国际化学奥林匹克竞赛试题汇编-第38届ICHO理论试题(中文版)答案

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。

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。

38届化学竞赛试卷

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

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

B. 共价键只存在于共价化合物中。

C. 非极性键只能存在于单质分子中。

D. 化学反应的实质是旧化学键的断裂和新化学键的形成。

3. 设N_A为阿伏伽德罗常数的值。

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

B. 用酒精萃取碘水中的碘。

C. 用托盘天平称取10.58gNaCl固体。

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

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

B. 同主族元素从上到下，金属性逐渐减弱。

2006年第38届IChO-预备试题中文

a — ｛）一（）—（）一（）—（）一（）一（）—（）—（） a 、夸克一质子、中子b 、 10M 个细胞——人类C 、H 、C 、N 、0——H" CH4、NH, HQ （在星际空间中） d 、质子、氮原子核+电子一中性的H 、He 原子 C 、蛋白质、核酸、膜一第一个细胞 f 、质子、中子一氨原子核g 、出、He 、CH, NH3、HQ 、粉尘——太阳系 h 、 H 、He 原子一去离子作用、第一代恒星和银河系i 、质子、氧原子核（轻元素）一重元素如C 、N 、0、P 、S 、Fc 、U ：超新星爆炸j 、 H r 、CH4、NH, H2O 等——地球上的氨基酸、糖、核薈酸、磷脂问题2：星际空间中的氢元素氢元素是宇宙中最丰富的元素，宇宙中大约有75%以上的元素是氢元素，其余的主要是问题1：宇宙生命的“簡史”化学是生命的语言。

