※ 引述《ddddccco30 (DC_Bank)》之铭言： 课程名称︰普通化学丙 课程性质︰系订必修 课程教师︰周必泰 开课学院：医学院 开课系所︰医学系 考试日期（年月日）︰990115 考试时限（分钟）：170 是否需发放奖励金：是，谢谢 （如未明确表示，则不予发放） 试题 : 医学系普化丙 General Chemistry (250分) Final Examination 2010/1//15 A.期中考以前之题目(60 分) ￣￣￣￣￣￣￣￣￣￣￣ 　A1. In one dimensional wave motion for electron along the x-axis coordinate where potential energy is assumed to be V=1/2 kx^2 (k 为常数) write down the corresponding Schrodinger equation for a particle with mass m , and also write down the Hamiltonian.(10) sol> ╭ -[(h bar)^2] d^2 ╮ HΨ =│ ────── ─── + 1/2 kx^2 │Ψ = EΨ ╰ 2m dx^2 ╯ Halmiltonian = -[(h bar)^2] d^2 ────── ─── + 1/2 kx^2 2m dx^2 A2. For the A4 family, CO2 gas is a prevailing species on the earth, however, SiO2 gas is rarely observed. Explain.(5) Why life is sustained by C(carbon) not Si, despite they are in the same family.(5) sol> Si的原子半径较大，其3p orbital亦较长。Si无法与O形成π bond乃是因为Si和O 的p orbitals相距太远。C的原子半径小，其2p orbital与3p orbital相较之下亦 较短，可与O形成π bond。由於Si无法形成π bond，也就无法形成 Si=Si,Si=O等 双键，其构形不似C能有变化，可以形成各种有机化合物，因此生物体的成份主要 还是由C来组成。 A3. OCN- and CNO- are isomers. Write down the most stable Lewis structure for OCN- and CNO-.(5) Determine and explain which one(OCN- or CNO-) is a more stable form.(5) sol> - - [ Ｏ－Ｃ≡Ｎ: ] [ :Ｃ≡Ｎ－Ｏ ] -1 0 0 -1 +1 -1 OCN-的formal charge的绝对值的总和小於CNO-的formal charge的绝对值的总和 因此OCN-为比较稳定的化合物。 A4. Draw the Lewis structures and geometry for the following molecules: a.XeF2 b.O3 c.I3- d.BeCl2 e.KrF4.(10) What type of hybridization orbital for Xe in XeF2, center I in I3- and Kr in KrF4.(5) sol> 图略 XeF2 为直线形 I3-为直线形 KrF4为平面四边形 XeF2 中的 Xe 为 dsp3 混成 I3- 中的 I 为 dsp3 混成 KrF4 中的 Kr 为 d2sp3 混成 A5. Draw MOs of B2 and O2 and Explain why both B2 and O2 are paramagnetic. (10) Arrange the bond order of O2, O2-,O2+ by increasing trend (由小到大)(5) sol> 图略 由 MO diagram 可以看出B2和O2都有未成对电子，因此具有顺磁性 Bond order: O2- O2 O2+ 1.5 2.0 2.5 B.期中考以後之题目(190 分) ￣￣￣￣￣￣￣￣￣￣￣￣ B-1. Definition and Explaination (70 分) B-1-1. (a)exothermic process,(b)endothermic process,(c)endergonic process, (d)exergonic process. (10) sol> (a) standard ΔH＜0 (energy is released in the form of heat) (b) standard ΔH＞0 (heat is absorbed from the surrounding) (c) standard ΔG＞0 (not spontaneous) (d) standard ΔG＜0 (spontaneous) B-1-2. First, Second and Third law of thermodynamics. (5) sol> First: 能量在各型式间转换时不灭 Second: 系统所吸收的热无法完全转换为功, or 自发反应的发生必伴随着乱度的增加 Third: 完美晶体在绝对零度时的乱度为0 B-1-3. Explain the differences between Schottky and Frenkel defects. (5) sol> Schottky defect: 是晶体结构中的一种因原子或离子离开原来所在的格点位 置而形成的空位式的点缺陷。在离子晶体中，各种离子形成的肖特基缺陷数 目符合晶体的元素构成比例，因为只有这样形成缺陷後的晶体才是电中性的。 Frenkel defect: 是指晶体结构中由於原先占据一个格点的原子（或离子）离 开格点位置，成为间隙原子（或离子），并在其原先占据的格点处留下一个空 位，这样的空位-间隙对就称为弗仑克尔缺陷。 B-1-4. What does state function mean? (5) Which of the following functions are state functions:(a)PV work,(b)heat,(c)entropy,(d)free energy (e)electrical work. (5) sol> State function: 处於平衡态的热力学系统，各宏观物理量具有确定的值，并 且这些物理量仅由系统所处的状态所决定，与达到平衡态的过程无关，因此被 称之为状态函数。 c, d, e 为状态函数 B-1-5. Which of the following compounds are with the higher standard entropy value than O2?(a)H2O,(b)He,(c)CH3COOH,(d)CO2,please explain. (5) sol> 在标准状态下，H2O和CH3COOH为液体，因此其entropy value应小於气体的O2。 