old_final - 14A2F01-F _ -1- Baur CHEMISTRY 14A LECTUH E...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 14A2F01-F _ -1- Baur CHEMISTRY 14A LECTUH E SECTION 2 FALL QUARTER, 2001 Final Examination Friday, December 14, 2001 Name TA: (check one) (Last) _ (First FINAL GRADE m counsa_____ Damian Aheme__ Kim Miller“ (place an 'X' here if you do not wishvyour Molly Cavanaugh___ Will Molenkamp‘ grade entered) Natalie Gassman . Giveunnstor all answers. unless dimensionless. . . Open book examination. Any written or printed materials may be used. ' . DO NOT LIFT THIS COVER SHEET AND BEGIN WORK UNTIL INSTRUCTED TO DO SOI WWW? H 1.008 O 16.00 Na 22.99 Cl 35.45 Cr 52.00 Br 79.90 He 4.003 F 19.00 . Mg 24.30 Ar 39.95 Fe 55.85 Ag107.87 C . 12.01 No 20.18 8 32.06 K 39.10 Cu 63.55 | 126.90 N 14.01 , Ca 40.08 Zn 65.39 NA (Avogadro's Number) = 6.022 x 1023 Ideal Gas Law: PV = nRT R = Gas Constant = 0.0821 lit~atm moi-1-K-1 T(K) = t(°C) + 273.15 = 0.0831 lit-bar moH‘K-I - ' h = Planck's Constant: 6.63 x 10-34J-s c = velocity of light in vacuum = 3.00 x 108 m/s Photon energy: a - hv - 110/)». Bohr formula: E, = -2.178 x 10-18(22/n2) J 1 e.v.(electron volt) = 1.60 x 10-19.! Coulomb energy between two charges 01 and 02;; EC = 2.31 x 10-19 J nm(Q102/r) (Q in multiples ot the elementary charge) The new and algalahlg are on the back of this sheet. THE BACK OF THE EXAM AND PIEIINTYOUFI MAILING ADDRESS. if you do not have your exam mailed back to you, inquire at WG Y-.3054 about availability for return after January 7, 2002. 14A2F01-F Name 1. (50 points). (a) (10 points) A laser light source emits a pulse of photons with wavelength 700 nm. If the pulse contains 0.100 mole of photons, calculate the energy of the pulse. ' A C fame“! : {1/00 MoLE XJJZK/ouprlo'mus MOI: x T : Lozwoflx 6‘ 3NDJ¢J5 X5.oo;</0 8M5“ 7'00K/0'7/n -: /.7/;s/0 KJ s/z/Aj (b) (10 points) Photon A has wavelength 600 nm. Photon B has wavelength 300 nm. lf photon A and photon B are combined in an optical device to make a new photon, C, whose energy is the sum of those of A and B, what is the wavelength of photon C? ha _ kc Eumy 0F c, _~: I- : En/wroFA-r EN££HOFB- ’“ I? C 3 _L AA’ .L—Js .L _ .J... ——L z ,- .3 go AO/AA'FAW- 6w+3w 60 20?) (c) (15 points) The ionization energy of a certain element is 402 kJ moi-1. A photon strikes and ionizes an atom of this element and the escaping electron has kinetic energy 4.55 x 10-19 J. Find the wavelength of the photon. (Assume the atom was in its ground state.) Ewflkl’ FM 1473." : 4-02 K/ongMUL" : X/g 47 TA‘MM—l -Ié .. WW1 7mm:- & = 6.68x/o""T+Ll.Y3—2</u Ti/A22x/o ’70' W W I, C A mine—mm.) Id/IVEFC git/up; .. a . ~3'+ . L“ ~i __7 A ’ "35X’0‘I’7 ' aux/l; L512)? 