生命是以原子、分子和涉及原子分子间的复杂化学反应为基础的。

一个很自然的问题就是^$种原子是从何而来。

按照一种被广泛接受的模型，宇宙起始于大约 150亿年前的一次大爆炸。

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

就现在所知，生命是地球上在一立适中温度下发生的特殊现象。

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

恒星是宇宙空间的特殊物体，因为在恒星形成的过程中温度不是下降而是升高的。

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

膨胀宇宙的温度可以用以下公式简单估计出来：7'= 10山""。

艮中T 是宇宙的平均温度（K ）, r 是时间（宇宙的年龄），以秒为单位。

回答问题1一1〜问题1一6。

保留一位有效数字。

当宇宙年龄为1秒时，质子和中子因温度太高导致熔融，不能聚变成氨原子核。

化学竞赛-第38届国家化学奥林匹克理论试题答案

1-1.The mass of a water droplet:m = V ρ = [(4/3) π r3] ρ = (4/3) π (0.5x10-6 m)3 (1.0 g/cm3)= 5.2x10-16 kg⇒10 marksAverage kinetic energy at 27o C:KE = mv2/2 = (5.2x10-16 kg) (0.51x10-2 m/s)2/2= 6.9x10-21 kg m2/s2= 6.9 x10-21 J ⇒15 marks*.The average kinetic energy of an argon atom is the same as that of a water droplet.KE becomes zero at –273 o C.From the linear relationship in the figure, KE = aT (absolute temperature)where a is the increase in kinetic energy of an argon atom per degree.a = KE/T = 6.9x10-21 J/(27+273K) = 2.3x10-23 J/K⇒25 marksS: specific heat of argon N: number of atoms in 1g of argonS = 0.31 J/g K = a x NN = S/a = (0.31 J/g K) / (2.3x10-23 J/K)= 1.4x1022 ⇒30 marksAvogadro’s number (N A) : Number of argon atoms in 40 g of argonN A = (40)(1.4x1022)= 5.6 x1023⇒20 marks2-1. ⇒ 30 marksmass of a typical star = (4/3)(3.1)(7x108 m)3(1.4 g/10-6 m 3) = 2×1033 g mass of protons of a typical star = (2×1033 g)(3/4 + 1/8) = 1.8×1033 g number of protons of a typical star = (1.8×1033 g)(6×1023/g) = 1×1057number of stellar protons in the universe = (1×1057)(1023) = 1×1080Partial credits on principles:Volume = (4/3)(3.14)radius 3×density; 4 marks 1 mole = 6×1023; 4 marksTotal number of protons in the universe = number of protons in a star ×1023; 2 marks Mass fraction of protons from hydrogen = (3/4)(1/1); 5 marks Mass fraction of protons from helium = (1/4)(2/4); 10 marks2-2. ⇒ 30 marks∆E(2→3) = C(1/4 - 1/9) = 0.1389 C λ(2→3) = 656.3 nm ∆E(1→2) = C(1/1 - 1/4) = 0.75 Cλ(1→2) = (656.3)(0.1389/0.75) = 121.5 nmNo penalty for using Rydberg constant from memory. 15 marks penalty if answered in a different unit (Hz, etc.)2-3.T = (2.9×10-3 m K)/1.215×10-7 m = 2.4×104 K ⇒ 10 marks2-4..⇒ 20 marksλ = 3 × 108 m/1.42 × 109 = 0.21 mT = (2.9 × 10-3 m K)/0.21 m = 0.014 K2-5. ⇒ 10 marks14N + 4He → (17O ) + 1HO-17, O acceptable1783-1.k des = A exp(-E des/R T)= (1x1012 s-1)(5x10-32) = 5x10-20 s-1 at T = 20 K ⇒10 markssurface residence time, τresidence = 1 / k des = 2x1019 s = 6x1011 yr ⇒20 marks(full credit for τhalf-life = ln2 / k des = 1x1019 s = 4x1011 yr)residence time = 2x1019s3-2.The distance to be traveled by a molecule: x = πr = 300 nm.k mig = A exp(-E mig/R T)= (1x1012 s-1)(2x10-16 ) = 2x10-4 s-1 at T = 20 K ⇒ 5 marksaverage time between migratory jumps,τ = 1 / k mig = 5x103 sthe time needed to move 300 nm= (300 nm/0.3 nm) jumps x (5x103 s/jump) = 5x106 s = 50 days ⇒15 marks(Full credit for the calculation using a random-walk model. In this case:t = τ (x/d) 2 = 5 x 109 s = 160 yr. The answer is still (b).)(a) (b)(c) (d) (e)10 marks3-3.k(20 K) / k(300 K) = exp[(E/R) (1/T1 - 1/T2)]= e-112 = ~ 10-49 for the given reaction ).) ⇒15 marks The rate of formaldehyde production at 20 K= ~ 10-49 molecule/site/s = ~ 10-42 molecule/site/ yr⇒10 marks(The reaction will not occur at all during the age of the universe (1x1010 yr).)rate = 10-42molecules/site/yr3-4. circle one(a) (b) (c) (a, b) (a, c) (b,c)(a, b, c)(15 marks, all or nothing)4-1.H PNumber of atoms ( 11.3 ) 1⇒ 10 marksTheoretical wt % ( 3.43 )⇒ 10 marks4-2.adenineN NN NN H H guanineNN N NO N HH HNN O N H H cytosineNN H O O thymine(10 marks on each)4-3. 7 marks each, 20 marks for threeadenineNNNNNHHguanine NN NNON HHH NNH OOthymineNNONHH cytosine NNH OOthymineguanine NN NNON HHHcytosineNNONHHcytosineNNON HHNNHOO thyminethymineNNHOONNH OOthyminethymine NNHOONNONHH cytosineadenineNNNNNHH adenineNNNNNHHadenine NNNNNHHguanineguanine NNNNON HHHNNNNONHHH4-4. 2.5 marks for each bracketadenineN NN N HNH 2guanine N NH N N HO NH 2Uracil N H NH O cytosineN H N NH 2OOHCN ( 5 ) ( 5 ) ( 4 )( 4 )H 2O ( 0 ) ( 1 ) ( 2 ) ( 1 )5-1.(20 marks)1st ionization is complete: H2SO4→ H+ + HSO4-[H2SO4] = 02nd ionization: [H+][SO42-]/[HSO4-] = K2 = 1.2 x 10-2 (1)Mass balance: [H2SO4] + [HSO4-] + [SO42-] = 1.0 x 10-7 (2)Charge balance: [H+] = [HSO4-] + 2[SO42-] + [OH-] (3)Degree of ionization is increased upon dilution.[H2SO4] = 0Assume [H+]H2SO4 = 2 x 10-7From (1), [SO42-]/[HSO4-] = 6 x 104 (2nd ionization is **plete)[HSO4-] = 0From (2), [SO42-] = 1.0 x 10-7 [5 marks]From (3), [H+] = (2 x 10-7) + 10-14/[H+][H+] = 2.4 x 10-7(pH = 6.6) [8 marks][OH-] = 10-14/(2.4 x 10-7) = 4.1 x 10-8[2 marks]From (1), [HSO4-] = [H+][SO42-]/K2= (2.4 x 10-7)(1.0 x 10-7)/(1.2 x 10-2) = 2.0 x 10-12[5 marks]Check charge balance:2.4 x 10-7≈ (2.0 x 10-12) + 2(1.0 x 10-7) + (4.1 x 10-8)Check mass balance:0 + 2.0 x 10-12 + 1.0 x 10-7≈ 1.0 x 10-7Species Concentration** x 10-12HSO4-** x 10-7SO42-** x 10-7H+** x 10-8 OH-5-2. (20 marks)mmol H3PO4 = 0.85 ⨯ 3.48 mL ⨯ 1.69g/mL ⨯ 1 mol/98.00 g ⨯ 1000 = 51.0 [5 marks]The desired pH is above p K2.A 1:1 mixture of H2PO4- and HPO42- would have pH = p K2 = 7.20.If the pH is to be 7.40, there must be more HPO42- than H2PO4-.We need to add NaOH to convert H3PO4to H2PO4-and to convert to the right amount of H2PO4-to HPO42-.H3PO4 + OH-→ H2PO4- + H2OH2PO4- + OH-→ HPO42- + H2OThe volume of 0.80 NaOH needed to react with to to convert H3PO4 to H2PO4- is:51.0 mmol / 0.80M = 63.75 mL [5 marks]To get pH of 7.40 we need:H2PO4- + OH-→ HPO42-Initial mmol 51.0 x 0Final mmol 51.0-x 0 xpH = p K2 + log [HPO42-] / [H2PO4-]7.40 = 7.20 + log {x / (51.0-x)}; x = 31.27 mmol [5 marks]The volume of NaOH needed to convert 31.27 mmol is :31.27 mmol / 0.80 M = 39.09 mLThe total volume of NaOH = 63.75 + 39.09 =102.84 mL , 103 mL [5 marks]Total volume of 0.80 M NaOH (mL) 103 mL5-3. (20 marks)p K = 3.52pH = pK a + log ([A-]/[HA])[A-]/[HA] = 10(pH-pKa) [5 marks]In blood, pH =7.40, [A-]/[HA] = 10(7.40-3.52) = 7586Total ASA = 7586 +1 = 7587 [5 marks]In stomach, pH = 2.00, [A-]/[HA] = 10(2.00-3.52) = 3.02x10-2Total ASA = 1+ 3.02x10-2 = 1.03 [5 marks]Ratio of total aspirin in blood to that in stomach = 7587/1.03 = 7400 [5 marks]** ( 103Ratio of total aspirin in blood to that