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Practice Exam 2 Answers 2005

Practice Exam 2 Answers 2005 - Exam 2 White Version Feb 25...

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Unformatted text preview: Exam 2 White Version Feb. 25, 2005 (15 pts) Indicate true or false for each of the following statements. Circle the answer. False b) An excited atom can return to its ground state by absorbing electromagnetic radiation. True 1 False i a) The energy of electromagnetic radiation increases as its frequency increases. c) An electron can be removed from a piece of metal if the light is sufficiently intense. True 1 False l ‘E False e) The frequency and wavelength of electromagnetic radiation are inversely proportional to each other. False (5 pts) For which of the following transitions does the light ave thi longest wavelength? Circle the answer. Y1 = 6 a) n=6 to n=4 —— VI: 6 T b) 71:4 to n=6 T ”n: ‘1 c) n=6 to n=3 d) An electron in the n = 4 state in the hydrogen atom can go to the n = 2 state by emitting electromagnetic radiation at the appropriate frequency. ._.rt:3 d) n=3 to n=6 DE : lA'V: l’t C/Afit 0“ Cr z; 5 mall c r A E (8 pts) It took 155 seconds for 50.0% of a particular substance to decompose. If the initial concentration is 0.060 M and the decomposition reaction follows second-order kinetics, what is the value of the rate constant? ”CW : FEE]. .L _ __l_ _L_ k — {l555r() (0ra6ai’l) =’ 0' l0? M 5;; _ ‘tl/r. [h], (8 pts) One pathway for the destruction of ozone in the upper atmosphere is: 03 (g) + 0 (g) -> 2 02 (g) The activation energy for the uncatalyzcd reaction is 14.0 kJ/mol. The destruction of ozone is catalyzed by Cl atoms. O3+Cl ->02+CIO C10 + 0 -—> 02 + C1 03 (g) + 0 (g) —' 2 02(3) The activation energy for the catalyzed reaction is 2.1 kJ/mol. Determine the ratio of the rate constant for the catalyzed reaction to that of the uncatalyzed reaction at 25°C. Assume the frequency factor A is the same for each reaction. ”new; (a (E. — Ea 7/” 6 0'2”“ 2 12.2 Exam 2 White Version 5. Feb. 25, 2005 For a certain process, the activation energy is greater for the forward reaction than for the reverse reaction. a) (4 pts) Draw the reaction profile, indicate the reactants, products, the forward activation energy E., and AE Reaction Coordinate b) (2 pts) Does the reaction have a positive or negative AE? Circle the answer: Q] Given the following reaction mechanism, k. 2N0 7—: N/zef2 fast 13102/4' 02 ’3’ N164 slow 3,01 5’» 2No2 fast a) (4 pts) What is the overall reaction? 2 N 0 + 02 —~> a? N 0:1 b) (4 pts) List the intermediate(s) in the reaction. N Z 0 z own 01 N2 0‘! c) (10 pts) Derive the rate law (eliminate intermediates from your answer). rwté; k2 [N202][01_i K : [Ml ~ 5' I [N0] Z [‘4 [M01]: 14. (“all (5 pts) The following two half-reactions take place in a galvanic cell. At standard conditions, what species are produced at each electrode? -— 9 E :w > 0 a Sn2*+2e’ —> Sn E°=—o.14v l. - - E . Cu2*+2e‘—>Cu E°= 0.34v 3,. —-9 Sr +ze +7.1‘IV z 4 . _, , + 0. 3 ‘l a) Sn is produced at the anode and Cu2+ is produced at the cathode. 0“ 4 l ( 7 6—“ b) Sn is produced at the anode and Cu is produced at the cathode. 0 - ‘18 V . d 2+ . . c) Sn is produced at the catho e and Cu is produced at the anode r ed ul C’ * ’ o M a} not} “an Cu is produced at the cathode and Sn“ is produced at the anode 0)“ M a) I 0 M a} 0 ML e) Cu is produced at the anode and Sn2+ is produced at the cathode. Exam 2 White Version Feb. 25, 2005 Exam 2 White Version Feb. 25, 2005 NOTE: Answer questions 8-13 by referring to the galvanic cell below (the contents of each half-cell are written beneath Answer question 14 using the following half reactions: each compartment) - e“:- K) E’)0 Pb“+2e'—> Pb(s) E"=—0.13V he — A13+ + 3e‘ —> A1(s) E" = — 1.66 14. (10 pts) Consider the cell described below: 0.10 M MnOr‘ 0.20 M Mn2+ 0130 M 0:07” A1 | A1“ (1‘00 M) ll Pb“ (1.00 M) | Pb 0.010 M H+ 0.010 M H+ Calculate the cell potential (at 25°C) afier the reaction has operated long enough for the [Pb2+] to have changed by 0.7 M. The standard reduction potentials are as follows: x 6 (Mnor + 8!? +5.5 _, Mn2+ + 41.120) 50:15“, X5 (Cr2072-+ 14H*+65_ —> 2Cr3+ + 7H20) 50:1.33V (—— re V€VS 5 rxVJ ——0-i3 V 8. (5 pts) Write the overall balanced cell reaction? 6W1n0.1'+ +MW— ——7 6mn*+2W LSIV __ _|.33V WC" +35H0 a 501701 + 70’?” +}/ 900010 0.18\/ F 0.7M 3W 2' i 6WM0LI' +l0Cr2++ ”H10 .9 6sz++ SCFLO; +22H+ 3mm 13W (4 pts) What is the value of E" cell? i 0 . I 8’ V i (4 pts) What' IS the oxidation state of Cr in (313072 "2 K4 +«é/l i + g i} Um’aiodram 7 - - (4 pts) When current is allowed to flow, which species is oxidized? Circle the answer. C r 2 i, —" 0 r2 07 ‘i 6 e a) Cr2072' c) Mno; d) Mn“ e) H+ rcOlUCJtlot/t (4 pts) When current is allowed to flow, which species is reduced? Circle the answer. W1 v1 9a., + 5 e ’ —-7 W‘ V‘ 1 a) Cr207z' b) 0:" d) Mn“ e) H+ (4 pts) In which direction do electrons flow in the external circuit? Circle the answer. a) left to right b) right to left 0) no current flows; the cell is at equilibrium ...
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