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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 particleina 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’éc/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"°'°”’ VfU =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/ccik‘c 9/5621 3
1v 01/ Ceé§40/X€ t l ‘3
ll
13 (2) (b) Name a particlelike property of em radiation, and a wavelike property
of electrons \,) [alwé—p/cd‘fc ~P/4c/ \) C///7(<;c7{"¥\ a; 64.67%»; i) cuffs73L; 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 dormscc 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, = 102741 = 3.9 x 1028 (/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 onedimensional box of length L =.40.0 pm (Note: pm — 10'12 m) 2 2 h 2 V 24112
En= “h ;AE=E5El= 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) =2,20x101°m— 022011111 1» AE 9.04x10161 ...
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 Spring '08
 Chiu
 Chemistry

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