W01 MT2 - CHEMISTRY 110A INSTR: R.N. SCHWARTZ MIDTERM EXAM...

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Unformatted text preview: CHEMISTRY 110A INSTR: R.N. SCHWARTZ MIDTERM EXAM #2 Wednesday, February 28, 2001 / NAME: fie SID #__i(€~_,{_.____ (1850 (first) INSTRUCTIONS QUESTION SCORE 1. This exam consists of 4 questions. Before starting, write your name on every page. 1 2. Indicate answers, including units, and show - your method of calculation and reasoning. - ‘ 2 3. No credit will be given for an answer alone or for an illegible answer. - 3 4. Use the conventions for significant figures and units. 4 5. If you run out of space working a problem, use the back of that page and indicate on the front that you have done so. - - - - TOTALS Some usefitl equations and values of fimdamental constants are given below: 6. You may use a calculator. No books or notes N A = 6.022 x 1023 particles-mol", R = 8.31451 JK'lmol" = 0.08206 L atm K'l mol'1 = 82.06 cm3 atm K"mol“, For f(x) = xa: (1% = coca—1, Ixndx = xn+1 /n+ 1, Ixnldx = lnx. ' dU=dq+dw, dwm =—PdV, as=dqrev T, HE U+PV, GE H—TS, AE U—TS, Cv = (alyarlv’ Cr = (afyarlp’ (alde)T = 7(3175'rlv " P’ (3%”)1 = V ' T(aV/3F)P’ ‘11an _ AH" dT er’ Partial molar Quantity: = (aY ani)T P , Y = 211,17,- . vnm i Vmean = Xi(deean dXi) + ‘7}; Lj = AB, AGO -RTln Kg, lnx = 2.303log10 x, For an ideal monatomic gas: CVJn = gR, Cp,m = 3-K; For an ideal diatomic gas: CW" = g-R, Cp,m = %R MAKE SURE YOUR EXAM A S l NUMBERED PAGES! CHEM 110A MIDTERM #2 28 February 2001 NAME: «'8: l . (25 points) a. Using the information give below for a single-component system draw and label as accurately as possible its phase diagram. Useful information: V515) = 17.60 cm3mol—1 , v5,” = 18.72 cm3mol‘1 at 220 K. i) P = Iatm, T = 195 K, [1(3) = Mg). ii) P = 5.1 atm, T = 217K; p“) = a“) = mg). iii) P = 67 atm, T=298 K; it“) = “(9. iv) PC = 73 atm, TC =304 K. b. Sketch and label the temperature dependence of the chemical potential ( u - Tisobar) for P = 1 arm. c. Calculate the difference in slope of the chemical potential against pressure ( p - P isotherm) on either side of the freezing point. CHEM 110A MIDTERM #2 28 February 2001 NAME:_V_&8/___ CHEM 110A MIDTERM #2 28 February 2001 NAME: 2. (25 points ) Consider the following data: B2H6(g) + 302(g) —> 3203(3) + 3H20(g) AH:m = —1941 k] C2,,n/J K‘Imol‘l 23(3) + £02 (g) —> 3203(5) Afon = —2368 k] B(s) 12.32 1 H2(g) + 502(g) —> H20(g) AH?” = —241.8 k] BZH6(g) 47.58 H2(g) 28.82 a) Determine AH} for B_,H6(g) at 298 K. b) Determine AU; for BzH6(g) at 298 K. c) Determine AH; for B_,H6(g) at 598 K. 7 A) JB@)+ «'5sz) "; BszQJ AH-BW: Min 5) “82036ng s—HzoCa) ——> Blimp + 301%) .Hq.