chem2Exam4 - Chemistry 1302-2 3 Exam 4 Name 11109110 M1166...

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Unformatted text preview: Chemistry 1302-2, 3 Exam 4 April 27, 2006 Name 11109110 M1166 9.00MWF ‘/ 11:00Mi1vri \lultiple choice Lpo oints each). Mark your answergn the scantron. Be careful to erase completely if you change an answer. Only the scanlron will be graded. Check ou ans ers; don' t be careless! Show your work on the exam booklet 1. A+2B+C—-—-’ 2D+E Experiments involving the reaction above yielded the following data; initial rate (mol/L—s) [A]Q (mol/L) 1319 (mol/L) LC1Q (mol/Ll 1 R1 1.407 ‘ 1.40';\ 1.00 "11 2.112: 1/2 R1 0.70: 1.40 s 1.00 3 R3— 1/2 R2 0.70; 0.70 1.00 f 1m 4. R4: 4R3— 3:; 1.403 140-" 0.5011: 11': 3‘17! “1 I ‘1, What IS the total reaction order? 3 C1393 lift "' I???“ A. 21/2 1 \ H .- 9% (E 2) I ‘ . M r \ I C. 3 i K C (13%“ Li} " D. 1 1/2 1 ‘ Lgyglp E. 1 Q’— am ”1:- .1 lizltflgfi ’1 {m :11“ 1 3“”71’M‘ ‘ ‘ 3* :1 L": 1" 3" e0 1 C "1 I" 39.: 2. 2H+ + H202 + 31— *#—’ 13- + 2 H20 {(3% 15‘ L— Experiments involving the reaction above yielded the following data: L45 mm; mm initial rate lmol/L—s) [H203101m01/L) {I}, (incl/L) [11“10 jmol/L) 1. 1.15 x 10‘6 ‘ C1010j ' O00005011L ° “"5 2. 4.37 x 10-6 (0.01038 .010 f 0.;00050 “’ ° CW 3. 1.45 x 10'5 0.0104 0.1262! 0.0005011 4. 5.51 x 10-5 0.038 0.126 0.00960 What IS the meowge reaction? HS 1110“" KL ”.5114 L 013“ C 0096/]? A 1.15 x10 L/mol n P B_ 115101110121 mam-S #1261131" C 1101 C «>133 —4 ,. . 1 1 ‘o 2 ,1" C. 2.30 x 10 /s \HSVMD-FJ‘M? “/1 MM ‘\ 3 /; D 1.15 x 10‘5 L/mol-s ”J" .15 ’ 1 DU 1.571134“ ‘v'b‘ like XED *1 ' - \ E. none of the above , ’ 51‘? m’ _ u ,q 51.51110 r" 1%»! >149" ~10qu 5 1" 1111241 11111199.. 31' 1 We .. ’ .— “-41 ‘ . _ 7 ”p {006119 \e- ‘ V 3. For the decomposition of gaseous dinitrogen pentoxide, \‘anfi ) '; “‘” Vet J 39‘ D ZNzostg) —’—“" 4N02(g) + 02(g) » the rate constant is k = 2.8 x 10'3 s’1 at 60 °C. If the initial concentration of N205 is 1.58mol/L, what is [N205] after 10.0 minutes? / - ‘ g A. 0.68M ‘0 CD ‘1, : -1.g_viujfl WW‘XCGO . ‘ 1 l L D. 1.56M E. 0.42M 4. The gas phase decomposition of H1 ,2 2,9016% 2: 4% Jr 2 Hltg) ——' H2(g) + 12<g) =0 teequfi‘tionglfilate— — k[HI]2. The rate constant k, is 300L/mol—mi at 443 C. ow much time oes 1%” concentration of H1 to drop from 0 010 mol/L to 0.0050 mol/L at443 °C? , \ ,L.» ‘\ 1:, (i 1\“’ mm“ :(090 'mnhxt A. 30 min to} .2 mal 00 3 6, 0‘ B 0.30 min L ) L’ j C. 6.5 min £52 Y @LW‘M : 3'73'7Jw1‘n m ml 1: mf the above W :NH” 5. The reaction S02C12(g) —"—‘“’ SOz(g) +C12(g) -; {93M isj first order in SO2C12 Using the following data, what 1s the rate constant? Wwe. 7121in time is! Eressure SOgCl2 (atm 1 “Maj 2-300 g 1. 