课程名称︰普通化学丙 课程性质︰系定必修 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）︰100／1／14 考试时限（分钟）：10:20~12:10(110min) 是否需发放奖励金：是 （如未明确表示，则不予发放） 试题： 医学系普化丙 General Chemistry Final Examination 2011/1/14 (250 分) A. 期中考前之题目: (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. 期末考相关考题: (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) (上课讲过) 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 (复选, 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 \.／ │ / (画得颇丑请见谅) │ / │ / │ / .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) 这是你们同学问我的一个问题，我原封不动的copy下来，可以试着回答加分喔(10) 老师，请问我们可以决定某物质的entropy，是依据第三定律: The entropy of a T2 perfect crystal at 0 K is zero，还有ΔS = n Cp ln(─) 而算出来的吧。就是 T1 可以从该物质在已知温度下的entropy绝对值推到任何一个温度下的entropy绝对值 ，但第三定律定义的不是0 K的环境吗？如此一来，T1=0，这样不能计算啊！ (分母为零)，哪里出错了呢？ -- --

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