0:0 M ) 1/177X/0 441 #774241 (d) (15 points) Identify the species with the electron configurations given below and show how the electrons are distributed over the valence atomic orbitals in the ground state of each. The first case is worked as an example. The +1 ion with configuration (He core)2s22p2 is fit Ground state: LL J. J. _ 28 2p 2p 2p The +1 ion with configuration (Ne core)3523 p6 is Ground state: ii 51 E g The -1 ion with configuration (He core)2$22p2 is Ground state: 3.5 gfgti g? The +2 ion with configuration (Ar core)3d9 is 53:. Ground state: a grail; 9 The +3 ion with configuration (Ne core)3323 p3 is Ground state: E: a 3: 3:3 The +7 ion with configuration 131 is _0: . Ground state: ’33 g??? 50? ? IS 14A2F01-F Name 2. (60 points) (a) (12 points) Consider the gas phase reaction in which chloroformyl fluoride reacts with elemental hydrogen to yield chlorofluoromethanol CIFC=O + H2 —> CIFHC-OH This can be considered a redox reaction. Assign oxidation numbers to each atom on each side of the reaction and state which atom is oxidized and which reduced: . a l . .. l Reagents. Cl__ _ Product. Cl_____ F 5" F -l O_"Z O — Z 0-13: C_iZ—_ H Q H it The atom oxidized is: H The atom reduced is: C (b) (16 points) Draw Lewis structures for the molecules of ClFC=O and for ClFHC-OH, using lines for bonding electron pairs and dots for nonbonding electrons. For each molecule, state what the skeletal geometry around the central C atom is and what the orbital hybridization of C is. W .Gcomejnt W .‘ o .‘ CIFC=O II n Tue/(war, FLA/VAR {F14 f7" ..‘rc IC ~4- fb’” H 'c/ 5 Pa ClFHC-OH .. / \ TemAHEVIZAu ‘F me ~. ( c) ( 10 points) The compound ClFHC-OH can occur in two different stereoisomeric forms. Draw these two forms below. Speciify whether these two isomers are geometrical isomers or optical isomers. or” 0“ l u: (1 M Trrase Am: M/lZeoa/MAPES') melt/CE, 0/9770}; 50 £25 14A2F01-F Name 2. (Can’t) (d) (22 points) The electronegativies of the second row elements are: Li: 1.0 Be: 1.6 B: 2.0 C: 2.6 N: 3.0 O: 3.4 F: 4.0 Consider the second row diatomic molecules LiB, BeN, CO and NF. Write out the ground state electron configurations for each (refer to the text, p. 148 for orbital energy sequence) and find the bond order for each: Also state whether the molecule is gammagnetig in its ground state. Molecule 92mm magma: Emmanstifl LIB 4e‘ : (2501 (2 so”? 0 N0 L 1 L ' BeN ge’: (LMDIEIS WW) (26F) ‘2 a co I be‘ .‘ (zarflzgfi‘fpmfiZWrJTzfoi' 3 M3 , L L x ' ' V55 NF / 26 -‘ 0502(25697" 1175397 W7") (2070’) W7)”; I”) 1 Which of the molecules is most stable? (0 5554.55 wanes—r EON P Oflpfifi Which of the molecules isjeaststable? L] 3 174/116 2 669 Ed”)? 0 a page). HTSNaT 97771315, With brief explanation, state which of the molecules has the largest electric dipole moment and which the smallest electric dipole moment. (Omit any molecule which is not stable.) 4 ’ / 5 /, “9967557” Dir—chE M/ fiascnuw/bm'n l//T'/.' 