw the structures (with stereochemistry) of the compounds A-C .A B CMeMeHHO4 pts (2 pts if wrong stereochemistry) 4 pts 4 ptsCode: Question 7a 7b 7c 7d 7e 7f Total ExaminerMark128 8 12 12 12 64Theoretical Problem 7 8.0 % of thetotalGradeThe acid C was subjected to iodolactonization using KI, I2 in aqueous. NaHCO3solution to afford diastereomeric iodolactones D and E (which differ in stereochemistry only at C3 ) in 70% yield.7b. Draw the structures (with stereochemistry) of the compounds D and E.The acid C was converted to diastereomeric iodolactones D and E (epimeric at the chiralcenter C3). Look at the number-indicated in the structure F in the next step.D E4 pts 4ptsThe iodolactone D was subjected to an intermolecular radical reaction with ketoneX using tris(trimethylsilyl)silane (TTMSS) and AIBN (azobisisobutyronitrile) in acatalytic amount, refluxing in toluene to yield the corresponding alkylated lactone F in72% yield as a mixture of diastereoisomers which differ only in stereochemistry at C7along with compound G (~10%) and the reduced product H, C10H16O2(<5%).7c. Draw the structures (with stereochemistry) of compound H and the reagent X.Because alkylated lactone F is known, we can deduce the reagent X as methyl vinylketone. H is the reduced product of D.X H2 pts 6 ptsThe keto group of F reacted with ethanedithiol and BF3•Et2O in dichloromethane(DCM) at 0o C to afford two diastereomers: thioketal lactones I and J in nearlyquantitative yield (98%). The thioketalization facilitated the separation of the majorisomer J in which the thioketal group is on the opposite face of the ring to the adjacentmethyl group.7d.Draw the structures (with stereochemistry) of the compounds I and J.The keto group of lactone F reacted with ethanedithiol and BF3·Et2O in dichloromethaneto afford thioketal lactones, I and the major isomer J.I J6 pts (3 pts if I and J are swapped) 6 pts (3 pts if I and J are swapped)The isomer J was further subjected to alkaline hydrolysis followed byesterification with diazomethane providing hydroxy methyl ester K in 50% yield. Thehydroxy methyl ester K was transformed into the keto ester L using PCC (P yridiumC hloro C hromate) as the oxidizing agent in dichloromethane (DCM).A two-dimensional NMR study of the compound L revealed that the twoprotons adjacent to the newly-formed carbonyl group are cis to each other andconfirmed the structure of L.7e. Draw the structures (with stereochemistry) of the compounds K and L .Hydrolysis followed by esterification of J provided hydroxy ester K .Oxidation of the hydroxy group in K by PCC resulted in the keto ester L in which two protons adjacent to the carbonyl group are cis-oriented.K L6 pts 6 ptsThe ketone L was subjected to a Wittig reaction with methoxymethyl triphenylphosphonium chloride and KHMDS (P otassium H exa M ethyl D i S ilazid - a strong, non-nucleophilic base) to furnish the required methyl vinyl ether M in 45% yield. Deprotection of thioketal using HgCl 2, CaCO 3 resulted in the key intermediate N (80%). Finally, the compound N was transformed into the target molecule Artemisinin by photo-oxidation followed by acid hydrolysis with 70% HClO 4.LMN323231. O 2, h υ4。

高中化学奥林匹克竞赛全真模拟试题（七）●竞赛时间3小时。

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第1题（6分）在液态SO 2溶液中，将硫磺（S 8）、碘和AsF 5混合加热，得到一种化合物A 。

经分析发现：A的阴离子为正八面体结构，阳离子与P 2I 4是等电子体。

1-1 确定A 的结构简式，写出生成A 的化学反应方程式。

1-2 根据路易斯电子理论，画出A 中环状阳离子的结构式。

1-3 预测A 中S －S 键与经典S －S 键的键能大小。

1.008Zr Nb Mo Tc Ru Rh P d Ag Cd InSn Sb Te I Hf Ta W Re Os Ir P t Au Hg Tl P b Bi P o At Ac-Lr H Li Be B C N O F Na MgAl Si P Cl S K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Rb Cs Fr Sr Ba Ra Y LaLu -6.9419.01222.9924.3139.1040.0885.4787.62132.9137.3[223][226]44.9647.8850.9452.0054.9455.8558.9363.5558.6965.3910.8126.9869.7212.0128.0972.61114.8204.4118.7207.2112.4200.6107.9197.0106.4195.1102.9192.2101.1190.298.91186.295.94183.992.91180.991.22178.588.9114.0116.0019.0030.9774.92121.8209.032.0778.96127.6[210][210][210]126.979.9035.454.00320.1839.9583.80131.3[222]He Ne Ar Kr Xe Rn 相对原子质量Rf Db Sg Bh Hs Mt第2题（11分）铜及其化合物在日常生活中应用极其广泛。