in stomach6-1. (5 marks)4 H2O + 4 e-→ 2 H2(g) + 4 OH- (or 2 H2O + 2 e-→ H2(g) + 2 OH-)6-2. (5 marks)2 H2O → O2 + 4 H+ + 4 e-(or H2O → 1/2 O2 + 2 H+ + 2 e- )6-3. (5 marks)Cu → Cu2+ + 2e-6-4. (20 marks)Reduction of sodium ion seldom takes place.It has a highly negative reduction potential of –2.710 V.Reduction potential for water to hydrogen is negative (water is very stable).But, it is not as negative as that for sodium ion. It is –0.830 V.Reduction of both copper ion and oxygen takes place readily and the reduction potentials for both are positive.In the present system, the reverse reaction (oxidation) takes place at the positive terminal. Copper is oxidized before water.Reduction potential for hydrogen ion is defined as 0.000 V.6-5. (15 marks)pOH = 14.00 – 4.84 = 9.16[OH-] = 6.92 x 10-10K sp = [Cu2+][OH-]2 = 0.100 x (6.92 x 10-10) = 4.79 x 10-206-6.E = E o Cu2+/Cu + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log (K sp / [OH-]2)= +0.340 + (0.0592/2) log (K sp) - (0.0592/2) log [OH-]2= +0.340 + (0.0592/2) log (K sp) - 0.0592 log [OH-],3 marksBy definition, the standard potential for Cu(OH)2(s) + 2e-→ Cu(s) + 2OH- is the potential where [OH-] = 1.00.E = E o Cu(OH)2/Cu = +0.340 + (0.0592/2) log (K sp)= +0.340 + (0.0592/2) log (4.79 x 10-20)= +0.340 - 0.5722 marks= -0.232 V10 marks-------------------------------------------------------------------------------------------------------------- One may solve this problem as following.Eqn 1: Cu(OH)2(s) + 2e -→ Cu + 2OH-E+o = E o Cu(OH)2/Cu = ?Eqn 2: Cu(OH)2(s) → Cu2+ + 2OH-E o = (0.05916/n) logK sp= (0.05916/2) log(4.79×10-20)= -0.5715 V3 marksEqn 1 – Eqn 2 : Cu2+ + 2e-→ CuE-o = E+o - E o = E o Cu2+/Cu = 0.34 VTherefore, E+o = E-o + E o = + 0.34 + (-0.5715)2 marks= -0.232 V10 marks-0.232 V6-7.Below pH = 4.84, there is no effect of Cu(OH)2 because of no precipitation.Therefore,E = E Cu2+/Cu = +0.340 + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log 0.1003 marks= +0.340 – 0.0296 = +0.310 V7 marks** V6-8.** g graphite = 0.0833 mol carbon6 mol carbon to 1 mol lithium; 1 g graphite can hold 0.0139 mol lithiumTo insert 1 mol lithium, 96487 coulombs are needed.Therefore, 1 g graphite can charge 96487 × 0.0139 = 1340 coulombs. 5 marks1340 coulombs / g = 1340 A sec / g = 1340 x 1000 mA × (1 / 3600) h = 372 mA h / g 5 marks372 mA h / g7-1. (10 marks)n/V = P/RT = (80 x 106 / 1.013 x 105 atm)/[(0.082 atm L/mol/K)(298K)] = 32 mol/L5 marksdensity = mass/volume = d = 32 x 2 g/L = 64 kg/m 3 5 marks64 kg/m 37-2.** or 0.23H 2(g) + 1/2 O 2(g) → H 2O(l); ∆H rexn-1 = ∆H f [H 2O(l)] = -286 kJ/mol = -143 kJ/g 7 marksC(s) + O 2(g) → CO 2(g); ∆H rexn-2 = ∆H f [CO 2(g)] = -394 kJ/mol = -33 kJ/g 7 marks(-∆H rexn-1) / (-∆H rexn-2) = 4.3 or (-∆H rexn-2) / (-∆H rexn-1)= 0.236 marks7-3. (a) (-)1.2 x 105 kJ, (b) (-)6.9 x 104 kJ** x 108 sec or 3.3 x 104 hr or 1.4 x 103 days or 46 month or 3.8 yrI = 0.81 AH 2(g) + 1/2 O 2(g) → H 2O(l)∆H c = -286 kJ/mol = -143 kJ/g = -143 x 103 kJ/kg 5 marksΔG = ΔH – T ΔSΔS c= 70 – 131 – 205/2 = -163.5 J/K/mol5 marksΔG c = -286 kJ/mol + 298K x 163.5 J/K/mol = -237 kJ/mol = -1.2 x 105 kJ/kg 5 marks(a) electric motor W max = ΔG c ⨯ 1 kg = - 1.2 x 105 kJ 5 marks (b) heat engine W max = efficiency x ∆H c 5 marks= (1 – 298/573) x (-143 x 103 kJ) = -6.9 x 104 kJ 5 marks119 x 103 kJ = 1 W x t(sec)t = 1.2 x 108 sec = 3.3 x 104 hr = 1.4 x 103 days = 46 month = 3.8 yr 5 marksΔG = -nFE n = # of electrons involved in the reaction F = 96.5 kC/molH 2(g) + 1/2 O 2(g) → H 2O(l) n = 2 5 marksE = - ΔG/nF = 237 kJ/mol / 2 / 96.5 kC/mol = 1.23 V5 marksI = W/E = 0.81 A5 marks8-1-1. (5 marks on each)①C②C③CO8-1-2.③ Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) 5marks① C(s) + O2(g) → CO2(g) ΔH①◦ = -393.51 kJ = ΔH f◦(CO2(g))② CO2(g) + C(s) → 2CO(g) ΔH②◦ = 172.46 kJFrom ① and ②,ΔH f◦(CO(g)) = (1/2){172.46 + (-393.51)} = -110.525 kJΔH f◦(Fe2O3) = -824.2 kJΔH③◦ = 3ⅹΔH f◦(CO2(g)) - ΔH f◦(Fe2O3) - 3ⅹΔH f◦(CO(g))= 3ⅹ(-393.51) – (-824.2) - 3ⅹ(-110.525) = -24.8 kJ 7 marks ΔS③°=2ⅹ27.28+3ⅹ213.74-87.4-3ⅹ197.674=15.36 J/K 3 marks ΔG③°=ΔH°-TΔS°=-24.8kJ-15.36J/Kⅹ1kJ/1000Jⅹ1473.15K=-47.43 kJ5 marksK = e(-ΔG°/RT)= e(47430J/(8.314J/Kⅹ1473.15K)) = 48 5 marksBalanced equation of ③:K = 48Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)8-2-1. (20 marks)One AB2O4 unit has available 4 (= 1 + (1/4)ⅹ12) octahedral sites.48-2-2. (20 marks)Since one face-centered cube in AB2O4 represents one Fe3O4 unit in this case, it has 8 available tetrahedral sites. In one Fe3O4 unit, 1 tetrahedral site should be occupied by either one Fe2+ (normal-spinel) or one Fe3+ (inverse-spinel). Therefore, in both cases, the calculation gives (1/8) ⅹ100% = 12.5% occupancy in available tetrahedral sites.**%8-2-3. (10 marks for d-orbital splitting, 10 marks for elec. distribution)9-1-1. 1 answer for 8 marks, two for 15 marksH 3CN NNH 3CNNN :::+_+::_:9-1-2. ( 10 marks)H 3CN::9-1-3.H 3CNCH 2CH 2:H 3CN HH CCH 2:(10 marks) (10marks )9-2-1. 5 marks eachHONN +_::ONN:H+:HH_O NN:H+:H_::::::9-2-2.( 10 marks)CH 2CO ::9-3-1.(40 marks)CH 3H 3CH 3C+BC H 2CCH 3CH 3CO 2DEOOO_9-3-2.(10 marks)O OH O n+F10-1. 10 marks eachNMLCH 2OHCH 2OHMeOOMeH HH HOMeMeO CHOCHOCH 2OHCH 2OHHHH H OHOMeMeO OH10-2. 8 marks each for correct structuresNumber of possible structures24 marks12OH(OH)OH(H)HH HHOMeOMeOH COOMeOH(OH)OH(H)HH HHOMeOMeOHCOOMe34OH(OH)OH(H)OH(OH)OHe(H)10-3. 10 marks eachGICH 2OHCH 2OHHHHHMeOOMeOHOMeCH 2OHCH 2OHHHHOMeOMeOMe10-4. 10 marksNumber of the correct structure for C from 10-2110-5.BOH(OH)OH(H)HHHH OHCOOHOHOH10 marks eachDJOH(OH)OH(H)HHHHOMeOMeCOOMeOMeOH(OMe)OMe(H)HHHHOMeOMeOMeCOOMe10-6. 20 marksHOOCOHHH OOOHOOH COOHOOHOHOH COOH11-1. 10 marks311-2. 30 marksCOOHHOOCOOH11-3. 2.5 marks eacha, c, d11-4 30 marksOOCOCOOOHTransition State11-5.For the enzyme-catalyzed reaction, Arrehnius equation could be applied.k cat/k uncat = A exp (-E a, cat/ RT) / A exp (-E a, uncat / RT)= exp [-∆E a, cat-uncat/ RT]= exp [-∆E a, cat-uncat(J/mol) / (2,480 J/mol)] = 106Therefore, -∆E a, cat-uncat = 34,300 J/mol 15 marksk uncat, T/k uncat, 298 = exp (-∆H≠ uncat/ RT) / exp (-∆H≠uncat / 298R)= exp [(-∆H≠ uncat/R)(1/T-1/298)]ln(k uncat, T/k uncat, 298 )= 13.8 = [(-86900/8.32)(1/T-1/298)]Therefore, T = 491 K, or 218o C 15 marks-E a, cat-uncat = 34,300 J/molT = 491 K, or 218o C。

