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課程名稱︰普通化學丙
課程性質︰系定必修
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）︰100／1／14
考試時限（分鐘）：10:20~12:10(110min)
是否需發放獎勵金：是
（如未明確表示，則不予發放）
試題：
醫學系普化丙 General Chemistry
Final Examination 2011/1/14 (250 分)
A. 期中考前之題目: (45 分)
A1. For the A4 family, CO2 gas is a prevailing species on the earth, however,
SiO2 gas is rarely observed. Explain.(5)
A2. Define Bosons and Fermions.(5)Which one, bosons or fermions, that electron
and photon, respectively, belong to?(5) Using the concept of bosons or
fermions to explain Pauli Exclusion Principle (5) and Bose-Einstein
Condensation.(5)
A3. (1) Write down the full name as well as explain the VSEPR model.(5)
(2) Draw the Lewis structures and geometry for the following molecules:
(a) XeF2 ; (b) SO2 ; (c) [I3]- ; (d) BeCl2 ; (e) KrF4 (10)
(3) What type of hybridizzation orbital for Xe in XeF2, center I in [I3]-
and Kr in KrF4.(5)
B. 期末考相關考題: (205 分)
B-1. Definition and Explanation
B1-1. C=C stretching frequency is around 1640-1680 cm^-1, while C-H stretching
is in much higher frequency of 2850-3300cm^-1. C=C double bond should be
stronger than C-H single bond. So, why the C=C stretching frequency is
lower? Explain.(5)
B1-2. For the rotation spectroscopy ΔE (J→J+1) = 2hB(J+1) where J is the
quantum number, h is the Planck's constant and B is the rotational
constant. According to the equation, it seems like as J increases to
infinite, ΔE is infinite. This apparently never happened in molecules.
Can you provide any explanation?(5)
B1-3. How many degrees of freedom in motion that CO2 has (5), among which how
many degrees of freedom are in rotation (5) and vibration (5)? Despite
CO2 has rotation, it however did not have rotational spectrum. Explain.
(5) (上課講過)
B1-4. (a) ＜0.1 kcal/mol ; (b) 0.1-0.5 kcal/mol ; (c) 3-6 kcal/mol
(d) 8-10 kcal/mol ; (e) 20-50 kcal/mol ; (f) 70-100 kcal/mol
(g) 150-500 kcal/mol ; (h) ＞1000 kcal/mol
Please locate each of the following energy level to the above evergy
range.
1. C-H stretching frequency 2. Van der Waals ploar-polar interaction
3. 1 eV 4. C-H bonding energy 5. rotational energy gap
6. C≡C triple bond energy. (20)
1
B1-5. Please qulitatively draw the H-NMR spectra of the following molecules
and spin-spin coupling.(10) (The spectra has to be in right order of
position relatively to TMS.)
(a) H H (b) H H H H
│ │ │ │ │ │
H─C─C─Br H─C─C─O─C─C─H
│ │ │ │ │ │
H H H H H H
B1-6. Define (a) exothermic process, (b) endothermic process
(c) endergonic process, (d) exergonic process
(e) chemical potential, (f) Schottky defect. (30)
B1-7. What does state function mean?(5) Which of the following functions are
state functions (複選, multiple choices): (a) work, (b) internal energy,
(c) heat, (d) temperature, (e) gravitational potential, (f) free energy,
(g) f(x,y)=xy+6x2+y^2-2x+3y (5)
B1-8. Arrange the standard entropy value of the following molecules from
lowest to highest one (assuming they are all in gas phase), please
explain. H2O, O2, He, CH3COOH, CO. (5)
B1-9. H = E +PV, using first law of thermodynamics and definition of entropy
to prove dH = TdS + VdP. (5)
B1-10. At 1500 K the process
I2 (g) → 2I (g)
10 atm 10 atm
is not spontaneous. However, the process
I2 (g) → 2I (g)
0.1 atm 0.1 atm
is spontaneous at 1500 K. Give explanation.(5)
B2. Cubic packing spheres are stacked on top of each other in successive
layers. Define
volume occupied by spheres in the unit cell
f_v = ───────────────────────
volume of the unit cell
prove f_v = 52.4% for a cubic packing.(10)
diamond o
B3. C(s) + O2(g) → CO2(g) ΔG = -397 kJ (298 K)
graphite o
C(s) + O2(g) → CO2(g) ΔG = -394 kJ (298 K)
(a) Which one is more stable, diamond or graphite at standard condition?
(5)
(b) If one likes to convert graphite to diamond, using
dΔG = ΔVdP－ΔSdT to explain favorable pressure and temperature.(5)
(c) Commercially, high temerature and high pressure are required to
produce artificial diamond. Explain the difference between this and
result (b).(5)
B4. The phase diagram for water in a close system is shown below, in which
points A-F represent the condition of water at a specific pressure and
temperature. Please answer the following question.
y↑B. .C
│ \ ／ x axis: Temperature
│ \ E ／ y axis: Pressure
│.A \.／
│ / (畫得頗醜請見諒)
│ /
│ /
│ / .D
│/ .F
└────────→
x
2.1 Which represents triple point? And define it.(5) How many unknown
parameters (i.e. degrees of freedom) need to be determined at triple
point?(5)
2.2 Which represents critical point?(5) At this point, what special
physical property can be observed?(5)
2.3 Using the formula dΔG = ΔVdP─ΔSdT for any reaction or phase change
to explain the negative slope for the line B-E.
2.4 Continuation of above question. The line B-E for the case of carbon
dioxide has the positive slope, explain.(5) Rationalize why the
critical temperature for water is greater than for CO2.(5)
B5. Consider 1.00 mol of CO2(g) at 300 K and 5.00 atm. The gas expands until
the final pressure is 1.00 atm. For each of the following conditions
describing the expansion calculate q, w, and ΔE. Cp for CO2 is
37.1 JK^-1‧mol^-1, and assume that the gas behaves ideally.(5)
(a) The expansion occurs isothermally and reversibly.
(b) The expansion occurs isothermally against a constant external pressure
of 1.00 atm.
(c) The expansion occurs adiabatically and reversibly.
B6. Considering two Bricks with two different temperatures and different heat
capacity (Cp) shown below:
┌─────┐┌─────┐
│ ││ │
│ Cp1, T1 ││ Cp2, T2 │
│ ││ │
└─────┘└─────┘
Assuming T1＞T2, using the second law of thermodynamics to predict the
direction of the flow of the heat when two bricks are contacted.(10)
Assuming Cp1 and Cp2 are both temperature independent, what is the final
temperature after connecting these two bricks?(5)
Extra Point (10)
這是你們同學問我的一個問題，我原封不動的copy下來，可以試著回答加分喔(10)
老師，請問我們可以決定某物質的entropy，是依據第三定律: The entropy of a
T2
perfect crystal at 0 K is zero，還有ΔS = n Cp ln(─) 而算出來的吧。就是
T1
可以從該物質在已知溫度下的entropy絕對值推到任何一個溫度下的entropy絕對值
，但第三定律定義的不是0 K的環境嗎？如此一來，T1=0，這樣不能計算啊！
(分母為零)，哪裡出錯了呢？
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