56W 50 F7 m; LHUF/f'r P/FOLE Manéo‘r D/WEWW swim/651’ ‘D/FFéefmg: a) /s 0,3,5MAwéxT , 14A2F01-F Name 3. (70 points) (a) (40 points total, 10 oints each part) A vessel with volume 50.0 liters contains only 1.00 mole of gaseous Hl (hydrogen iodide) at 300 K. Find the pressure in the vessel (bar or atm are acceptable units). ? 3 2'51”; [Loom 0,06% mm min-0mm) {0.0 L :: @4172 47m ($477830 ‘The volume of the vessel is reduced to 35.0 iters with the temperature kept at 300 K. The Hl remains entirely in the gas phase. Find the new pressure in the vessel (bar or atm are acceptable units). Tl; CMUJWHW’ §0 PIViz' ‘77va ?,_c ‘71— c_L/J..50.oo_ 'PT a-qewflm 14/ 53¢» A4; 731/: @703 km (0.7” 23616) The volume of the vessel is kept at 35.0 liters and it is cooled. When the temperature reaches 225 K, liquid Hl begins to condense out. Find the vapor pressure of liquid HI at 225 K (bar or atm are acceptable units). flaw V15 co/Vsflh/i/AMD T15 pacegA5£D~ KM) cw : 22,3”) 0,75, Paco) ' 0.703 ATM 2900 BM Paar) ; 0 my; 0.70; km soc/27 ALT/74 0.37! ) VIWOK maxing = 057/7 “7'4 Finally. the volume of the vessel is reduced to 17.5 liters with the temperature kept at 225 K. Find the pressure in the vessel (bar or atm are acceptable units) and the quantity of liquid Hl which has condensed out. You may assume that the volume occupied by the condensed liquid is negligible, that is, that all 17.5 liters of volume is available to the remaining gaseous HI. 77113 PKEfo/ZF/ fire/Mm; con/SWT M 777557 Mame mess was am) 3:47)", T prose amt/5W. 774E maul/3% A185 71+}: l/owme Mpmu Gee ca: mode/s /4/ fiffifié 7349155- #7/1/‘05 44,; 44b _—; V; } I'll." 2 0I>PJD #9,” A71 hwmob 7, 5m" 447/; 0.3‘00 MDL/ " M6(0MPEA£€D. 5lfllc£ 0.5“JDMDL Ré/Ir/HA/fi MI 77%. 6A5 Wfi’SE/OADOMOL -5- 14A2F01-F Name 3. (Con’t.) (b) (30 points) The equilibrium constant for the gas phase reaction 2H|(9) ¢> H2(9) + |2(9) is K = 1.26 x 10-3 at 298 K, 6.25 x 10-3 at 500 K and 1.85 x 10-2 at 700 K. (This is a slightly modified form of data given on p. 469, Table 9.2, of your textbook.) -At 500 K, a 100.0 liter vessel contains 1.00 mole of Hl(g), 0.080 mole of H2(g) and 0.040 mole of l2(g). Evaluate the reaction quotient Q and on this basis (give brief explanation) state in which direction the reaction will take place to reach equilibrium. (10 points) a = r». We . ; “w’EELM-WW’Z fry?“ ("Mm/v)" ms Mo EMILE Q=3.ano"3< K (Coo K)/mi€£flr€l‘lw mum—H) 7‘0 77W, RIG-PH” (Hurry) TB WH EQUIL/6KIUM -In a different experiment at 500 K, Hl(g), H2(g) and l2(g) are at equilibrium in a 500.0 liter vessel. The vessel contains 0.100 mole each of H2(g) and 12(9). Find the quantity of Hi in the vessel. (10 points) K: 6.2453(4)‘3 3 MHz/"1:, : (Oi/av)(o./w) t.