第38届化学奥林匹克试题

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

国际化学奥林匹克竞赛——第38届国际化学奥林匹克选训营初选试题(台湾)

第三十八届国际化学奥林匹克选训营初选试题(台湾)考试时间：120分钟成绩满分：100分(100%)注意!!!※ 学生证(或身份证)置于桌面右上角备查。

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※ 本试题共有二大部分：第一部分，选择题共60分(60%)，单选题与多选题各10题，每题3分。

单选题答错倒扣1分;多选题条错不倒扣.考生不得带电子计算机.第二部分，非选择题共40分(40%)。

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※ 考试时间共120分钟。

第一部分: 选择题(占60%)一、单选题(每题答对得3分,答错倒扣1分,共30分)1-4为题组化学反应常伴随着能量的变化，而以能量的形式放出或吸收的能量通称为化学反应热(△H)。

反应热又因化学反应的分类给予不同的名称。

物质1摩尔,由其成分元素化合生成时的反应热称为生成热(△H f )，例如甲烷的生成热：)(4)(2)(2g g s CH H C →+ mol kJ x H f -=∆而物质1摩尔完全燃烧时，所产生的热量称为燃烧热(△H C )，例如甲烷的燃烧热：)(2)(2)(2)(422g g g g O H CO O CH +→+ mol kJ y H c -=∆而要切断分子内特定的结合键时，所吸收的能量称为键结能(△H b )，例如：)()()(44g s g H C CH +→ z H =∆在甲烷要切断其四个C-H 键中的每一个C-H 键所需的能量各不同,而在实用上常取其平均值,因此C-H 的键结能是上式中总键结能Z 的四分之一,亦即4z H b =∆。

下徇两个图中的数据分别表示水与二氧化碳各1mol 分解时,能量变化的示意图,其中的各数据是以kJ 为单位所表示者.试根据此两图回答题1-4(答案要选取数据最接近者):1. 求H 2O(g)的生成热为几kJ/mol?(A) -243 (B) -247 (C) -436 (D) -463 (E) -6792. 求CO(g)的燃烧热为几kJ/mol?(A) -109 (B) -285 (C) -394 (D) -494 (E) -6033. 求O-H 的键结能为几kJ/mol?(A) 243 (B) 436 (C) 463 (D) 679 (E) 9264. 求)(2)(2)(2)(g g g g H CO O H CO +→+的反应热为几kJ/mol?(A) -42 (B) 67 (C) -67 (D) 436 (E) -9265. 工业上制造硫酸常使用五氧化二钒为催化剂,而制造中的有关过程,都可用化学反应式来表示.试问下列哪一反应最难进行,须用五氧化二钒来催化?(A))(2)(2)(888g g s SO O S →+(B))(3)(2)(222g g g SO O SO →+(C))(42)(2)(3aq l g SO H O H SO →+(D))(722*)(42)(3l conc g O S H SO H SO →+(E))(42)(2)(7222l l l SO H O H O S H →+* 表示98%的浓硫酸6. 一氧化氮是大气污染物,严重危害人体健康.倘若化学家有能力研究出某种催化剂,使其在适当的温度下,能使NO 与过量的甲气体在密闭的反应室作用,将NO 转变为无害的气体,并且可避免二次环境污染.试问最有可能的甲气体是下列的哪一种?(A) SO 2 (B) H 2S (C) CH 4 (D) NH 3 (E) NO 27. 物质甲会有下列的现象或反应:(1). 物质甲溶于稀盐酸得乙溶液(2). 乙溶液中加入硝酸银并搅拌后,过滤得丙溶液(3). 丙溶液中加入不足量的锌粉,搅拌后过滤得沉淀丁(4). 沉淀丁与氧反应即得物质甲试问:物质甲是什么?(A) Cu (B) CuO (C) ZnO (D) MgO (E) Mg8. 下列分子化合物中,何得的沸点最高?(A) 丙醛(MW=58) (B) 丙酮(MW=58) (C)2-丙醇(MW=60)(D) 醋酸(MW=60) (E) 丁烷(MW=58)9. 分子式为C7H7Br 的芳香族化合物,共有几种异构物?(A) 2种 (B) 3种 (C) 4种 (D) 5种 (E) 6种10. 某酯类C 8H 16O 2经水解后，所得的醇再用KMnO 4氧化，结果得到的酸与原水解得到的酸相同。