0 1.000 in (""3” “4 “‘4 (1 > 2. 2,500 0.947 23 l 3. 5,000 0.895 [(2 2 1‘?) 3:31;)” 4. 7,500 0.848 S 5. 10,000 0803 A. 8.9 x 10'3 Is - B. 1.1X102/s C. 1.83:1,05/5 {1); 2 2 x10'5/s} E. 1 none of the above ’i 7. 8. The decomPositlgnjeaeaeP, 2 N205(g) _’ 4 N02(g) + 02(9 has the rate equation: 1Rate - k[N205]. E/igflpggcgegt of the N205 remains after 6.0 h at 300K. hen e at 300m, .33.. Mater . (A. with m(£9_%:b flijtgw‘) f: \00 B. 2.6h C 7 0.27 h 0 o 43"! 3.1; D. 0.037 h E 15 h h - £2,939.22. { i 'L I XL, R I For the elementary process, N205 (g) —* N02(g) + N03(g), the reverse reaction activation energy (Ea ) and overall are18 kJ/mol and 136 kJ/mol, respectively What 13 the activation energy for the forwar reaction? A. \«290kJ/mol AH AH A019 irxo’r We? eyganJ CV3 Nitrosyl bromide, N OBr, can be made by treating nitrogen monoxide with bromine: 2 N0(g) + BI2(g) —" 2 NOBr(g) Experiments involving the reaction above yielded the following data: M \‘q’ * ’ “ .1319; wmmowmrflomgm 1 24 0.10 1. 0.20 ‘ Li ‘W99 2. 150 .25 0.20 1.1 X10 3. 60 .10 0.50 4 735 0.35 0.50 met 0k . m W O¥€ 1,4 1 (.l0\ L;5 m 1°50 .153 ., 4L0 ”LL41,“ M31 B. 11.21 .2- 7940'}:mele . . 2 n G’WBO—MBS‘W/moP-s £99 ”I @231, 31:...“ 13.1.... 1-2.x 104umo1—s 1515 (.131 a3 ' E. none of the above ”WWW 1,. a mix" it meet/k, Which is false for a balanced elementary process equation? A 1/ Unimolecular and bimolecular processes are more likely than termolecular ones. B./ Elementary processes may be reversible. if/C‘ {l {5 S I C. \/ The slowest elementary process in a mechanism is the rate determining step. D. y/Rate law concentration term exponents are the same as stoichiometric coefficients E None of the above statements is false purify D 19613 “$551215; \C :1 was 1;; 11 item/b 10. In the reaction, A ~——,-’__ products, the fOHOV in0 data are obtained: t= 0s, [A]: 0.88M 1255, 0.;174M 50s 0.62M; (755, 0.5%52M11005 044M ‘1‘.” w_.w 125 s, 0.37 ,in, 150 s, 0.31 M What 13 the instantaneous rate oij reaction at t: 125 s? g, Aaassc—ieiM/s 1, , (is ,1, , 1 é Bwas M/S' ’i’j , w; .00 S? to ”‘4 82 0D 1L; OQAI/ Cw;. .4 2x- 10’: M/s- :1 15 3'5 (D 2.61110“ M/s> _ W \ W“ E. Not enough information Is given to answer the question. “Ea-’V/‘t "$.130" ”0' 90‘4”!) “1%: degassEm As (71%” ¥~ i: A3“ ' S ”1 ?%:»1.1sr1 1121100" (011110113 ' Yea-7”} [11113127. 111111: ‘5 11. The gas phase decomposition of 7N02 at 383 C, gives the following data. V 12. In the reaction, A w products, the fol] exp. 1 [A}O : 1.00 M, t1”: 50 min 0.10% exp. 2 [A10 2 0 50 M, tm- _ 25 min [Cr/q," W exp. 3 ML: 2. 00 M owing data are obtained: , stant for this reaction? A.,,;::.0.02.0. 