“ k3 __ I93 )5: a ) 9 8C5) + 30163) -> 8930‘) 23 3x {—2413} ‘25 03;) f V/// a ._ 125 = W 9% + BHKQ) —> 8350(3) A35: Nina, b) H:u+P\/5 AH=AM+A(PV)3+M *0 81“} Afizbw'} Ame“ Mtge/J n(<‘r . M M 7-66) +3H2bj'962-H5992dl‘w *4 .4: , " 29P:—1/¢ ,éu Ad‘EM’ + 2.67: ~3sz , 2xfizlv’me-(X #z,‘ ’é i” ‘4’?“2‘ 14”» f“ 5 [do - o . O = ‘ A = . t n 0M - am): Ffifira 46,47 .. Mfr») 571:)? a: inf/“111w”; 2%” ,l Blackoe 62mg) J MM Afp- '53 25 J .. A; m o o 5’” » - “15255—433745” ’79 7.“ , 59,: : 29mm -/-fl—- 63.6 24d?" - p / 0 Aggy. 2,; =3 ’é‘J d "‘ o = O '- A CHEM 110A MIDTERM #2 28 February 2001 NAME: 3. (25 points) The volume of two completely miscible liquids A and B is represented by : V 2 . Vmean = = a + [3X3 — yXB; X3 = molfractzon ofB. nA + 713 a) Calculate the molar volume of pure A and that of pure B. b) Calculate the partial molar volumes of A and B at X B = 0.4. W Vfis 0L / ,— _../— \/ gveflif Mm Xa=> 0) X5594 ’. W ' WVO V6: 03-!- %" 2% //..//':/-_‘ 01 max—.- "‘ lo; :43 b) M VW: X,L '32 +9 J .4 dV‘ :4“ VA 3 V‘mM—r— X6(’Z"Tn<"& waneh (3+ >< ~6‘x7‘ .—. -mn/R (CW5, — d/X5( E 5 § 5 AVW - 7’ Keg We) _ W?“ 2W‘5 / / L T! v-Vm‘bM ” 043— “’95): d*9*fi‘bflxe"%5+1%yl ‘ A” 2— X :04 /: 0k +®"‘°> :1 5 B VA/ifl CHEM 110A MIDTERM #2 28 February 2001 NAME—M.— .— vf): \fm" dxfl w VrYw—aw = d + @CA‘YA)” X‘M-KASL K é. 0} (4,1 vg‘(4,X‘D"> K3714 )” CW» < + g “3 3 ~? + zx‘u—xfi) Xfi (dig—2M = -§XA -\-2A/(A"x4)’<4 W B 1 A ._ —- .. xA 1—1254U'K4 “a VE" C (ii—“)1. (Pay W422 L X .: 0k+$xfn£><5 +§>><A~m§xe A .= 0k + g— yflu—xfir—J—gz— 2mm =2 0M (-13 —— 29 (A—gxn+KAz)—- 9‘6"XA+ aw.“ / a. q, 2 AA- @‘Y‘ +‘Q/X‘XA‘DAXfi " 2§¥g+ 1‘6/‘XA CHEM 110A MIDTERM #2 28 February 2001 NAME: %: 4. (25 points ) For the reaction A2(g) + 32%): 2AB(g), the measured Kg values in the range 450 to 700 K follows the equation: 4.83x103K 10 K0 = 7.55- glO P T Find AGO, AH”, ASO, and AC?) for this reaction at 500 K. Assume that the gases are ideal. 0 .3 .. I , 5x0 ( = "L 40.2- M // ° ° ° A//° '- 9 d/rWV A” - 0’ [9 / W 0 4m x1036) a AL " ( a7; 755" 2503/37" 3,2: J _ «fax/03K 4H0 "4/3x/9dr/7' " -———;,—:,:" 3 AH; fi.%bx1€x fiérsx/o g 3 = 2303;459:641“ X94731” /< KM M” = +425 HM fl 7 CHEM 110A MIDTERM #1 31 January 2001 NAME: :5 a END OF THE EXAM ...
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This note was uploaded on 04/03/2008 for the course CHEM 110A taught by Professor Schwartz during the Winter '06 term at UCLA.

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W01 MT2 - CHEMISTRY 110A INSTR: R.N. SCHWARTZ MIDTERM EXAM...

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