Chemy化学竞赛联赛试题集答案

Chemy化学竞赛联赛试题合集第一辑（第一届~第八届）参考答案Chemy化学竞赛团队两周年献礼2018年5月本版试题合集仅用于非商业用途交流使用，婉拒商业机构等盈利组织未经授权的转载行为。

版权归全体试题作者共有，侵权必究。

第一届Chemy化学奥林匹克竞赛联赛答案（2016年7月24日19:00 ~ 21:00）·竞赛时间3小时。

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草稿纸在最后一页。

不得持有任何其他纸张。

·姓名、报名号和所属学校必须写在首页左侧指定位置，写在其他地方者按废卷论处。

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第1题（8分）1-1 将明矾置于河水中搅拌，生成载带着悬浮物的氢氧化铝，然后沉降，使水澄清。

但若用蒸馏水代替河水，便不会发生以上现象，请解释该过程的原因，并写出生成氢氧化铝的方程式。

程式。

出化学反应方程式。

第2题（8分）有一个经验规则可以用于估算基元反应的活化能。

①反应中没有键的断裂的反应；②自由基与分子的放热反应；③反应中键不完全断裂（如协同反应）的反应。

④反应中键完全断裂的反应。

遵循上述情况时，分别通过对应下表中计算方法计算活化能：BE是反应中断掉的键的键能之和。

根据上述经验规则及所提供的键能信息，估算氢气与氯气反应中各步基元反应的活化能，并与所给出的实验值比较。

Cl2 = 2Cl·E a1 = 242.8 kJ·mol-1Cl· + H2 = HCl + H·E a2= 25.1 kJ·mol-1H· + Cl2 = HCl + Cl·E a3= 12.6 kJ·mol-1-1-1-1第3题（8分）3-1 Cr2，Mo2与W2分子中存在着六重键，金属的ns轨道与(n-1)d轨道参与成键。

2024化学奥林匹克竞赛试题

2024 化学奥林匹克竞赛试题一、试题有一化学反应 A + B → C，在一定温度下，当A 的浓度为0.5mol/L，B 的浓度为1mol/L 时，反应速率为0.2mol/(L·s)。

若将 A 的浓度增大到1mol/L，B 的浓度不变，此时反应速率变为多少？解析根据反应速率方程v = k[A]^m[B]^n，设该反应中 A 的反应级数为m，B 的反应级数为n。

1. 首先求反应级数：-当 A 的浓度为0.5mol/L，B 的浓度为1mol/L 时，反应速率v1 = 0.2mol/(L·s)，可得方程①：0.2 = k×0.5^m×1^n。

-当A 的浓度增大到1mol/L，B 的浓度不变时，设此时反应速率为v2，可得方程①：v2 = k×1^m×1^n。

-用方程①除以方程①可得：v2/0.2 = (k×1^m×1^n)/(k×0.5^m×1^n)，化简得v2/0.2 = 2^m。

-由于只改变了A 的浓度，B 的浓度不变，且反应速率变为原来的倍数只与A 的浓度变化有关，所以可以通过设特殊值来确定m 的值。

-假设m = 1，则v2/0.2 = 2，解得v2 = 0.4mol/(L·s)。

-假设m = 2，则v2/0.2 = 4，解得v2 = 0.8mol/(L·s)。

-假设m = 3，则v2/0.2 = 8，解得v2 = 1.6mol/(L·s)等，依次类推，可通过给出的选项来确定m 的值，进而确定反应速率v2。

二、试题已知在25①时，水的离子积常数Kw = 1×10^(-14)。

在该温度下，某溶液的pH = 3，求该溶液中氢氧根离子的浓度。

解析1. 因为pH = -lg[H①]，已知pH = 3，则[H①]=1×10^(-3)mol/L。

2. 又因为在任何水溶液中，Kw = [H①][OH①]。

第38届ICHO理论试题(中文版)

理論競賽2006. 7. 7 Gyeongsan, KoreaChemistry for Life, Chemistry for better Life一般規定-每一頁的答案紙上都必須寫上你的名字和代碼。

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常數與有用的公式氣體常數 R = 8.314 J K -1 mol -1 法拉第常數 F = 96485 C mol -1 標準壓力: p = 1.013∙105 Pa 標準溫度: T = 25°C = 298.15 K 亞佛加厥數 N A = 6.022∙1023 mol -1 普朗克常數 h = 6.626∙10-34 J s 光速（真空） c = 3.00∙108 m s -1∆G = ∆H - T ∆S ∆G = - nFE∆G 0 = - RT∙lnK∆G = ∆G 0 + R T∙lnQ with Q =)()(reactands c of product products c of product∆H(T 1) = ∆H 0 + (T 1 - 298.15 K)∙C p (C p = constant)Arrhenius (阿瑞尼士) 方程式 k = A ∙TR E a e⋅-理想氣體定律 pV = nRT Nernst 方程式E = E 0 +redox c c ln nF RT⋅ Beer- Lambert 定律 A = logPP 0= ε∙c∙dV (圓柱體積) = πr 2h A( 球表面積) = 4πr 2 V (球體積) =34πr 31 J = 1 N m 1 N = 1 kg m s -2 1 Pa = 1 N m -2 1 W = 1 A V = 1 J s -1 1 C = 1 A s1. 亞佛加厥數(5 分)現有相同大小之球形水滴分散於氬氣之中。