49届国际化学奥林匹克竞赛预备试题中文译本2017-02-03"Bonding the World with Chemistry"49th INTERNATIONALCHEMISTRY OLYMPIADNakhon Pathom，THAILAND第一部分：理论题第一题 醋酸的二聚乙酸（醋酸，CH 3COOH)在气相时部分二聚。

在298 K 下，总压力0.200 atm 时有92.0%的乙酸发生二聚。

升温至318 K 时，二聚体含量下降，此时反应的平衡常数K p = 37.3。

1.1)计算二聚反应的焓变和熵变，假设它们不随温度变化。

1.2)根据勒夏特列原理，增加压力（选择一个正确答案) ⃝ 对二聚有利 ⃝ 对二聚不利1.3)上接1.2)题，二聚的程度会（选择一个正确答案) ⃝ 随着温度的升高而降低 ⃝ 随着温度的升高而增加第二题 方解石的溶解度方解石是一种碳酸钙（CaCO 3)的稳定形态。

它的溶度积（K sp )随着温度的升高而降低， 在0℃和50℃时分别为9.50 × 10-9和2.30 × 10-9。

计算方解石溶解过程反应的焓变。

第三题 理想气体的膨胀和液体混合的热力学3.1)已知0.10 mol 理想气体A 的初始温度为22.2℃，从0.200 dm 3膨胀到2.42 dm 3。

假设该过程等温且不可逆，外压为1.00 atm ，计算该过程的功（w )、热（q )、内能变（ΔU )、系统熵变（ΔS sys )、环境熵变（ΔS surr )和总熵变（ΔS univ )。

3.2)将3.00 mol A 冷凝为液态并与5.00 mol 液体B 混合，假设混合物是理想液态混合物，计算在25℃下混合过程的熵变和吉布斯自由能变。

第四题 双原子分子的振动频率对于双原子分子的振动运动，根据谐振子模型，允许的振动能级用如下公式描述：1 ; 0,1,2,...2E h υυνυ⎛⎫=+= ⎪⎝⎭其中υ是振动量子数，ν是振动频率。

全国高中化学奥林匹克竞赛初赛试题(满分100分时间120分钟)也许用到旳相对原子质量：H-1 C-12 N-14 O-16 Na-23 Mg-24 Al-27 S-32Cl-35.5 K-39 Fe-56 Cu-64 Zn-65 Ag-108 Ba-137一、选择题(本题涉及18小题，每题2分，共36分，每题有．1．～．2．个．选项符合题意。

多选错选不得分，有两个答案旳只选一种且答对者得1分)1．两位美国科学家彼得·阿格雷和罗德里克·麦金农，由于发现细胞膜水通道，以及对离子通道构造和机理研究作出旳开创性奉献而获得诺贝尔化学奖。

她们之因此获得诺贝尔化学奖而不是生理学或医学奖是由于( )。

A．她们旳研究和化学物质水有关B．她们旳研究有助于研制针对某些神经系统疾病和心血管疾病旳药物C．她们旳研究进一步到分子、原子旳层次D．她们旳研究进一步到细胞旳层次2．为了摸索月球上与否有生命存在旳痕迹，就要分析月球岩石中与否包藏有碳氢化合物(固然这仅仅是摸索旳第一步)。

科学家用氘盐酸(DCl)和重水(D2O)溶液解决月球岩石样品，对收集旳气体加以分析，成果只发既有某些气体状态旳碳氘化合物。

这个实验不能用一般盐酸，其理由是( )。

A．一般盐酸旳酸性太强B．一般盐酸具有挥发性C．一般盐酸和月球岩石中旳碳化物无法反映D．无法区别岩石中本来具有旳是碳化物，还是碳氢化合物3．下列与“神舟五号”载人飞船有关旳说法不对旳．．．旳是( )。

A．飞船旳表面覆盖旳一层石墨瓦之因此能起到保护作用是由于石墨可以耐高温B．宇航员旳食物做成大小可口旳“一口酥”，目旳是避免飞船内产生飘尘C．飞船在宇宙中运营时，船舱内温度低、氧气少，无法划着火柴D．在飞船上使用LiOH吸取多余旳CO2而不使用NaOH4．下列解释不科学．．．旳是( )。

A．“水滴石穿”重要是溶解了CO2旳雨水与CaCO3长期作用生成了可溶性Ca(HCO3)2旳缘故B．长期盛放NaOH溶液旳试剂瓶不易打开，是由于NaOH与瓶中旳CO2反映导致瓶内气体减少形成“负压”旳缘故C．严格地讲，实验室使用“通风橱” 防污染是不负责任旳，由于实验产生旳有害气体没有得到转化或吸取D．“雨后彩虹”与“海市蜃楼”都是自然界旳光学现象，也与胶体旳知识有关5．12月31日，世界上第一条商业磁悬浮铁路在上海投入运营。

