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exam_answers_152A_Exam_2_Key_Version_A

exam_answers_152A_Exam_2_Key_Version_A - {Ma/Q7 E CHEMISTRY...

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Unformatted text preview: {Ma/Q7) E CHEMISTRY 152A HOUR EXAM ll (Version A) E Winter 2008 Wednesday, February 27, 2008 E Name: ETA Section: E Student Number: TA Name: I E Score (page) 1 E 2 GOOD LUCK! 3 E Watch sig. figs and units! 4 E 5 E Total (x/ 100) |. CONCEPTUAL I SHORT ANSWER: Answer all questions (subtotal =52 pts) (M) (4) 1. (a) Give the sign (+ or - ) of the cell voltage 8 and of AG for both an (2) {3) electrolytic cell and a galvanic cell. Galvanic: 8; AG; Electrolytic: 8 - AG + (b) (Circle correct) In an electrolytic cell, the electrons move toward the E -,anode), while in a galvanic cell, the electrons move toward the E (anode, (0) (Circle correct Oxidation always occurs at the (..cathode), while reduction always occurs at the (anode, -. 2. (a) For the solution of the wave equation for the particle-in-a box, why is n=0 not an allowed quantum number? 41- a may»: 6': O is: 4“ ”w W gexmmw’é“ EIMM) ~ ”(/40 Rev 0 nay/d i/lfl/gr/ by, w'c/cv MISMLPI’ Llsncn IVA/op . g— 4’: O ‘% {bro} 5'6 $2.20 44%;; ,éflcfl/E/p p’hc/x‘fléax/ VIE/5,41 Aah’v/efj (b) What happens to the spacing between energy levels AE= E(n)- E(n- 1) as L (the length of the ox) isa wed to growtOI finity? 296:2“; A55 rz'né fey c/e bra”; J Wham/J m; ém’, cl»); 6 jéff’é-c/f4 5:135! e// f /a:r/ 142 3. In the photo— electric effect, the kinetic energy of the electron ejected from the metal surface depends on: (a) (DF) The energy of the incoming photon in excess of the work function of the metal. (b) (T® The intensity (brightness) of the light source. (0) (BF) The chemical element of the metal /5‘ZJ ff)! @ M) (4) 4. For the chemical reaction A —9 B, give the appropriate conditions for spontaneity as a function of temperature for each of the following cases (Possibilities: spon at all temperatures, becomes spont at higher temperature, becomes spont at lower temperature, neverspont at any temperature: (a) AS<0 and AH >0 ”my” 5m74 rev/qr; %279. (b) AS<0 and AH <0 yon/- e/j/ay >474 (c)AS>0 and AH >0 Spa» -e /1 PA”; (d)AS>O and AH <0 5/0»/ 9// 705- 5. For the f0llowing values of AG° , give the corresponding values for the 3 equilbrium constant K (K >1, or K<1 or K=1) lf AG°=0, then K =? 1 If AGO > 0, then K = ? <1 lf AGO < 0, then K=? >1 7 (.2) 6. (a) @F) The change in the Gibbs free energy AG represents the maximum possible useful work obtainable from a process at constant temperature and pressure. (‘2) (b) Which of the following can be assigned an absolute value (circle correct): @H (36) (6) 7. Balance the following redox reaction, showing the separate oxidation and reduction half reactions, then the combined net redox reaction: v02+(aq> + H*<aq> +Zn(s) ------ > voz+<aq> + H200) + Zn2+(aq) *2 (Va: +2/a+c' —s Wyn/71,0) reg/g QM: ?1n ”—5 3447:" 2 f * OX/J/G 0/» 9 Ghao/C 7, 7, 2+ 27‘ [XS/.17 ’7 V0} ('4!) 4' 4’flfef) '7" Zia/,9 "fiQVU/éf) “f 2%?!) +31%?) (-1435. 