2150-prelim-ii-2011

2150-prelim-ii-2011 - Chemistry 2150 Fall 2011 Prelim ll Hi...

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Unformatted text preview: Chemistry 2150 Fall 2011 Prelim ll Hi. There are six problems on this prelim each on a separate page (problems 3 — 6 are all spread over two pages). There are five pages at the end of this exam giving a list of useful formulas and constants. (l have added the formulas for spheres and circles to this list.) Please make sure you have all the above pages of the exam. Regular calculators can be used on this exam; graphing calculators are not permitted. The TAs proctoring this exam have been instructed not to answer any questions during the exam. Nonetheless, rarely, questions posed to the TAs are answered to the class as a whole during the exam time. If you finish the exam more than fifteen minutes before the exam time is up, you may turn in your exam. After this time, could you please wait at your desk for all exams to be picked up. Name: Day of lab/TA name: Problem 1: Problem 2: Problem 3: Problem 4: Problem 5: Problem 6: Total: 1. (12 points) For the reaction shown below. the values of AH"f and 8° have been tabulated. This table is also given below. 2N02 (9) ‘3 N204 (9) cmpd AH“f (kJ/mol) 8° ( J K'1 mol'l) NO2 33.1 240.4 N204 9.16 304.3 Asealed flask initially contains 10 atm of NO2 and 1 atm of N204 and is at 250 K. Calculate the initial AG for the reaction. Please also state the value of AG for the reaction once equilibrium has been obtained. (The value of R is given in the printed table of constants.) 2. (16 pts)Consider two different flasks both at STP. One flask contains one mole of ammonia, 14N‘H3, where all the nitrogen atoms are ”N and all hydrogen atoms are 1H. In the second sample, 75% of the ammonia molecules are 15N1H22H1 and 25% of the ammonia molecules are 14N‘H22H1, Assume that these two isotopically labeled ammonia molecules have exactly the same motional behavior as do normal ammonia. Please calculate the difference of entropy of the two flasks. You may use either R or k in your answer. IQ” may leave your answer as a mathematical amraaalgn. l o O 9. 7 5' 9o 2' 5.6)“ q :5 l p; N “--..I N H-.,2.H N “~..z|_ H vs. \ 4 / \ I 'H/ \'H 'H/ 1H [Fl IH 3. (a) (12 points) In Chapter 4 of Fermi, the derivation of the formula 5&0 is given. Please in concise language repeat this proof in the space below. Note Fermi uses results from Chapter 3 (including but not limited to the formula for reversible Carnot cycles, 0le1 = TZIT1 and the formula comparing reversible to irreversible processes, 02/01 >,QZ'/Q1') without proving again these earlier results. Please do the same in your work. extra Space for problem 3 if needed (b) ( 8 points) Fermi needs to use the Second Law to complete the above proof. What idea or device is central in this proof to allow Fermi to use the Second Law? Explain how this idea or device allows Fermi to apply the Second Law. 4. One mole of monoatomic ideal gas undergoes a cycle shown in the pV diagram below, The cycle has the form of a perfect circle. At the point of maximum pressure, point A, gas pressure is 10 atm. At the point of minimum pressure, point C, gas pressure is 1 atm. At point B, the maximum volume of 10 L is reached, while at point D the minimum volume of 1 L is obtained. Motion on the cycle is COUNTERCLOCK-WISE. a)( (8 points) In units of L-atm what is the total work done BY the system after one full cycle? b) (12 points) For each of the four points, state if at that point of the cycle, if the WORK done BY the system is positive, negative or zero. Also state for each of these four points if the HEAT received BY the system is positive, negative or zero. For credit BRIEFLY explain your thinking used to arrive at these answers. extra space for problem 4 Plus-z Kg W AQ‘:AHFTAS 5. (20 pts)The following problem may look difficult, but it can be done quite simply. Remember that at the phase boundary in a pT diagram, AG between the two phases is zero. And remember to use the Clapeyron equation. It's really just a couple of steps ( ie 2 steps) to the answer. In class we looked at liquid-gas transitions, solid-gas phase transitions, and the Clapeyron equation. The arguments developed there apply equally to the geological transformation of Mg(OH)2, brucite, to MgO, periclase, and H20: Mg(OH)2(s) ‘:. MgO(s) + H20 Brucite is solid Mg(OH)2 and periclase is solid MgO. To a very good approximation: AS °< T AV or if we include a proportionality constant, a : AS = a T AV AS is the difference of entropy between one mole of brucite and one mole of periclase plus one mole of water. Brucite has the lesser entropy. AV is the difference in volume of one mole of brucite and one mole of periclase plus one mole of water. Bructite has the lesser volume. Use the Clapeyron equation to express the phase boundary between the two phases as a possible function of AS, AV, a, T, and a constant of integration . Draw a corresponding pT phase diagram for brucite vs. periclase plus water. extra space if needed for problem 5 6. (20 points) A piston is situated as shown in the diagram below. The piston is placed in the wall connecting two very large rooms. The room on the left is at 3 atm pressure, while the room on the right is at 1 atm pressure. As shown in the diagram there are two valves which can let gas very slowly into the chamber where the piston is located. There are 2 moles of monoatomic ideal gas at initially 2 atm pressure. The volume of the chamber is 22.4 L Initially both valves are closed. the balloon is deflated and the piston wall is on the extreme left. The valve on the right is then slowly opened. Only after the gas has expanded into the shown balloon, the valve on the left is slowly opened. All walls and the balloon are perfectly thermally insulated. Assuming the left large room + the piston chamber + balloon (in fact everything but the right large room) is the system, what is the total amount of work done by the system? 2_ moles mdkoa’l'm" i‘le‘l 3’6" 22.0: L Calflad'hbaf‘ extra space if needed for problem 6 Some formulas and constants pV = NRT EK = (m v2)/2 F = m a F = ~ dEP/dx F = dpldt ET = EK + EP Work done = - Work received AE = Q + W dW = - p dV EK = (3NRT)/2 (for a monoatomic ideal gas) R 8.3 J/(moi K) = 0.082 L atm/(moi K) A0 = 6.0 x 1023 Joule = J = Kinetic energy of2 kg moving at 1 m/s 1foot = 3hands ldanL fiasco P\/-.- NRT E =MCJT CV = 3-K I Moll do mmav‘wic EM 344 CVI£%R [mean-La 5W iota-L1 36L.) (lmole\ q. = CMK 3 = («{qu +R‘n‘v H1} Evfiaqma, AQrw AS : 3(3)-SLH3 ’ T A45 Fun. 3 = hlmfl as .. laud—Ea» Fi/MA Law: é d _ AH .23 DH' CF E (—3)? -(F)r ’Zhgmflm 4°wa [gm CV E (fig-L : ($7.61?» “Ll-IQ. E Lo ”(”4" 5° ’ A 4%va .1) TLV. Xv. X 4,5 cl): = M(%) 1"I M l Vl‘l'l c S x on ‘ 32:? x "‘ he“ *2. I 11+! 1““ 5 XMAX -'- :1" { x; "x: 5 ”I L S 'X, aLx 1’ '3‘?" + Q S ’XT’I' : 5 X. : -I?C +8.: 16 1' N ! N ! . . H . Mam aim-J4 I L C'J‘Meuhu c'WJL : Tl'f‘?‘ = 257‘" snafu. W 4, ‘PM'Q Velma 6k 3’51“” ...
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