— Z— 11”: 4,”; .l 1,1 «#375,, (0.1”)(0 03)... .2. fié 6 7"TX/0‘} -Suppose Hl(g), H2(g) and l2(g) are at equilibrium at 500 K in a vessel of volume V. Could you 1mm the amounts of H2(g) and of l2(g) present (that is, shift the equilibrium to the right) by changing the volume? Explain briefly. (5 points) gar/ca 77415 A/UM 66/6 oFMouas Dogs A/o'r C/Mfl/fé M/wemcmg CPfMé/W' WVoLu/ur; OHM/07* MaTWd/WUFmU/HKE/om, Could you jncLeasethe amounts of H2(g) and l2(g) by changing the W? Explain briefly and state whether you would need to raise orlower the temperature. (5 points) [ M/CEEMF/fi (W 73” mpmwm) So /A/o€545/A/é I. WfEMPEZA—‘Nw («i/ILL xA/cflééé E MIA/mum -7- 14AFO1-F Name 4. (70 points) Tartaric acid, C2H402(COOH)2 or H2Ta for short, is a diprotic acid with Ken = 6.0 x 10-4 and Kag = 1.5 x 10-5 ((pKa1 = 3.22, pKaz = 4.82). This data can be found on p. 560, Table 10.9, of your textbook. Each of the parts below involves aqueous solution of tartaric acid and/or the conjugate bases HTa- and Ta2- derived from it. The temperature is assumed to be 25°C throughout. 10 points each part. (a) 50.0 ml of tartaric acid solution of unknown concentration M is titrated with 0.150 M NaOH solution. Both hydrogens in tartaric acid are titrated. The stocihiometric point (equivalence point) is reached when 22.0 ml of the NaOH solution has been added. Find the concentration of the tartaric acid solution. 5001141. X lit/mm Y5 = 22.0 ML x0,/>’DMX l D/Pen-zc Ac: p MA'CI'D > %x% ons‘omz 0,0 330M (b) CalCUIate the pH of a 0.100 M tartaric acid solution. The relevant equilibrium is “27“ ¢> H“ + Hfa‘ Ka1=6.0x10'4 0.1w-"X 7C 76 _ + DFIZJ _. 1.5:)" [H12 =0./w-1)£ZJ,:, ,3;— -7! —— —5 M -5 _ X“ .1 . -‘/’ L:(.o)(/o 7K - 7.7Y/0 my mfiklwéo o M0 2 X 2 (#717 27.7X/o‘3 m) fer/7L: ~4060 [fl-1']: '7'//' 77¢ S'AECOA/P 015500477042 M41955 #1 Mm/a/Kze (Mm 3mm 72 [6”]. (c) Find the molar concentration [T a2-] of tartrate ion in a 0.100 M tartaric acid solution. The relevant equilibrium is HTa' © W + Ta2' K82 = 1.5 x 10'5 L57 [73".] : :3; £552 [H733 = 77>w'3 HEA/Cfi ,7X/(67 : £41,:/,rx/o'5_ (d) Calculate the pH of a 0.100 M solution of the salt sodium hydrogen tartrate, NaHTa (which dissociates in solution to give Na:(aq) and HTa-(aq). You may wish to refer to p. 563 and equations to be found there for assistance with this calculation. Fez Anna/mam 94073, Wfi‘z—f/(FKQ, +F542> '4: (522441485) > 51,0; -8- 14A2F01-F Name 4. Con’t. . Ee) 50.1)) ml of 0.200 M tartaric acid solution is titrated to the stioichiometric point (equivalence point).by the addition of 50.0 ml of 0.400 M NaOH solution. Caicuate the pH at the equrvalence pornt. Remember: the solution at the equivalence point may be considered as- 100.0 mi of 0.100 M Ta2- solution. The relevant equilibrium is Ta2- + H+ 4: HTa- + OH' Kb2=KWIKa2=10-14/1.5x10*5 :1 7X,0~/0 Z [ff—72‘] Eo/fj] (4T 2’; Cfifi‘J_ 52:22 )LRZ‘J=0:W‘7 3 0W” M " m A M ,_// _ _6 may 07; 2 (-67X/d’m/ 7L; gum / 74— 81/7200 Mali/cf, Cal-F3 = 81/7 X/o‘é/ ,vOA’: .507) 7,9 :/¢,ov 70/7“ 37/ What would be an appropriate indicator to use for this titration? (You may want to refer to p. 601 of your textbook.) awe/a THY/"0L 640E {P/v‘ RAM/(ff 3.0 7» 7,6) 06 FHEAML PhLWLl-LM/ (FHflAA/éfl K, 2 7a m .0) (f) it is desired to prepare a buffer solution with pH = 4.50 by dissolving sodium hydrogen tartrate, NaH'lUn 1.00 liter of 0.100 M sodium tartrate, Na2.Ta, solution. Calculate the number of moles of NaHTa. which needs to be added to obtain the desired pH. The relevant equilibrium is HTa' <=> H+ + Taz- Kaz=1.5x10_5 M55? [#27: /0"/.)'D._. X/O__5— M .