11-1 7 ”of w! ~ mu . L we x I. 0‘ a 1,, in. \B' ‘ 0.010 moi/Lfmm “1 L m“ A ' f ’LL 7 W133i ~m1n * " vi" “5:. I, 02 X/L' t fl "'— o D. 0 010 L/mol~min F A— . b Et—"N6Te‘nough information Is given to answer the question . ,, An: Awe L .. _._... a (0351 ’L t 13 For a zero order reaction What f four haif— lives ‘2 i , 1 » _____ 1 ”‘ PM“, A. one—half {5:1 1 - 3, 3:; 2 :9 w“ B. one-fourth { 3.! M45“ C. one~eighth <1 :3 \ \ .5 1 ,L one-sixteenth ' a ‘Q MM. ‘ w E. . 14. For a zero order reaction, A N products, a plot ofC versus 15 linear. L '( . "a “7.. ,, D} (0091 2 N'z/OGA t .1 ., I}; [2,, E. none of the above Q“ 15. The followmg mechanism has been proposed for the gas phase reaction of H7 and NO to produce N7 and H20 "Va 2 NO ”fi— W—M— -N2Q73t~n - -- afiasggguilibnum (Hz/+3120; N N20 + HcO E193...) N2 4L H20 ‘ “““““ fast It is possrble to predict the rate expressron and show that the total reaction orderls g 1 am i» an waL 4: :wlo 17. If the mechanism for the reaction Hm) + I2_(g) “H 2 HI'rg) is I204) 2/1?gi fast equilibrium e apparent overall order for {.3121 (”e-“'13” ’5” :1E ”IL 9 ' T " E. 3} l: 1 H2 fr; — TH; C204'2 4* 2 HgCIz M 2 C0; + 2 Cl‘1 + HggClz Experiments involving the reaction above yielded the following data: initialrategmollL—min) ‘ [C204'12]0 (moi/L! IHgClg]D (mol/L) 1. 1.8x 10-5 0.15 2, 0.104~ ‘ l 2. 7.2 x 10'-5 0.30 . 0.104 3. 3.6 x 10—5 0.30 . 0.052 4. X 0.21 0.020 What is the value of X in experiment #4? W «.5 Va Livro'w820wxifi'3’ 333,523. : . 3 L3 ‘; 2m {23; I If the mechanism for the redox reaction ' ng2+ (aq) 19. A sample of uranium was found to have 0.72 % 73 ago did the sample contain 4.0 % 23 reactor. The halfulife for uranium— 5U. (A. 4,5210% ii: 0'19“?) B "”.;§:57X:1087 :flghihaq‘g L" (MAJDES D "1.5x108y m (3:1. : _. K—@ E. ‘ 1.5 x109y ‘03» 20. A A A 10 ears ”‘3 B.V‘__<,}123§:ears ;n( «w E “a: Z) 0T1 fif‘fifyaarg """" " i E. 25 years 00mg}, {1: w ........ n 1-: '3 [an 5U, a level high enou 235 = 7.04 x 108 years? involving an aqueous solution of Hga2+ and Tl3+ is as follows, A Approximately how many years gh to sustain a natural nuclear €1.8"Yw'm .12; w w i: in :00)" L .15 came"? 21. 22. 23. Boron—12 is a radioactive isotope. Predict the decay mode. m 1m 1.. €)_ . m; l :LE _.,_7 (H HQ” W* $9.," C. positron-only ti ;2 b . L . . » .. . D. ,ezeapture-onlym .1 1‘, 5 (3‘ 1. t1. ‘.- E. ‘ either positron or e—capture SE7 '7 F t a LG ,1"; i ' 2) lb 1’; {b 7 g t H Em A sample containing Ra—224 decays by 0: emission and is observed to have the following disintegration rate in dpm (disintegrations per minute). At t = 0 (first observance), 648 dpm; at t = 3 hour, 633 dpm; at t = 12 hours, 590 dpm; and at t = 18 hours, 563 dpm. When will the activity have dropped to 25 % of its initial Value? A. 87 hours ln (“7333 —; v \Q (’5 hr) B. 133 hours 2%? 