Re: [試題] 98上 周必泰 普通化學丙 期末考

[liltwnboiz (TCL)]

※ 引述《ddddccco30 (DC_Bank)》之銘言：
課程名稱︰普通化學丙
課程性質︰系訂必修
課程教師︰周必泰
開課學院：醫學院
開課系所︰醫學系
考試日期（年月日）︰990115
考試時限（分鐘）：170
是否需發放獎勵金：是，謝謝
（如未明確表示，則不予發放）

試題 :

                  醫學系普化丙 General Chemistry (250分)

                        Final Examination 2010/1//15


A.期中考以前之題目(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 為常數)
      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的原子半徑較大，其3p orbital亦較長。Si無法與O形成π bond乃是因為Si和O
      的p orbitals相距太遠。C的原子半徑小，其2p orbital與3p orbital相較之下亦
      較短，可與O形成π bond。由於Si無法形成π bond，也就無法形成 Si=Si,Si=O等
      雙鍵，其構形不似C能有變化，可以形成各種有機化合物，因此生物體的成份主要
      還是由C來組成。

  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的絕對值的總和小於CNO-的formal charge的絕對值的總和
      因此OCN-為比較穩定的化合物。


  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>

      圖略

      XeF2 為直線形
      I3-為直線形
      KrF4為平面四邊形

      XeF2 中的 Xe 為 dsp3 混成
      I3- 中的 I 為 dsp3 混成
      KrF4 中的 Kr 為 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>

      圖略

      由 MO diagram 可以看出B2和O2都有未成對電子，因此具有順磁性

      Bond order:  O2-   O2    O2+
                   1.5   2.0   2.5


B.期中考以後之題目(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:  能量在各型式間轉換時不滅
         Second: 系統所吸收的熱無法完全轉換為功, or
                 自發反應的發生必伴隨著亂度的增加
         Third:  完美晶體在絕對零度時的亂度為0


  B-1-3. Explain the differences between Schottky and Frenkel defects. (5)

         sol>

         Schottky defect: 是晶體結構中的一種因原子或離子離開原來所在的格點位
         置而形成的空位式的點缺陷。在離子晶體中，各種離子形成的肖特基缺陷數
         目符合晶體的元素構成比例，因為只有這樣形成缺陷後的晶體才是電中性的。

         Frenkel defect: 是指晶體結構中由於原先佔據一個格點的原子（或離子）離
         開格點位置，成為間隙原子（或離子），並在其原先佔據的格點處留下一個空
         位，這樣的空位-間隙對就稱為弗侖克爾缺陷。


  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: 處於平衡態的熱力學系統，各宏觀物理量具有確定的值，並
         且這些物理量僅由系統所處的狀態所決定，與達到平衡態的過程無關，因此被
         稱之為狀態函數。

         c, d, e 為狀態函數


  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>

         在標準狀態下，H2O和CH3COOH為液體，因此其entropy value應小於氣體的O2。
         He為單原子氣體，相較於O2並沒有轉動與振動自由度，因此其entropy value亦
         應較小。CO2在標準狀態為氣體，自由度較O2為高，entropy value亦較高。


  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)
       ( 圖為一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. (習題 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>

       圖略

       如果幾個原子集合成分子，這會產生與原子數量成比例的分子軌道。當大量
       （數量級為10^20或更多）的原子集合成固體時，軌道數量急劇增多，軌道相互
       間的能量的差別變的非常小。形成所謂的band。

       物質的導電性決定於價帶與傳導帶之間的能量差，也就是所謂的band gap。
       band gap越小則導電性越大。金屬沒有band gap；半導體的band gap大約為
       1.1 eV；絕緣體則有相當大的band gap。

       (p-type 和 n-type 從略)

       金屬的導電性隨著溫度上升而下降，因為溫度的上升會導致金屬陽離子振動的增
       加，阻礙自由電子的運動。在半導體中，溫度的上升導致一些電子從價帶被激發
       到傳導帶，使得價帶中有電洞而傳導帶中有自由電子，因此導電性上升。


  B-4. (習題 chapter 16-95 衍生) Compare and contrast the phase diagrams of
       water and carbon dioxide(CO2) shown below. (圖為兩者之三相圖)
       (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,或翻成解釋) why the critical temperature for
          H2O ia greater than that for CO2. (5)

       sol>

       (1) 對於液態CO2與氣態CO2而言，在一大氣壓下其分子間作用力大小相近，因此
           固態CO2會直接昇華成氣態CO2，不會經過液態。
       (2) 平衡狀態時 dG = VdP - SdT = 0 => VdP= SdT => dP = S/V dT
              ΔH             ΔH
           = ─── dT => P = ── lnT + C
              TΔV            ΔV
           由以上式子得知對P-lnT作圖，則其斜率為ΔH/ΔV。
           溶解時，ΔH＞0，水的ΔV＜0，斜率為負，
           但是CO2的ΔV＞0，因此斜率為正。


  B-5. (a)由 G=H-TS 的定義，試導出 G 和 pressure(P) and temperture(T) 的相關性
          為 dG = VdP - SdT, in which V denotes the volumn and S is system
          entropy at 25度C. (10)
       (b)利用你學的微積分，試導出

             /δS \       /δV \
             |──|  =  －|──|    (5)    (這裡δ指偏導數符號 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 (過程省略)
                             V1

       (b) Adiabatic: ΔE = q + w = w , P1V1^γ = P2V2^γ

           ΔE = n(Cv)ΔT (過程省略)


  B-7. 化合物如下: (15)

                     O
                    ∥
       CH3 ─ CH2 ─ C ─ CH3
      ￣￣    ￣￣        ￣￣
        a      b           c

       (a)其中氫原子在NMR中可以區分出 a, b 和 c 三群，請問 a 和 c 哪一個
          chemical shift 較大?  請務必解釋. (5)
       (b)請解釋 a, b, c 各有幾個因為 spin-spin coupling 分裂的吸收峰? (5)
       (c)NMR的英文全名為何? 那麼 MRI 呢? (5)

       sol>

       (a) c的chemical shift比較大，因為其氫原子上的電子遭受到鄰近氧原子的
           吸引，使氫原子核受到的遮蔽較少，chemical shift比較大。

       (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. (挑戰題, 14-78, 沒列入習題)
       For each chemical formula below, an NMR spectrum is described,
       including relative overall area (intensities) for the various signals
       given in the parentheses (括號). 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()內數字表示其氫原子數，即積分所得
       而 quartet 為四裂    quintet 為五裂    heptet 為七裂

       sol> 太複雜了 Orz
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