He为单原子气体，相较於O2并没有转动与振动自由度，因此其entropy value亦 应较小。CO2在标准状态为气体，自由度较O2为高，entropy value亦较高。 B-1-6. (a)How many rotations and vibrations for He,O2,H2O,CO2 and O3. (5) (b)These rotation and vibration may not be detected by light absorption. Give the general requirement for the observation of rotational and vibrational spectra. (5) (c)Among these five molecules which have both rotational and vibrational spectra? (5) sol> (a) He: no rotaion nor vibration O2: 2 rotation, 1 vibration H2O and O3: 3 rotation, 3 vibration CO2: 2 rotation, 4 vibration (b) Rotational spectrum: the molecule has to have a permanent dipole moment. Vibrational spectrum: the molecule has to have a temporary change of dipole moment during vibration. (c) H2O and O3 have both rotational and vibrational spectra. B-1-7. (Chapter 16, pages 833-835) (1)Give a general definition of nanoscience (or nanotechnology) (10) (2)What is the general size of a cell? Or the diameter of blood vessels(血管)?, to which the nanoparticles can be applied. (5) (3)Give the explaination of "quantum dots"(see page 835) (5) sol> (1) The field of nanotechnology is typically defined as dealing with particles approximately 1 to 100 nm in size. These particles behave quite differently from normal-sized particles because of their incredible surface area. (2) A cell has a diameter of 10~20 μm. Nanoparticles smaller than 50 nm can enter the cell easily. The diameter of small blood vessels is about 5 μm. Nanoparticles smaller than 50 nm can pass through the blood vessels. Questions(120 分) B-2. Silver crystallizes in a cubic closest packed structure shown below.The radius of a silver atom is 1.44埃, calculate the density of solid silver.(molecular weight of silver = 107.9 g/mol) (10) ( 图为一cubic closest packed structure ) sol> Mass 4×107.9(g/mol)/(6.02×10^23)(1/mol) D = ──── = ────────────────── = 10.6 g/cm^3 Volume [4(1.44×10^(-8))(cm)/√(2)]^3 B-3. (习题 chapter 16-51 衍生) (a)Using the band gap theory to explain the insulator, conductor, semiconductor, p-type semiconductor and n-type semmiconductor. (10) (b)Upon increasing temperature, predict the trend of conductivity (increase of decreade) of conductor and semiconductor.Explain. (5) sol> 图略 如果几个原子集合成分子，这会产生与原子数量成比例的分子轨道。当大量 （数量级为10^20或更多）的原子集合成固体时，轨道数量急剧增多，轨道相互 间的能量的差别变的非常小。形成所谓的band。 物质的导电性决定於价带与传导带之间的能量差，也就是所谓的band gap。 band gap越小则导电性越大。金属没有band gap；半导体的band gap大约为 1.1 eV；绝缘体则有相当大的band gap。 (p-type 和 n-type 从略) 金属的导电性随着温度上升而下降，因为温度的上升会导致金属阳离子振动的增 加，阻碍自由电子的运动。在半导体中，温度的上升导致一些电子从价带被激发 到传导带，使得价带中有电洞而传导带中有自由电子，因此导电性上升。 B-4. (习题 chapter 16-95 衍生) Compare and contrast the phase diagrams of water and carbon dioxide(CO2) shown below. (图为两者之三相图) (1)Why doesn't CO2 have a normal boiling point (meaning a boiling point at 1 atm), whereas water does? (5) (2)Why are the slopes different the solid/liquid lines in the phase diagrams between H2O and CO2? (5) (3)Rationalize(合理化,或翻成解释) why the critical temperature for H2O ia greater than that for CO2. (5) sol> (1) 对於液态CO2与气态CO2而言，在一大气压下其分子间作用力大小相近，因此 固态CO2会直接昇华成气态CO2，不会经过液态。 (2) 平衡状态时 dG = VdP - SdT = 0 => VdP= SdT => dP = S/V dT ΔH ΔH = ─── dT => P = ── lnT + C TΔV ΔV 由以上式子得知对P-lnT作图，则其斜率为ΔH/ΔV。 