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CHEM 132 final.08.key

CHEM 132 final.08.key - CHEM 132 FINAL EXAM I.Idndenberg...

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Unformatted text preview: CHEM 132 FINAL EXAM I{.Idndenberg VtherZOOS NAME g. awe-mama STWJIHENUTLD 1 /20 4. /14 2 /15 5 /18 3 /18 6. /15 TOTAL FINAL EXAM/ 100 TOTAL COURSE / 100 (3()IJI{SIBILIBTTIWBI{ (}I{}&I3E} TOTAL NUMBER OF PAGES INCLUDING THIS COVER PAGE: 16. CHEM 132 1. 20 points maximum; -10 points minimum The following 20 True/ False questions cover a range of chapters. I just want a T or F answer in each, but be very careful! A correct answer is worth +1 point, a blank is 0 points, and an incorrect answer counts —0.5 points. Hence the maximum (+20) and minimum (-10) indicated above. léja. T m 7’ The thermal conductivity K. of a phase is an intensive property that depends on T, P, and the composition of the phase. The heat flow rate dq/dt across the yz plane is proportional to the temperature gradient dT/da: at that plane. As temperature increases, the viscosity of liquids usually decreases and the viscosity of gases usually increases. The root mean square net displacement of diffusing molecules is proportional to the diffusion time. The molar conductivity Am for a strong electrolyte decreases as the electrolyte concentration increases. The viscosity of a liquid decreases as intermolecular attractions increase. Partial orders in a chemical rate law are never negative. For an elementary reaction step A <—> 2B the reaction rates are related as d[A]/dt = -—2d[B]/dt. If r = k[A]2[B] then the initial rate is multiplied by 8 if the initial concentration of A is tripled. The rate law for the elementary reaction A + B —> Prods must be r = k[A][B]. If we know the rate law of a reaction, we can deduce what its mechanism must be. CONTINUED ON NEXT PAGE CHEM 132 Because the Arrhenius equation contains the gas constant R, the Arrhenius equation applies only to gas-phase reactions. When an overall reaction is at equilibrium, the forward rate of a given elementary step must equal the reverse rate of that same step. In homogeneous catalysis, doubling the catalyst concentration will not change the rate of the reaction. In homogeneous catalysis, doubling the catalyst concentration will not change the equilibrium composition of the system. In the partition function sumXZ = 23: e‘fiEfi the sum runs over all the possible energies of the system. For a pure ideal gas of a very large number of molecules, In Z is directly proportional to the number of molecules present. For a system in thermodynamic equilibrium, the total population of states of a given energy always decreases with increasing energy. For a thermodynamic system in equilibrium, molecular states that have the same energy must have the same population. A paremeter—free determination of the rate coefficient based on transition state theory does not require detailed knowledge of the activated complex. w . EXM‘J‘L ‘53 CHEM 132 2. 15 points, 5 points each part A typical electric—field strength for an electrolysis experiment is 10 V/cm. Useful info: R = 8.3145 J mol‘1 K’1 = 8.3145 X 107 erg mol”1 K‘I. Should you need any of these, the atomic weight of magnesium is 24, the molecular weight of water is 18. I also provide here a table of experimental electric mobilities extrapolated to infinite dilution for ions in water at 25°C and 1 atm as provided in Levine: (a) Calculate the drift speed for Mg2+ ions in this field in dilute aqueous solution at the given temperature and pressure. Wit/LE— : (es-.0 i Iii/m2 \j” g") (10 \/ W”) _m (b) Find the rms speed of random thermal motion of these ions, and compare with the result of (a) by giving the ratio rms speed /drift speed. 3: lkgT-ZI is: 4‘“ (Vi) l/L —;_> mes: (gflT/M) : /=_ : Ems I r/mi 10(29291) ; mom/5 (0.022s hag/mi CHEM 132 ((3) Compare the distance traveled by Mg2+ ions in one second due to the electric field with the diameter of a solvent molecule by giving the ratio of the distance to the solvent diameter, which is 3.2 Angstroms. 5 CHEM 132 3. 18 points, 6 points each part l 1'3; (a) Initial rates 7'0 for the overall reaction 2A + C —> Prods at 300 K at various sets of initial concentrations are as follows (where c" E 1 mol/dm3): [Ab/0° 0.20 0.60 0.20 0.60 [B]0/c° 0.30 0.30 0.90 0.30 [C]o/c° 0.15 0.15 0.15 0.45 100T0/(c°/s) 70.60 1.81 5.38 1.81 Assume that the rate law has the form r = k:[1i1]'5"[B]’3[C']’l and determine the partial orders. meWskwm ‘ ifijnaf' (fight/(W :Ejoauwl [C1, M "111;: J 12> 06:: WWW 5.0 MW Sm Wm E33. afm—S‘I‘W in]. M [a]. W mm MEL—l __,_.