第38届ICHO理论试题及答案

Theoretical Test2006. 7. 7 Gyeongsan, KoreaChemistry for Life, Chemistry for better LifeGeneral Directions-Write your name and code number on each page of the answer sheet.-You have 5 hours to finish the task. Failure to stop after the STOP command may result in zero points for the task.-Write answers and calculations within the designated box.-Use only the pen and the calculator provided.-There are 23 pages of Problems and 19 pages of Answer Sheet.-An English-language version is available.-You may go to the restroom with permission.-After finishing the examination, place all sheets including Problems and Answer Sheet in the envelope and seal.-Remain seated until instructed to leave the room.Constants and useful formulasGas constantR = 8.314 J K -1mol -1Faraday constant F = 96485 C mol -1Use as standard pressure:p = 1.013∙105PaUse as standard temperature: T = 25°C = 298.15 K Avogadro ’s number N A = 6.022∙1023mol -1Planck constant h = 6.626∙10-34J sSpeed of lightc = 3.00∙108 m s-1∆G = ∆H - T ∆S ∆G = - nFE∆G 0= - RT ∙lnK∆G = ∆G 0+ RT ∙lnQ with Q =)()(reactands c of product products c of product∆H(T 1) = ∆H 0+ (T 1 - 298.15 K)∙C p (C p = constant)Arrhenius equationk = A ∙TR E a e⋅-Ideal gas law pV = nRTNernst equationE = E 0 +redox c c ln nF RT⋅ Beer- Lambert Law A = logPP 0= ε∙c ∙dV(cylinder) = πr 2h A(sphere) = 4πr 2 V(sphere) = 34πr 31 J = 1 N m1 N = 1 kg m s -21 Pa = 1 N m -21 W = 1 A V = 1 J s -11 C = 1 A s1. Avogadro's number(5 pts)Spherical water droplets are dispersed in argon gas. At 27o C, each droplet is 1.0 micrometer in diameter and undergoes collisions with argon. Assume that inter-droplet collisions do not occur. The root-mean-square speed of these droplets was determined to be 0.50 cm/s at 27o C. The density of a water droplet is 1.0 g/cm3.1-1. Calculate the average kinetic energy (mv2/2) of this droplet at 27o C. The volume of a sphere is given by (4/3) π r3 where r is the radius.If the temperature is changed, then droplet size and speed of the droplet will also change. The average kinetic energy of a droplet between 0o C and 100o C as a function of temperature is found to be linear. Assume that it remains linear below 0o C.At thermal equlibrium, the average kinetic energy is the same irrespective of particle masses (equipartition theorem).The specific heat capacity, at constant volume, of argon (atomic weight, 40) gas is 0.31 J g-1 K-1.1-2. Calculate Avogadro's number without using the ideal gas law, the gas constant, Boltzmann’s constant).2. Detection of Hydrogen (5 pts)Hydrogen is prevalent in the universe. Life in the universe is ultimately based on hydrogen.2-1. There are about 1023 stars in the universe. Assume that they are like our sun (radius, 700,000 km; density, 1.4 g/cm3; 3/4 hydrogen and 1/4 helium by mass).Estimate the number of stellar protons in the universe to one significant figure.In the 1920s, Cecilia Payne discovered, by spectral analysis of starlight, that hydrogen is the most abundant element in most stars.2-2. The electronic energy of a hydrogen atom is given by -C/n2relative to zero energy at infinite separation between the electron and the proton (n is the principle quantum number, and C is a constant). For detection of the n=2→ n=3transition (656.3 nm in the Balmer series), the electron in the ground state of the hydrogen atom needs to be excited first to the n=2 state. Calculate the wavelength (in nm) of the absorption line in the starlight corresponding to the n=1→n=2 transition.2-3. According to Wien's law, the wavelength (λ) corresponding to the maximum light intensity emitted from a blackbody at temperature T is given by λT = 2.9×10-3 m K. Calculate the surface temperature of a star whose blackbody radiation has a peak intensity corresponding to the n = 1 → n = 2 excitation of hydrogen.The ground state of hydrogen is split into two hyperfine levels due to the interaction between the magnetic moment of the proton and that of the electron. In 1951, Purcell discovered a spectral line at 1420 MHz due to the hyperfine transition of hydrogen in interstellar space.2-4. Hydrogen in interstellar space cannot be excited electronically by starlight.However, the cosmic background radiation, equivalent to 2.7K, can cause the hyperfine transition. Calculate the temperature of a blackbody whose peak intensity corresponds to the 1420 MHz transition.2-5. Wien generated hydrogen ions by discharge of hydrogen gas at a very low pressure and determined the e/m(charge/mass) value, which turned out to be the highest among different gases tested. In 1919, Rutherford bombarded nitrogen with alpha-particles and observed emission of a positively charged particle which turned out to be the hydrogen ion observed by Wien. Rutherford named this p article the “proton”. Fill in the blank in the answer sheet.14N + 4He → ( ) + 1H3. Interstellar Chemistry (5 pts)Early interstellar chemistry is thought to have been a prelude to life on Earth. Molecules can be formed in space via heterogeneous reactions at the surface of dust particles, often called the interstellar ice grains (IIGs). Imagine the reaction between H and C atoms on the IIG surface that forms CH. The CH product can either desorb from the surface or further react, through surface migration, with adsorbed H atoms to form CH2, CH3, etc.Depending on how energetically a molecule “jumps” from its anchored site, it either leaves the surface permanently (desorption) or returns to a new position at the surface (migration). The rates of desorption and migratory jump follow the Arrhenius formula, k = A exp(-E/R T), where k is the rate constant for desorption or migratory jump, A the jumping frequency, and E the activation energy for the respective event.3-1. Desorption of CH from the IIG surface follows first-order kinetics.Calculate the average residence time of CH on the surface at20 K. Assume that A = 1 x 1012 s-1 and E des = 12 kJ mol-1.3-2. Consider the shortest time it would take for one CH unit to move from its initial position to the opposite side of an IIG by successive migratory jumps. Assume that the activation energy for migration (E mig) is 6 kJ mol-1, and the IIG is a sphere with a 0.1 μm radius. Each migratory jump laterally advances the molecule by 0.3 nm. Show work and choose your answer from (a)-(e) below.(a) t≤ 1 day (b) 10 day ≤t≤ 102 yr (c) 103 yr ≤t≤ 106 yr(d) 107 yr≤t≤ 1010 yr (e) t≥ 1011 yr3-3. Consider the reaction of CO with H2 to form H2CO. The activation energy on a metal catalyst is 20 kJ mol-1, which produces formaldehyde at a rate of 1 molecule/s per site at 300 K. Esitmate the rate of formaldehyde formation per site if the reaction takes place at 20 K.3-4. Which is a set of all true statements? Circle one.(a) Most CH species desorb from the IIG surface before encountering other reactants by surface migration.(b) IIGs can assist transformation of simple molecules to more complex ones in interstellar space.(c) For a reaction on the IIG to occur at an appreciable speed during the age of the Universe (1 x 1010 yr), the reaction energy barrier must be absent or negligible.(a) (b) (c) (a, b) (a, c) (b, c) (a, b, c)4. The Chemistry of DNA (5 pts)4-1. In 1944 Oswald Avery isolated a genetic material and showed, by elemental analysis, that it was a sodium salt of deoxyribonucleic acid. A segment of DNA with formula mass of 1323.72 is shown.Assuming that equimolar amounts of the four bases are present in DNA, write the number of H atoms per P atom. Calculate, to 3 significant figures, the theoretical weight percentage of H expected upon elemental analysis of DNA.4-2. Chargaff extracted the separated bases and determined their concentrations by measuring UV absorbance. The Beer-Lambert law was used to obtain the molar concentration. Chargaff discovered the following molar ratio for bases in DNA:adenine to guanine = 1.43 thymine to cytosine = 1.43 adenine to thymine = 1.02 guanine to cytosine = 1.02Chargaff’s discovery sugge sted that the bases might exist as pairs in DNA. Watson and Crick mentioned in their celebrated 1953 paper in Nature : "It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material."Draw structures of the specific pairing found in DNA. Indicate hydrogen bonds. Omit the sugar-phosphate backbone.4-3. Mutation can occur through base pairings different from the above. Draw structures of any three alternative base pairs.4-4. The plausibility of the formation of purine and pyrimidine bases in the prebiotic atmosphere of the Earth from HCN, NH 3, and H 2O has been demonstrated in the laboratory. Write the minimum number of HCN and H 2O molecules required for formation of the following compounds.adenineN NN N HNH 2guanine N NH N N HO NH 2Uracil N H NH O cytosineN H N NH 2OO5. Acid-Base Chemistry(5 pts)5-1. Calculate [H +], [OH -], [HSO 4-], and [SO 42-] in a 1.0 x 10-7 M solution of sulfuric acid (K w = 1.0 x 10-14, K 2 = 1.2 x 10-2 at 25o C). In your work you may use mass- and charge-balance equations. Answer with two significant figures.5-2. Calculate the volume of 0.80 M NaOH solution that should be added to a 250 mL aqueous solution containing 3.48 mL of concentrated phosphoric acid in order to prepare a pH 7.4 buffer. Answer with three significant figures. (H 3PO 4 (aq), purity = 85 % wt/wt, density = 1.69 g/mL, FW = 98.00) (p K 1 = 2.15, p K 2 = 7.20, p K 3 = 12.44). 5-3. The efficacy of a drug is greatly dependent on its ability to be absorbed into the blood stream. Acid-base chemistry plays an important role in drug absorption.Assume that the ionic form (A -) ofa weakly acidic drugdoes not penetrate the membrane, whereas the neutralform (HA) freely crosses the membrane. Also assume that equilibrium is established so that the concentration of HA is the same on both sides. Calculate the ratio of the total concentration ([HA] + [A -]) of aspirin (acetylsalicylic acid, p K = 3.52) in the blood to that in the stomach.+ Stomach pH = 2.0BloodpH = 7.4 H + + A-HAHAH + A-Membrane6. Electrochemistry (5 pts)Water is a very stable molecule, abundant on earth and essential for life. As such, water was long thought to be a chemical element. However, soon after the invention of a voltaic cell in 1800, Nicholson and Carlyle decomposed water into hydrogen and oxygen by electrolysis.6-1. Water can be thought of as hydrogen oxidized by oxygen. Thus, hydrogen can be recovered by reduction of water, using an aqueous solution of sodium sulfate, at a platinum electrode connected to the negative terminal of a battery. Thesolution near the electrode becomes basic. Write a balanced half-reaction forthe reduction of water.6-2. Water can also be thought of as oxygen reduced by hydrogen. Thus, oxygen can be recovered by oxidation of water at the Pt electrode connected to the positiveterminal. Write a balanced half-reaction for the oxidation of water.6-3. When copper is used at both electrodes, gas is generated only at one electrode during the initial stage of electrolysis. Write the half-reaction at the electrode thatdoes not generate gas.Another species in solution that can be reduced is sodium ion. The reduction of sodium ion to metallic sodium does not occur in aqueous solution, because water is reduced first. However, as Humphrey Davy discovered in 1807, sodium can be made by electrolysis of fused sodium chloride.6-4. Based on these observations, connect the half-reactions with the standard reduction potential (in volts).Reduction of copper ion (Cu2+) · -------------------- · +0.340Reduction of oxygen ·· -2.710Reduction of water ·· -0.830Reduction of sodium ion (Na+) ·· 0.000Reduction of hydrogen ion ·· +1.230The electrode potential is affected by other reactions taking place around theelectrode. The potential of the Cu2+/Cu electrode in a 0.100 M Cu2+ solution changes as Cu(OH)2 precipitates. Answer with 3 significant figures for the following problems.The temperature is 25o C. Note that K w = 1.00 x 10-14 at 25o C.6-5. Precipitation of Cu(OH)2 begins at pH = 4.84. Determine the solubility product of Cu(OH)2.6-6. Calculate the standard reduction potential for Cu(OH)2(s) + 2e-→ Cu(s) + 2OH-.6-7. Calculate the electrode potential at pH = 1.00.Lithium cobalt oxide and specialty carbon are active ingredients for the positive and negative electrodes, respectively, of a rechargeable lithium battery. During the charge/recharge cycles, the following reversible half-reactions occur.LiCoO2Li1-x CoO2 + x Li+ + x e-C + x Li+ + x e-CLi xThe total amount of