- -* [722.] g 0- N? Kaz:/.rX/0 > 2 [9+]- [fizz] — 3M6 x/ E7211] / g‘ o [#723' 3.4,. 0 4 Amour- or: 7;“ I5 mom” 0. Mom : /ov.0MmL£/5 44510055 :_ M Lg: p \ 2’0 Armor; :0, mommy or— AA #7; (9) Suppose 100 ml of 1.00 M HCl solution were added to the buffer solution described in part (f) above. Calculate the pH of the resulting solution. you“ Mom #aé == wo “Mug 6P #CgM/V/Vé/Mmmws 47s 77% /5 54/0054! 7‘0 JusTCoA/umr 4M. 77+£72.’"7v H—RT 50 A? we” /5 kDPfl), 7771.5 1041/7714; (M QAW?Q M JosT‘ kféLUWoll/ OF #761“. WWW} 15 JoST‘rW Hz ,4 jaw-mu .oP/M/ Amp/#572017; {M7715 M/P'VZT (0’3 fifl/o hi, 4.4 L -9- 14A2FO1-F Name 5. (50 points) KSp for the salt lead chloride, PbClg, is 1.6 x 10-5 at 25°C. (a) (10 points) Calculate the maximum quantity (moles) of PbClg which will dissolve in 1.00 liter of distilled water at 25°C. X ; CszTj/fi‘ Z X; [adj/M 22/.6 A/O’S 413 :>/.6 X/O’i— Z X3 3‘ 4.0 X/D— 2+ :- ‘VC /0-M 7’ ; fir? )1/0‘2', [Pb J 7% Hgll/CE /, fl wo‘L mow/5. (51> Pb Cflle/ILL DISJoLVt /N No l 0 p IVA-715K. (b) (10 points) Calculate the maximum quantity (moles) of PbC12 which will dissolve in 1.00 liter of 0.100MHClsolutionat25°C. 4/0“; [(1174,] ; @,/00 f 2% \ L [pbz’j/M: X (Z)(O./oo+g() :M,X,o—s 7i ’ /-é X/0”-7> 3 ' [PH/fl: /, éx/o'gM/ 50 LOW MIL55 or waLthpwwfim[hula/town - / ( c) (15 points) 100.0 mi of 0.0400 M Pb(N03)2 solution is mixed with 100.0 ml of 0.0200 M NaCl solution at 25°C to give a final volume of 200.0 ml. Does a solid precipitate of PbClg form? Show WMMOLE/fi Vb“ : woo ML x o-ovoou : 4.00 MMOLES MMoLeS 00‘ ~_ “(0.0 m, xo.oLeo/vl : £02) MMdbfi/g. L ‘ S/mg “mg PZoDvc-r [175:1 (53%;? )LIS LL55 W 155?, £40 PKECIP/M’JEW/LL flvgm. (d) (15 points) 0.100 M NaCl solution is added at 25°C to 100.0 mi of a solution which contains Pb2+ ion. A precipitate of PbCl2(s) begins to form when 50.0 ml of the NaCl solution has been added. The volume of the mixed solution at this point is 150.0 ml. Calculate the molar concentration of Pb2+ which was present in the original solution before any NaCl solution was added. 5 ’ =— M @953)” mom)" 5'0”? M) W 9/ IVA/L 304077011/ [fig/+3} é ‘omLsLM' ISAOML IN’I’i‘i’E OEIFINALSOLUWA/ 5W1} = 0 “WM " M/ovo/M ~.—. 0 .0 2/6 14 ...
View Full Document

This note was uploaded on 02/08/2012 for the course CHEM 14A 142042200 taught by Professor Lavelle during the Spring '10 term at UCLA.

Page1 / 9

old_final - 14A2F01-F _ -1- Baur CHEMISTRY 14A LECTUH E...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online