___> :2 Leila afl' Wilma Wig], Lu ”WW wWW; CHEM 132 V}. 3 CL (b) Evaluate the rate constant for the problem in part (a). CHEM 132 I?“ 3:5 (C) NOW consider the reaction A + B a C + D. A run with [AL] = 400 mmol ' dm’3 and [B]0 = 0.400 mmol dm‘3 gave the following data: Us 0 120 240 360 00 104[C]/c° 0 2.00 3.00 3.50 4.00 and a run with [A]0 = 0.400 mmol dm’3 and [B]o = 1000 11111101 (31111—3 gave Us 0 69 208 435 00 104mm 0 2.00 3.00 3.50 4.00 Find the rate law. The numbers have been Chosen to make detennination of the 1:0 m QMrw/IL [PF] >> [.630 ‘30 [A] is MMHOJLK ”‘51.,” TL. E) mW+®s Rah-L as m W W CE? = 0.qu 0.31.00 0.10.: 0,050 4714:4143 i” : 0 ram 2‘10 Tabby—Lila Ls cart-Jung :3) MAM, 0311+ g [‘51. 1D» WWW E61, 7> [111. so (£1 00W [0—) 1 O,L{OO 0.2100 0400 0,050 k t = 0 64 go? L13 5 :3 1.11-13... .1 H.131 0.1. m. WWW “W “i“ mkwfé fled-“1612x137 14¢, 2’ CHEM 132 4- 14 POintS W: Consider the following reaction of methane with molecular eme- CH4(9) + 012(9) ——> CH3Cl(g) + HCl(g) Experimental studies have shown that the rate law for this reaction is one-half order with respect to 012. Find the rate law for the formation of HCl as determined by the following mechanism, and tell me Whether the result is consistent with the experimen- tally observed behavior. 012 —> 20] 01+ CH4 a HCl + CH3 201 —> 012 The steady state approximation applies to the intermediates Cl and CH3, which should of course not appear in the final rate law. I strongly suggest you implement this for CH3 first. THERE IS ADDITIONAL WORK SPACE ON THE NEXT PAGE. USE IT INSTEAD OF CRAMMINC IN THIS LITTLE SPACE. SWIM shiroqaw. {ya 0143'- CIICHQ Z 0 :_ hmmcmslfiicmiag an: g EClAsj: kaia—JECHWJ *3 {all CHEM 132 ADDITIONAL WORK SPACE FOR PROBLEM 4. shaman“ 3}on 6&de fig CL: 0%? [an a: o _—. M. W; —k1$aj$cm +423 {a Ha] [cm '2.— .— Q [h M: SEBS‘l-{wfia {-ww Aoneufiows W 10 CHEM 132 5. 18 points, 9 points each part 39:9? (a) Suppose a system consists of N noninteracting1 indistinguishabble, identical particles and that each particle has available to it only two quantum states, Whose energies are 51 = 0 and 52 = a. Find expressions for 2, Z , and U. Note: as you no doubt remember from this problem in the homework, the UN! should be omitted from Z since the 1 /N ! belongs as part of the translational factor in Z, which is not being considered in this problem. Be sure that I can clearly recognize your three answers respectively [or 2, Z, and U. -o __a_/,k,T‘ __ kze 4.3 it‘ll—e ai/‘k‘r 1] CHEM 132 (b) For NO, the ground electronic level and the first excited electronic level are each doubly degenerate. The separation between these two electronic levels is a mere 0.0149 eV. There are no other low—lying electronic levels. For NO(g), calculate 2.31 at 150 K. Possibly useful values and conversions: 1 eV = 1.6022 x 10-19 J; k3 : 1.38065 x 10-23 J/K = 1.38065 x 10-16 erg/K. E _ :52. :COIOWMV)CW”3><W '7: _— ' h Cragogfxm I) 12 CHEM 132 6. 17 points, 5 points for (a) and 12 points for (b) 19 g; (a) The hardvsphere collision theory for the elementary reaction NO+O3HN02+OQ leads to a prefactor A equal to 3 x 1014 cm3 mol’l s’l. The experimental A for this reaction is 8 x 1011 (31113 mol‘1 5*. Calculate the steric factor. Show}. 10min F =— M 13 CHEM 132 (b) Celeuiatc (give a numerical value as an answer) the equilibrium constant at T = 300 K for the reaction 2A : A2 given the following information. o The mass of A is 102 g/mole. o The moment of inertia of A2 is 10‘2 g cmz/mol. I The vibrational frequency of A2 in units of degrees Kelvin (i.e. biz/kg) is 300 K. o The energy difference between the lowest electronic energy of A2 and that of 2 A’s is Ale/kg 2 1500 K, and the lowest electronic energy states are non—degenerate. o The partitition functions (which you probably also have on your cheat sheet) are ztrans = (gfifnkBjj/hxg)3/2/CHI3 zmt = 471'21’ kBT/ h? Zvib = (1 — ekhy/kBTYI . h : 6.62607 x 10-34 J s : 6.62607 x 10-27 erg s - k3 = 1.38065 x 10-23 J K-1 = 1.38065 x 10-16 erg K-l. ADDITIONAL WORK SPACE ON NEXT PAGE. 14 I ~1om7— WV :2 CHEM 132 ADDITIONAL WORK SPACE ON THIS PAGE AND ALSO ON NEXT PAGE. AJ‘ __ (L'senonK/K) (30014) _____._._——- k-L. 1’. 6'!HB X1038}! (Mlh-lwo‘fi WE’- 53/54“; 3/2. ‘flfiM‘ET C I, Mx 10—2103) (@5fo {059% lg 3/2 _ “I o A. _ [0x19 u _ Om] CHEM 132 ADDITIONAL WORK SPACE. T. ZR). .8 1; \3 : afinffims Efiiflmf 253.9165 2%:de —-(A€+é!1 )/uk6T— 2. 2_ 8. 35mm anti-ob; W :2 C2-1éxf0n)(é,18’xl0 N15?) 3 3 4-09)! )0— cm (0. 9?6)1 (10“) 16 ...
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