energy a battery can store is rated in mAh. A battery rated at 1500 mAh can power a device drawing 100 milliamps for 15 hours.6-8. Graphite has lithium intercalation sites between its layers. Assuming a maximum 6:1 carbon-to-lithium intercalation stoichiometry, calculate the theoretical charge capacity of 1.00 gram of graphite to intercalate lithium. Answer in mAh/g with 3 significant figures.7. Hydrogen Economy (4 pts)Hydrogen is more energy-dense than carbon, by mass. Thus, historically there has been a move toward fuel with higher hydrogen content: coal → oil →natural gas → hydrogen. Cost-effective production and safe storage of hydrogen are two major hurdles to the successful inauguration of a hydrogen economy.7-1. Consider hydrogen in a cylinder of 80 MPa at 25 o C. Using the ideal gas law, estimate the density of hydrogen in the cylinder in kg/m3.7-2. Calculate the ratio between heat generated when hydrogen is burned and heat generated when the same weight of carbon is burned. The difference comes to a large extent from the fact that the most abundant isotope of hydrogen has noneutron and hydrogen has no inner electron shell. ∆H f o [H2O(l)] = -286 kJ/mol,∆H f o [CO2(g)] = -394 kJ/mol.7-3. Calculate the theoretical maximum work produced by the combustion of 1 kg hydrogen (a) from the electric motor using hydrogen fuel cell and (b) from the heat engine working between 25 o C and 300 o C. The efficiency (work done/heatabsorbed) of an ideal heat engine working between T cold and T hot is given by [1 –T cold/T hot].S o298[H2(g)] = 131 J/(K mol)S o298[O2(g)] = 205 J/(K mol)S o298[H2O(l)] = 70 J/(K mol).If the fuel cell is working at 1 W and the standard potential difference, how long will the electric motor run at what current?8. Chemistry of Iron Oxides (5 pts)The nucleus of iron is the most stable among all elements and, therefore, iron accumulates at the core of massive red giant stars where nucleosynthesis of many elements essential for life (such as C, N, O, P, S, etc.) takes place. As a result, among heavy elements iron is quite abundant in the universe. Iron is also abundant on Earth.8-1. Development of a technology for reducing iron oxide to iron was a key step in human civilization. Key reactions taking place in the blast furnace are summarized below.C(s) + O2(g) → CO2(g) ΔH◦ = -393.51 kJ(/mol) ----- ①CO2(g) + C(s) → 2CO(g) ΔH◦ = 172.46 kJ(/mol) ----- ②Fe2O3(s) + CO(g) → Fe(s) + CO2(g)ΔH◦ = ? ------------------- ③8-1-1. Indicate the reducing agent in each reaction.8-1-2. Balance reaction ③and calculate the equilibrium constant of reaction ③at 1200 o C. (ΔH f◦(Fe2O3(s)) = -824.2 kJ/mol, S°(J/mol/K): Fe(s) = 27.28, Fe2O3(s) =87.40, C(s) = 5.74, CO(g) = 197.674, CO2(g) = 213.74)8-2. In the manufacture of celadon pottery, Fe2O3 is partially reduced in a charcoal kiln to mixed oxides of Fe3O4and FeO. The amount of thedifferent oxides seems to be related to the“mystic” color of celadon ceramics.Fe3O4 (magnetite) itself is a mixed oxide containing Fe2+ and Fe3+ ions and belongs to a group of compounds with a general formula of AB2O4. The oxide ions form a face-centered cubic array. The figure shows the array of oxygens (gray circles) and representative sites for divalent A and trivalent B cations. The dark circle represents a tetrahedral site and the white circle an octahedral site.8-2-1. How many available octahedral sites for iron ions are there in one AB2O4 unit?Certain sites are shared by neighboring units.AB2O4 can adopt a normal- or an inverse-spinel structure. In normal-spinel structure, two B ions occupy two of the octahedral sites and one A occupies one of the tetrahedral sites. In an inverse-spinel structure, one of the two B ions occupies a tetrahedral site. The other B ion and the one A ion occupy octahedral sites.8-2-2. What percentage of available tetrahedral sites is occupied by either Fe2+or Fe3+ ion in Fe3O4?8-2-3Fe3O4 has an inverse-spinel structure. Draw the crystal field splitting pattern of Fe2+ and fill out the electrons. The electron pairing energy is greater than the octahedral field splitting.9. Photolithographic process (5 pts)Photolithography is a process used in semiconductor device fabrication to transfer a pattern from a photomask to the surface of a substrate. In a typical photolithography process, light is projected, through a mask that defines a particular circuitry, onto a silicon wafer coated with a thin layer of photoresist.9-1. The earliest photoresists were based on the photochemistry that generates areactive intermediates from bis(aryl azide). Patterning becomes possible through the cross-linking reaction of the nitrenes generated from the azides .N 3N 3Bis(aryl azide)reactive intermediate called as nitrene+ 2 N 2SO 3- Na ++Na -O 3S9-1-1. Draw two possible Lewis structures of CH 3-N 3, the simplest compound havingthe same active functional group of bis(aryl azide). Assign formal charges.9-1-2. Draw the Lewis structure of nitrene expected from CH 3-N 3.9-1-3. Draw the structures for two possible products, when this nitrene from CH 3-N 3reacts with ethylene gas (CH 2CH 2).9-2. Photoresists consisting of Novolak polymers, utilizes acid to change theirsolubility. The acid component can be produced photochemically from diazonaphthaquinone. In fact, “Novolaks” have been the representative “positive” photoresists of the modern microelectronic revolution.CH 3OHnNovolakWhen irradiated, diazonaphthaquinone undergoes photochemical decomposition followed by rearrangement eventually producing a carboxylic acid.ON 2S O OOR+ N 2+ H 2ODiazonahpthaquinonederivativeCO 2HS OOORcarbene intermediaterearranged intermediate9-2-1. Draw three Lewis structures of diazoacetaldehyde (see below), the simplestcompound having the same active functional group of diazonaphthaquinone. Indicate formal charges.H-C-CHN 2Odiazoacetaldehyde9-2-2. Draw a Lewis structure of the rearranged intermediate, A (see below),generated from diazoacetaldehyde after losing N 2. A satisfies Lewis’ octet rule and reacts with water to form acetic acid, CH 3CO 2H.HCHN 2O carbene intermediate_N 2ACH 3COOHH 2O9-3. Advanced photoresists were invented in 1982 based on chemical amplification.The most popular chemical amplification for positive-tone involves the acid catalyzed deprotection of poly(p -hydroxystyrene) resin protected by various acid-sensitive protecting groups such as t -butyloxycarbonyl (t -BOC).OnOThe thermal decomposition of carbonate ester itself normally occurs well above 150℃.9-3-1. Two plausible mechanisms have been suggested for this decompositionreaction having relatively high activation energy. Draw expected intermediates and products from this reaction.OOOCH 2H CH 3CH3OOOOH+O+BD++pericyclic trans. stateheterolytic cleavageE+B+H +OH+_CC9-3-2. In the presence of a trace amount of acid, the reaction temperature can bereduced to below 100℃. Draw expected intermediate F from the following chemical amplification process based on using t -BOC.O OOnH ++OHn+ H +OOHOnFDC+B10. Natural Products – Structural Analysis (9 pts)Licorice (Glycyrrhizia. Uralensis) Licorice RootThe flavor extracted from the licorice root is 50 – 150 times sweeter than table sugar. The most important and abundant compound responsible for the sweetness and medicinal effects of licorice is glycyrrhizin (C 42H 62O 16).Glycyrrhizin requires three equivalents of NaOH to effect neutralization.``When glycyrrhizin was subjected to acid hydrolysis, Glycyrrhizinic acid (A (C 30H 46O 4)) and B (C 6H 10O 7) were obtained in a 1:2 molar ratio (figure 1).hydrolysis, hydrolysis produced A’ (methyl glycyrrhizinate), C and D (figure 2). B, C and D exist as mixtures of anomers.Methylation of C and D with MeI produced the same isomeric mixture of compounds, J (figure 3.)C was reduced with LiAlH 4 to give K , and L was produced by the reduction of K . Oxidative cleavage of vicinal diol of L with NaIO 4 produced M and two equivalents of formaldehyde. Reduction of M produced N . The structure and stereochemistry of N was confirmed by the synthesis of N from D-(-)-tartaric acid through methylation followed by reduction (figure 4). A 1H-NMR spectrum of L showed two distinct peaks for methyl groups. (There is no symmetry in L)10-1. Complete structures for L , M, and N in the answer sheet.10-2. How many structures for C are possible? Complete possible structures for C .To determine the correct structure of C , following set of reactions were performed.J was reduced to E, and acid hydrolysis of E produced F . Reduction of F generated G, and G was oxidized with NaIO4 to H with formation of one equivalent of formaldehyde. I was obtained from H through reduction. Among all compounds from A to I , only I was optically inactive (figure 5).10-3. Complete structures for G and I .10-4. Which one is the correct structure for C among ones you have drawn in 10-2?10-5. Complete structures for B, D, and J .10-6. Complete the structure for Glycyrrhizin.11. Enzyme Reaction (7 pts)Shikimic acid biosynthesis is an important pathway for amino acids, alkaloids and heterocyclic natural product production. Nature converts shikimic acid to chorismic acid through a cascade of enzymatic reactions. Then chorismate mutase catalyzes the conversion of chorismic acid to prephenic acid at the branch point for the biosynthesis of aromatic amino acids such as tyrosine and phenylalanine.OCOOHChorismic AcidPrephenic AcidChorismate mutase2COOHOShkimic Acid11-1. During the transformation ofshikimic acid to chorismic acid, dehydrationisoccurring. Choose the hydroxyl group in shikimic acid that is lost through above dehydration among all possible reactions.11-2. Chorismate mutase rearranges chorismic acid into prephenic acid withoutchanging the molecular formula. Chorismic acid becomes prephenic acid through the Claisen rearrangement, a concerted pericyclic process like the Cope rearrangement as shown below:DD DDBased on the following spectral data, propose the structure of prephenic acid.1H-NMR (D 2O, 250 MHz): δ 6.01 (2H, d, J = 10.4 Hz), 5.92 (2H, dd J = 10.4, 3.1 Hz), 4.50 (1H, t, J = 3.1 Hz), 3.12 (2H, s). Note that there are three protons, which have been exchanged by D 2O very fast, and two protons at δ 3.12, which are exchanged slowly in prephenic acid. 13C-NMR (D 2O, 75 MHz): δ 203, 178, 173, 132 (for two identical carbons), 127 (for two identical carbons), 65, 49, 48.δ, chemical shift; H, integrals; d, doublet; dd, doublet of doublet; J, coupling constant; t, triplet; s, singletChorismate mutase is believed to stabilize the transition state of Claisen rearrangement. Thus it is an interesting target for inhibitor design. Inhibitors, called transition state analog (TSA)s that resemble the transition state (TS, e.g., the species in brackets “[ ]” above) of the reaction are designed to occupy the active site. Several inhibitors were designed and synthesized, and among them eight turned out to be potent inhibitors of the enzyme. The lower the IC 50 (inhibitor concentration of 50% of the enzymatic activity) value, the better the inhibitor.OHCO 2HCO 2HOHCO 2HCO 2HOHCO 2HCO 2H1IC 50 = 2.5 mM 2IC 50 = 1.3 mM3IC 50 = 0.78 mMOHO 2COHCO 2HOCO 2H OHCO 2HOHO 2COHCO 2H8IC 50 = 0.00015 mM6IC 50 = 0.017 mM 7IC 50 =0.0059 mM OHCO 2H4IC 50 = 1.1 mMHOCO 2HCO 2H5IC 50 = 5.3 mMHaHa11-3. Choose all correct statements based on the structures and IC 50 values of aboveinhibitors. Increase of factor 5 is considered to be important.(a) Configuration of the hydroxyl group plays an important role in the TS and inhibitor design.(b) The presence of both carboxylic groups is important in the TS and inhibitor design.(c) Transition state of the reaction contains two six-membered rings with one chair and one twist-boat conformation.(d) 7 and 8 can be distinguished on the basis of the 1H-NMR of H a .11-4. Draw the transition state of the transformation of chorismic acid to prephenic acid based on the TSA structures and their IC50 values.11-5. Compared with the uncatalyzed thermal conversion, chorismate mutase accelerates conversion of chorismic acid to prephenic acid 1.0 x 106 fold at 25o C by lowering the activation energy of the reaction. Calculate the decrease in activation energy of chorismate mutase at 25o C.∆H≠uncat is 86,900 J/mol for the thermal conversion of chorismic acid to prephenic acid. At what temperature will the rate of the uncatalyzed thermal conversion be the same as that of the enzyme-catalyzed conversion at 25o C, assuming that E a =∆H≠..。