7 \ (a) C. 167 hours lb 7' U .001 “8 1T D. 172 hours 0 m 2a,) y at t: w z , t t {37 1 L9 The age of some rocks can be determined by the ratio of the number of atoms of 206Pb to the number of atoms of BSU in the rock (assuming that all of the Pb-206 comes from the decay of U—238). The age determined is when the rock solidified. If the ratio of atoms of 206Pb to atoms of 238U in an igneous rock is 0.33, what is the age of the rock? I0 (t 1,2 for mU = 4.5 x 109 years) 2})”, my 132% j :37 l) . «3’73 1, \ [km—fl years ' a .L— B._§.~2».9~x , years L- 9A9 (333 k L6H X10.— (5/ r“ M 9D; 1.9 x109 yeéfi\ E.;~‘:;;r+5~x—109years’ 1‘.“ 3;» \ ~«_, “‘ if" "(V (L035! {: __ \‘E'CQYiOO‘ u. '1 1W! 8E — 328i is‘ an unstable inuclide. Predict the stable isotope which resfil’es fro"‘ sequence starting with silicon—32. A WW2” 93?}: ”A Lf M JV” 1? i B. magnesium—28 / Dir” sulfur-3B '53 , O 3,; ,- i . W“ W e r E. “43‘ \ .. (5 +* 36 PLFCMtDCt neon—24 I; V reg; x o a) e e M New 3&5 M (q; “”1- ! 3 Kb . lv 4 i J E m 333; -3 gm «r .4 ‘5 9‘ 3 3;. I i 359$; * 1% a; mag 25. The age of some fossils can be determined by the ratio of the number of atoms of 87Sr to the number of atoms of 87Rh in the fossilized material. If the ratio of atoms of 8731~ to atoms of 87111b in a fossil is 0.0549, what is the age of the fossil? 3 (t mforszb=4Bx lawyers) , «315v 6:0 A. m y 2e B. 2.7 X 109 years /1, (”CW7 x' ears mm D. 1.9 x 1010 years («LE ) he f6 2.2x 10“ years 0 (9% 9H msyro'” \ V L“, M. 0 _ 2; ‘ I ‘\ L ‘Z'ZL‘OOQ'L (wuuaamama 'srn aw mama's 'uu’ 19d ‘9“): OlLDHH/G SlHi 033:1 ——‘> El S‘HSWOLSHU Elma NOILVHBdHOO‘ llllllrllI-I 401019qu 1 I ‘ .n .L —l e 0? Cl ZZ— awn... ID 93 __ can :3: :9: CV: 92 .— 8 178 — .5; :3: :9: :V: I73 .— 98— Wa L8 — a no: :83 EV:l 98 — :V: E 3 :V: ’F :V: fit :‘a: :Cl: :3: :3 M a n > u r. U! u :9: :V: :3: :fl: :0: :92! :9: :V: :3: ca: :0: 05V: :3: :a: :V: :3: :a: :0: null =V= mg: :0: :0: :‘a: q. :0: :9: :V: 33 3: Q: 3: c :0: :0: CE: EV: ca: :3: :3: CV: ee- :3: m :3: E03 IG- -=a: :0: :9: M13 ca: :3: :9: :V: :0: :5): Ca: :V: n It: as: :3: =17: uh :15“: =8: no: :3: can 36" :3: :0: :3: t :03 EH: EV: :El: n > u n to u n O u n U u n II n cl: can :0: =E= CV: 38 .— cl: :0: can :83 “ch L9 .— =an :0: :02 IE: EV: OB _ =3: =03 :3: :9: av: 62 _ can :0: 50:: :g: EV] gz __ :5]: can :9: :9: EV: Lz _ :El: m C03 593 EVJ ea —— :3: :0: :0: t9] :35 E03 m :3: can aw =33 *9. cfi: can lib : Li; a: ...
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