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exam_answers_Exam_2_Version_A_Key - kfl’ CHEMISTRY 152...

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Unformatted text preview: kfl’ @ CHEMISTRY 152 HOUR EXAM ll (Version A) 5 Winter 2007 Friday, February 23, 2007 Name: TA Section: l Student Number: TA Name: 4 Score (page) 1 2 l GOOD LUCK! 3 l Watch sig. figs and units! 4 5 Total (x/100) I. CONCEPTUAL I SHORT ANSWER: Answer all ten (10) (~4 pts each) R “DAWN..." ,A....._.r,.WMWWW. q 1. Give the sign (+ or -) of the cell voltage sand of AG for both an electrolytic cell and a galvanic cell. 8 AG ( l) I) m) Galvanic + __ Electrolytic .. + 2. (Circle correct wording) In an electrolytic cell, the electrons move toward H the w anode), while in a galvanic cell, the electrons move toward the l ‘ (2, Q) (anode, @1529. Q 3. (a) For the solution of the wave equation for the particle-in-a box, why is n=0 not an allowed quantum number? 2 . Via/ab éoone/ca renal/’5“ I 1150’ ‘y sol/Grics/E WW4 ’3‘ i'? % (3) W04}: Ké/k'eaZf/jfi' ”#0012413; Iflrflvo’. f, few/{'c/e 9/hefi/ (IX-=0, d/JO, JO/flID/AO)£ i L (b) What happens to the spacing between energy levels AE = E(n)-E(n-1) ~ as L (the length of the box) is allowed to grow to infinity? ., . i (’3) ‘ A! a’mfi (7M9) kwéfsfecwj 09/4/2512:- 9” fcrgb/c JPCWI 2’9555'64/; ho AV,” yum/5w kw, 57w m mag/1‘9»? pole-n/{u/ (g x 217;) (’{ 4. (Circle all correct quantum numbers): a (9.3 (a) The energy levels of the H atom depend on: @ I, ml [0 (D) (b) The energy levels of the Li 2+ ion depend on: @ I, m; 5. How many possible orbitals (wavefunctions) are available for eachof the L( following states of the hydrogen atom: C 3d 4p n=4 23 ”7 r I 6. In the photo-electric effect, the kinetic energy of the electron ejected from the Ll metal surface depends on: («3 Q (a) @U The energy of the incoming photon in excess of the work " function of the metal. 0°) 2 . (b) ([email protected] intensity of the light source. ’ 7. For the following values of AG°, give the correspondin values for the |fAG° >1, then K=. lfAG° <1, then K=? >1 8. In solving the Schroedinger wave equation for the H atom, spherical polar coordinates (r, o, o) are chosen, rather than Cartesian (x,y,z) coordinates. . Why? Ari/544; 57/6/31 sv'hr'e o4» K‘q/ 5e:- 4 5’0‘on'cq/ 5‘9 ”9%; 1/ «P4— & "mwéak (fl/qr; 91/ 9. Balance the following redox reaction, showing the separate oxidation and i; (a reduction half reactions, then the combined full redox reaction: Mno4'<aq) + I‘<aq) < ------ > I2 (s) +Mn2*(aq) Q2) flPrdur/IA) (5}- + 5w" .L MM; ——» M9744 4% a) A: z A)/0‘x:a/¢Ifa) (1.2" ——-> jE/Zr‘l 4' f - .l 75"?” %+/6#“+2W7, ——§ :7me +9450 @2‘ "7‘73 73/5/ (23 AW.- x; //" 4 Mm; + /a.z‘ —> 2/222? :22 +840 l" 10. Next to each of the following founders of quantum theory, write the equation most associated with their name in this context: l Heisenberg (4 WI)? l Schroedinger fl’ltzfy \ De Broglie 3;: 2y ‘ Einstein (not E= mcz) /o .60) 5/382 ll. QUANTITATIVE: Do all four (4) of the following: (15 pts each) 1. (a) Using the thermodynamic data given below, calculate AGo at 25 °C for the reaction, with all gases at 1.00 atm pressure: 2302(9) f 02(9) -‘——--> 2303(9) AGf°(S65)é-’-3"é71.1'kdlmole - AGf° (302) = - 300 kJ/mole AG° =Z‘.n AG; mm —Z‘.n AG;react ti,AG G°=2(-371) [2(-300.)]=-142kJ P2 ‘ AG = AG° + RT 1n Q -—j -142 k] + RT 1n 7 303 ;_Note: AG= AG° when all gases are at PS-‘Ozxpoz 100 atm. (b) Repeat the calculation of AG0 with all gases at 10.0 atm pressure. (8 3145 J mol"K'1) (10.0)2 = _ ' . : =-142kJ+———————————- 298K)1n ——.—————— 148k] At 10 0 atm AG 1000 J/kJ ( _(10.0)2(10.0) ,1, 2. The reaction and equilibrium constant for the H3, fuel cell at 25 °C (298 K) are: 2H2(9) + 02 (g) —-—-->,13;;,2.H29 (I) K = 1.28x1083 3 ‘5 (a) Calculate the standard cell voltage 8° and AG0 at 298 K for this reaction. 2 H2(g) + 02(g) -+ 2'H20(1); Oxygen goes-from the zero oxidation state to the -2 oxidation state in H20. Since two mol 0 appears in the balanced reaction, then 11 = 4 mol electrons transferred. a. EEC“: 0.0591 log K = 00591 10g(1.28 x 1083), E26“ = 1.23 V n . AG" = —nFE;;,,— = (4 mol e')(96,485 C/mol e')(1.23 J/C) = 4.75 x 105 J = 475 H (b) Predict the signs of AH" and AS° for this reaction. -.b 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 (0) How does the maximum amount of work obtainable from this fuel cell vary as the temperature increases? (increases, decreases, remains the same) c. AG= WW =AH— TAS. Since AS is negative, then as T increases, AG becomes .more positive (closer to zero). Therefore, Wm will decrease as T increases. (“'66) f; 1.)”). 4V lav/7.3 /" “a 3. Consider the following galvanic cell at 25 °C: (Pt / Cr 2* (0.30 M), Cr 3+(2.0 M) II 002* (0.20 M) lCo) for which the overall reaction and equilibrium constant value are: ZCr2+(aq) + 002*(aq) -——-> zcr3*(aq) + Co (s) K = 2.79 x107 Calculate the cell potential 8 and AG for these concentration conditions. (Cr2+ —* CP” + e’) x 2 Co2+ + 2 e" —> Co 2 Cr2+(aq) + Coz+(aq) _, 2 Cr“(aq) + Co(s) ~E2’eu= 0 0:91 10 0g K - 2%?1010 ‘ (2-793 107) = 0.220 V 3+ 2 2 . E =E° - 09591 log—ici—l— = 0.220 V - 00591 log ________(2.0) = 0.151 V 1‘ [Cr2*]2[Coz*] 2 (030)2(020) AG.= -nFE = {2 mol e‘)(96,485 C/mol e')(0.151 1/0) = —2.91 'x 104 J = 29.1 k] 4. Calculate the energy AE ( in Joules) and wavelength '7‘. ( in meters) of a photon capable of exciting an electron in a hydrogen atom (H) from an initial ‘state n.= 2 to a final state nf- — 6 Ali: ~217mzo~‘~"6z‘°‘> ‘1 ...
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