高中化学奥林匹克竞赛预赛试题（必修模块试题）一、选择题（本题包括16小题；每题有1～2个选项符合题意。

）1．2008年夏季奥运会将在北京举行；届时要突出“绿色奥运、人文奥运、科技奥运”理念。

绿色奥运是指（）A．加大反恐力度；并讲求各国运动员的营养搭配；使他们全身心投入比赛。

B．严禁使用兴奋剂；使运动员公平竞争。

C．把环境保护作为奥运设施规划和建设的首要条件D．奥运场馆建设均使用天然材料；不使用合成材料2．类推的思维方法在化学学习和研究中常会产生错误的结论；因此类推出的结论最终要经过实践的检验才能决定其正确与否。

下列几种类推结论中正确．．的是（）A．第二周期元素氢化物稳定性顺序是HF＞H2O＞NH3；则第三周期元素氢化物稳定性顺序也是HCl＞H2S＞PH3B．Fe3O4可写成FeO·Fe2O3；Pb3O4也可写成PbO·pb2O3C．可电解熔融MgCl2来制取金属镁；也能电解熔融AlCl3来制取金属铝D．晶体中有阴离子；必有阳离子；则晶体中有阳离子；也必有阴离子3．t℃时；将一定量A(不含结晶水)的不饱和溶液均分为三份；分别加热蒸发；然后冷却为t℃；已知三份溶液分别蒸发水10g；20g；30g；析出A晶体的质量依次为ag；bg；cg.则a；b；c三者的关系是（）A.c=a＋bB.c=2b－aC.c=2b＋aD.c=2a－b4．下列离子方程式书写正确的是（）A．小苏打中加入过量的石灰水Ca2++2OH-+2HCO3-=CaCO3↓+CO32-+2H2OB．氧化铁可溶于氢碘酸Fe2O3+6H+=2Fe3++3H2OC．过量的NaHSO4与Ba(OH)2溶液反应：Ba2++2OH-+2H++SO42-=BaSO4↓+2H2O D．明矾溶液加入Ba（OH）2溶液至沉淀质量最多+3Ba2++6OH-=2Al（OH）3↓+BaSO4↓2Al3++3SO-245．2002年诺贝尔化学奖表彰了两项成果；其中一项是瑞士科学家库尔特·维特里希“发明了利用核磁共振技术测定溶液中生物大分子三维结构的方法”。

46th国际化学奥林匹克July 25, 2014Hanoi, Vietnam理论考试中文译本物理常数，单位，公式和方程Avogadro's 常数 N A = 6.0221 ⨯ 1023 mol –1 普适气体常数 R = 8.3145 J∙K –1∙mol –1 光速 c = 2.9979 ⨯ 108 m∙s –1 Planck's 常数 h= 6.6261 ⨯ 10–34 J∙s 标准压力 p ︒ = 1 bar = 105 Pa大气压 1 atm = 1.01325 ⨯ 105 Pa = 760 mmHg 摄氏零度** K电子质量 m e = 9.1094 ⨯ 10–31 kg1 纳米 (nm) = 10–9 m ; 1 埃 (Å) = 10–10 m 1 电子伏特 (eV) = 1.6022 ⨯ 10–19 J = 96485 J∙mol –1波长为λ的光子的能量 E = hc / λ 一摩尔光子的能量 E m = hcN A / λ Gibbs 自由能G = H – TS平衡常数与标准Gibbs 自由能的关系= exp GK RT ⎛⎫∆ ⎪-⎪⎝⎭积分形式下的van’t Hoff 方程 ⎪⎪⎭⎫⎝⎛-∆=2101211ln T T R H K K内能与热、功的关系 ∆U = q + w摩尔定容热容vm v dT dU C ⎪⎭⎫⎝⎛= ,假设 C v,m 不变，温度从T 1到T 2时，内能的改变U (T 2)=U (T 1)+nC v,m (T 2–T 1),仅考虑自旋，未成对电子数与有效磁矩的公式B.M.)2(+=n n eff μ第一题.箱中微粒: 多烯量子力学中，π 电子沿碳原子中性共轭链的移动可以仿照 “箱中微粒”的方法。

π 电子的能量在下面的方程中给出：2228mLh n E n = n 是量子数(n = 1, 2, 3, …)，h 是Planck’s 常数，m 是电子质量， L 是势箱的长度，并可近似地用 L = (k + 2)×1.40 Å (k 是分子碳链上共轭双键的数目)表示。