第38届国际化学奥林匹克选训营初选试题

第三十八届国际化学奥林匹克选训营初选试题(台湾)考试时间：120分钟成绩满分：100分(100%)注意!!!※ 学生证(或身份证)置于桌面右上角备查。

※ 必须在答案纸上作答。

※ 本试题答案纸的每一页，都要写上姓名，试卷连同试题一起缴交监考老师。

※ 请将呼叫器、行动电话、计时器必须关机。

※ 作答时请用蓝、黑色原子笔，可使用立可白涂改，如修改不清楚不予计分。

※ 本试题共有二大部分：第一部分，选择题共60分(60%)，单选题与多选题各10题，每题3分。

单选题答错倒扣1分;多选题条错不倒扣.考生不得带电子计算机.第二部分，非选择题共40分(40%)。

作答必须分开在第10-12页标明题号处作答,题目提供必要之常数与资讯，考生不得带电子计算机。

※ 考试40分钟后才可以开始交卷。

※ 考试时间共120分钟。

第一部分: 选择题(占60%)一、单选题(每题答对得3分,答错倒扣1分,共30分) 1-4为题组化学反应常伴随着能量的变化，而以能量的形式放出或吸收的能量通称为化学反应热(△H)。

反应热又因化学反应的分类给予不同的名称。

物质1摩尔,由其成分元素化合生成时的反应热称为生成热(△H f )，例如甲烷的生成热：)(4)(2)(2g g s CH H C →+ m o l kJ x H f -=∆而物质1摩尔完全燃烧时，所产生的热量称为燃烧热(△H C )，例如甲烷的燃烧热：)(2)(2)(2)(422g g g g O H CO O CH +→+ m o l kJ y H c -=∆而要切断分子内特定的结合键时，所吸收的能量称为键结能(△H b )，例如：)()()(44g s g H C CH +→ z H =∆在甲烷要切断其四个C-H 键中的每一个C-H 键所需的能量各不同,而在实用上常取其平均值,因此C-H 的键结能是上式中总键结能Z 的四分之一,亦即4z H b =∆。

下徇两个图中的数据分别表示水与二氧化碳各1mol 分解时,能量变化的示意图,其中的各数据是以kJ 为单位所表示者.试根据此两图回答题1-4(答案要选取数据最接近者):1. 求H 2O(g)的生成热为几kJ/mol?(A) -243 (B) -247 (C) -436 (D) -463 (E) -679 2. 求CO(g)的燃烧热为几kJ/mol?(A) -109 (B) -285 (C) -394 (D) -494 (E) -603 3. 求O-H 的键结能为几kJ/mol?(A) 243 (B) 436 (C) 463 (D) 679 (E) 926 4. 求)(2)(2)(2)(g g g g H CO O H CO +→+的反应热为几kJ/mol?(A) -42 (B) 67 (C) -67 (D) 436 (E) -9265. 工业上制造硫酸常使用五氧化二钒为催化剂,而制造中的有关过程,都可用化学反应式来表示.试问下列哪一反应最难进行,须用五氧化二钒来催化?(A))(2)(2)(888g g s SO O S →+ (B))(3)(2)(222g g g SO O SO →+ (C))(42)(2)(3aq l g SO H O H SO →+ (D))(722*)(42)(3l conc g O S H SO H SO →+ (E))(42)(2)(7222l l l SO H O H O S H →+ * 表示98%的浓硫酸6. 一氧化氮是大气污染物,严重危害人体健康.倘若化学家有能力研究出某种催化剂,使其在适当的温度下,能使NO 与过量的甲气体在密闭的反应室作用,将NO 转变为无害的气体,并且可避免二次环境污染.试问最有可能的甲气体是下列的哪一种?(A) SO 2 (B) H 2S (C) CH 4 (D) NH 3 (E) NO 2 7. 物质甲会有下列的现象或反应: (1). 物质甲溶于稀盐酸得乙溶液(2). 乙溶液中加入硝酸银并搅拌后,过滤得丙溶液 (3). 丙溶液中加入不足量的锌粉,搅拌后过滤得沉淀丁 (4). 沉淀丁与氧反应即得物质甲试问:物质甲是什么?(A) Cu (B) CuO (C) ZnO (D) MgO (E) Mg8. 下列分子化合物中,何得的沸点最高?(A) 丙醛(MW=58) (B) 丙酮(MW=58) (C)2-丙醇(MW=60)(D) 醋酸(MW=60) (E) 丁烷(MW=58)9. 分子式为C7H7Br的芳香族化合物,共有几种异构物?(A) 2种(B) 3种(C) 4种(D) 5种(E) 6种10. 某酯类C8H16O2经水解后，所得的醇再用KMnO4氧化，结果得到的酸与原水解得到的酸相同。

第38届中国化学奥林匹克初赛试题及答案

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

2) 凡题目中要求书写反应方程式，须配平且系数为最简整数比。

3) 只有题1-3和题2-1-1的计算结果要求修约有效数字。

4) 每个解释题的文字不得超过20个。

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

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

后来，英国化学家Robert Boyle 验证了他的实验。

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

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

“矿渣”难溶于水和稀盐酸，其化学式中两种元素的计量比为1。

每得到1.000g 灰白色金属需要“吞噬”0.7826g 铜。

1-1写出“灰狼”A 、灰白色金属B 和“矿渣”C 的化学式。

A Sb 2S 3B SbC CuS 各2分，共6分。

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

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

第38届化学竞赛初赛试题

选择题
下列哪个分子是极性分子？
A. CO₂
B. CH₄
C. H₂O（正确答案）
D. N₂
在标准状况下，1摩尔任何理想气体的体积约为：
A. 11.2 L
B. 22.4 L（正确答案）
C. 44.8 L
D. 1 L
下列哪个反应是放热反应？
A. H₂O(l) → H₂O(g)
B. 2H₂(g) + O₂(g) → 2H₂O(l)（正确答案）
C. CaCO₃(s) → CaO(s) + CO₂(g)
D. N₂(g) + 3H₂(g) → 2NH₃(g) (注：此反应在特定条件下为吸热)
下列哪个元素的电离能最大？
A. Na
B. Mg
C. Al
D. F（正确答案）
在配位化合物中，中心原子或离子与配体之间形成的化学键主要是：
A. 共价键
B. 离子键
C. 配位键（正确答案）
D. 金属键
下列哪个反应是氧化还原反应？
A. AgNO₃ + NaCl → AgCl + NaNO₃
B. CaCO₃ → CaO + CO₂
C. 2H₂O₂ → 2H₂O + O₂（正确答案）
D. NaOH + HCl → NaCl + H₂O
下列哪个是强酸？
A. HClO
B. H₂CO₃
C. HNO₃（正确答案）
D. CH₃COOH
下列哪个是芳香烃？
A. C₆H₁₂
B. C₆H₆（正确答案）
C. C₄H₁₀
D. C₂H₆
在实验室中，下列哪种方法可以用来制备纯净的氯气？
A. 加热浓盐酸
B. 电解饱和食盐水
C. 用二氧化锰与浓盐酸反应并加热（正确答案）
D. 用铁与稀盐酸反应。

第38届ICHO实验试题(中文版)

2006. 7. 5 Gyeongsan, Korea實作競賽Chemistry for Life, Chemistry for better Life一般規定●你有五小時來做完本實驗，請將時間做有效分配。

建議你用1小時於實作一(佔總分10分)，2小時在實作二(佔總分15分)，及2小時在實作三(佔總分15分)。

●每一頁的答案紙上都必須寫上你的名字和代碼。

●本試卷共有13頁的試題(和3頁附圖) 和7頁的答案紙。

●把你的答案和計算過程寫在規定的地方。

●只可以使用大會提供筆、尺和計算機。

●你可以要求英文題目。

●解釋光譜儀、C18 (逆相管柱) 和安全吸取管使用方法的圖形在另3頁紙上。

●若需要額外的藥品、試劑或玻璃器具需要扣分，一樣東西扣總分一分。

蒸餾水可以無限提供。

●若要上廁所，要告訴助理人員。

●做完後，將所有的紙張(題目和答案) 全部放進信封袋裡，並封好信封。

●請留在座位上，直到宣布可以離開。

●你可以將鉛筆盒、筆、尺、計算機和C18管柱帶回家。

安全和廢液●實驗室裡，一定要戴安全眼鏡。

●本實驗中沒有特別危險的藥品，所有的酸鹼都是稀釋的。

但若沾到，仍應立刻用溼的拭淨紙擦掉。

●不要聞任何藥品。

●將用過的藥品倒到貼有”DISPOSABLE”的白色塑膠瓶，將用過的試管或破玻璃丟到”WASTE BASKET”的籃子裡。

器材與藥品實作1, 2 (白色籃子內)實作3 (黑色籃子內)本實作中，有三樣未使用過之儀器，使用手續如下，若有任何疑問，都可請助教示範給你看。

如何使用光譜儀光譜儀分為三部份：光源、偵測器和樣品槽座。

你會發現樣品槽座處的蓋子(銀色部分)是打開的，在實驗中就讓它開著，不要蓋上。

有一塑膠樣品槽放在樣品槽座內，樣品槽有一面貼有標籤，此面應面向光源，在做實驗時，方向固定如此擺。

(塑膠樣品槽只有兩面是光學面，另兩面為較不透明的。

測光譜時一定要用光學面。

) 參考圖A。

光譜儀已全部準備好，可以使用。

用下列方法測光譜。