52A Ké‘[email protected] o )‘N (4) 8. Next to each of the following founders of quantum theory, write the equation most associated with their name in this context: Heisenberg (AA) *fifl) 9- éi Schroedinger H471 : [V De Broglie /7 = % u Einstein (not E= mcz) 6/7”, I!a :7 A? " 41$ (é) 9. For a concentration cell with 0.1 M Ag+ (anode) vs 1.0 M Ag+ (cathode), give an expression for the cell voltage 8 in terms of the standard cell voltage 80 and these concentrations in the reaction quotient Q = / ._ c . E '2‘ ~ 0.05/“9/v/03(0) E :- 0,00 0 J 0'} I ””1 ”s'mw’ 6:5 “0.0591V/03I773 : O"°'°”’ Vf-U =io.ar9lul l i (Z) 10. (a) Name the two physical phenomena for which the explanation led to the l overthrow of classical physics and the beginning of quantum mechanics in the early 20th century. I, /0 4 é—e/cc-ik‘c 9/5621 3 1v 01/ Ceé§40/X€ t l ‘3 ll 13 (2) (b) Name a particle-like property of e-m radiation, and a wave-like property of electrons \,) [alwé—p/cd‘fc ~P/4c/ \) C///7(<;c7{"¥\ a; 64.67%»; i) cuffs-73L; LVg‘cc (2) 11. (a) In the Bohr model for the energy levels of the H atom, why is the sign of the energy negative? _ . fi‘ - - Mfiwvre 57425 [rd/Is A/ +e/ 4,0 4// éouh/ s /t’; /m<'CU) AG“ ”547%,” WV}; @024 ”f; Nflwe. (2:? (b) |_-low does the energy spacing between energy levels AE vary with l - Increasrng n? g .13 z i E : , 2.47.9200 f/%) 5.0 A5 dorms-cc i 4” / M/Z /'M(‘r~e45‘/-rj 47/ 65‘ /é‘70/‘;‘© ! , /;m/7'//4¢=cv)/L< 9///29042/ l X l rs) . l /SZ&’ K52?) (25) ll. QUANTITATIVE: Do all four (4) problems (12 pts each, subtotal = 48) oz) 1. Calculate the value of the solubility product ( Ksp ) for cadmium sulfide CdS, given the following reduction potentials: CdS +2e" ————— > Cd +52' 3° =—1.21v Cd2+ +2e' ----- > Cd 2° = -o.402v CdS+2.e‘->Cd+SZ‘ E°'=-1.21V Cd—+Cd2++2e' ~E°= 0402V CdS(s) ~> Cd2+(aq) + S2'(aq) E° cell =-0.81V. K=K,,-—;? For this overall reaction, E56“ = 00591 log Ksp 11 long— — — = ~3LQ§Q— = -27.41, Ks, = 10-27-41 = 3.9 x 10-28 (/2) 2. The reaction and equilibrium constant for the H2 fuel cell at 25 °C (298 K) are: 2H2 (9) + 02 (g) ------- > 2H20 (I) K = 1.28x1o83 (a) Calculate the standard cell voltage 8° and AG° at 298 K for this reaction. 2 H2(g) + 02(g) -> 2 H200); Oxygen goes from the zero oxidation state to the -2 oxidation state in H20. Since two mol 0 appears in the balanced reaction then n= 4 mol electrons transferred. _0.0591 0.05911 logK= =1.23V a. E° og(1.28 X 1083),E cell: cell 11 AG° = -nFE:eu = -(4 mol e')(96,485 C/mol e')(1.23 J/C) = -4.75 x 105 = —475 k] (b) Predict the signs of AH0 and 138° for this reaction Since mol of gas decrease as reactants are converted into products, then AS° will be negative (unfavorable). Since the value of AG° is negative, then AH° must be negative (AG°— " ‘WO 7 A‘ (c) How does the maximum amount of work obtainable from this fuel cell vary as the temperature increases? (increases, decreases, remains the same) AG = Wmax = AH - TAS. Since AS is negative, then as T increases, AG becomes more positive (closer to zero). Therefore, Wm will decrease as T increases. ‘ 52,9 K£@ (fig) (/7) 3. Considerthe reaction: Fe203(s) + 3H2(g) 9 2Fe(s) + 3H20(g) Assuming that AH° and A30 do not depend on temperature, calculate the temperature at which K=1.00 for this reaction. AH° (kJ/mol) AS° (J/Kmol) F6203 -826 90. H2 131 Fe 27 H20 —242 189 AG°=—RTan; WhenK 1=00, AG° =0sinceln1.00= 0. AG°=0= AH°- TAS°,s AH°= TAS". AH° = 3(—242 kJ) - [-826 kJ] = 100. kJ; AS" = 20.7 J/K) + 3(189 J/K) - [90. J/K + 3031 J/K)] = 138 J/K AH°=TASO AH° 2 100. k] , T= -—— =725 [33" 0.138 kJ/K K ( /z) 4. Calculate the wavelength M in meters) of a photon capable of exciting an electron from the ground state (ni = 1) to a final state nf = 5 of a one-dimensional box of length L =.40.0 pm (Note: pm — 10'12 m) 2 2 h 2 V 24112 En= “h ;AE=E5-El= h (52—12): 2 BmLZ 8mL2 SmL AB ___ 24(6.626 x 10‘34 J s)2 = 9.04 x 10.16 J 8(9.109 x10 31 kg) (40.0 x10 12 m)2 AE—E 7.: __ (6.626x1034Js)(2.998x103m/s) =2,20x10-1°m— 022011111 1» AE 9.04x10161 ...
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