溶解时，ΔH＞0，水的ΔV＜0，斜率为负， 但是CO2的ΔV＞0，因此斜率为正。 B-5. (a)由 G=H-TS 的定义，试导出 G 和 pressure(P) and temperture(T) 的相关性 为 dG = VdP - SdT, in which V denotes the volumn and S is system entropy at 25度C. (10) (b)利用你学的微积分，试导出 /δS \ /δV \ |──| = －|──| (5) (这里δ指偏导数符号 partial) \δP /T \δT /P (c)On the above vasis and common sense, explain why you need to make diamond from graphite at high temperature and high pressure, giving diamond 0 CO2 (g) → C(s) + O2 (g) △G = 397kJ graphite 0 CO2 (g) → C(s) + O2 (g) △G = 394kJ at 25度C. (5) sol> (a) G =H - TS => dG = dH - d(TS) = dw + dq + d(PV) - TdS - SdT = -PdV + TdS + PdV + VdP - TdS - SdT = VdP - SdT ╭δG ╮ ╭δG ╮ (b) dG(P,T) = │──│ dP + │──│ dT = VdP - SdT ╰δP ╯T ╰δT ╯P Since G is a state function, ╭ δ ╭δG ╮ ╮ ╭ δ ╭δG ╮ ╮ ╭δV ╮ ╭δS ╮ │──│──│ │ = │──│──│ │ => │──│ =－│──│ ╰δT ╰δP ╯T╯P ╰δP ╰δT ╯P╯T ╰δT ╯P ╰δP ╯T (c) At high temperature, ΔG = ΔH - TΔS＞0 (not spontaneous). However, it provides the activation energy needed to make the reaction happen. At high pressure, the process of graphite becoming diamond is favourable because the molar volume of diamond is smaller than that of graphite. B-6. (Chapter 10-85) Consider 1.00 mol of CO2(g) at 300K and 5.00atm. The gas expands until the final pressure is 1.00atm. For each of the following conditions describing the expansion calculate q, w, and △E. Cp for CO2 is 37.1 JK^(-1)mol^(-1), and assume that gas behaves ideally. (10) (a) The expansion occurs isothermally and reversibly. (b) The expansion occurs adiabatically and reversibly. sol> (a) Isothermal: ΔE = q + w = 0 => q = -w V2 w = -∫P dV = -nRT∫ 1/T dT (过程省略) V1 (b) Adiabatic: ΔE = q + w = w , P1V1^γ = P2V2^γ ΔE = n(Cv)ΔT (过程省略) B-7. 化合物如下: (15) O ∥ CH3 ─ CH2 ─ C ─ CH3 ￣￣ ￣￣ ￣￣ a b c (a)其中氢原子在NMR中可以区分出 a, b 和 c 三群，请问 a 和 c 哪一个 chemical shift 较大? 请务必解释. (5) (b)请解释 a, b, c 各有几个因为 spin-spin coupling 分裂的吸收峰? (5) (c)NMR的英文全名为何? 那麽 MRI 呢? (5) sol> (a) c的chemical shift比较大，因为其氢原子上的电子遭受到邻近氧原子的 吸引，使氢原子核受到的遮蔽较少，chemical shift比较大。 (b) a: 3 b: 4 c: 1 (c) NMR = Nuclear Magnetic Resonance MRI = Magnetic Resonance Imaging B-8. Please fill in the sign (>0 or <0) of △S and △H (15) Sign of △S and △H Results ΔS＞0, ΔH＜0 Spontaneous at all tempertures. ΔS＞0, ΔH＞0 Spontaneous at high tempertures. ΔS＜0, ΔH＜0 Spontaneous at low tempertures. ΔS＜0, ΔH＞0 Process not spontaneous at any temp. B-9. (挑战题, 14-78, 没列入习题) For each chemical formula below, an NMR spectrum is described, including relative overall area (intensities) for the various signals given in the parentheses (括号). Draw the structure of a compound having the specific formula that would give the described NMR spectrum. (20) a. C2 H3 Cl3:NMR has one singlet signal b. C3 H6 Cl2:NMR has a triplet(4) and a quintet(2) signal c. C3 H6 O2:NMR has a singlet(1), a quartet(2), and a triplet(3) signal d. C5 H10 O:NMR has a heptet(1), a singlet(3), and a doublet(6) signal e. C3 H6 O:NMR has a triplet(3), a quintet(2), and a triplet(1) signal 其中()内数字表示其氢原子数，即积分所得 而 quartet 为四裂 quintet 为五裂 heptet 为七裂 sol> 太复杂了 Orz --

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