Exam2_Spring06

Exam2_Spring06 - Chem 105b Exam 2 Print Name Last Thursday...

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Unformatted text preview: Chem 105b Exam 2 Print Name Last, Thursday February 20, 2007 Last 4 d1 gits SID # ‘u Professor Stephen Bradforth TA's Name m uestions Points Score Grader 1-5 20 6 15 7 20 8 20 Please Sign Below: I certify that I have observed all the rules of Academic Integrity while taking this examination. Si gnature: \ INSTRUCTIONS 1. You must show your work to receive credit. You must show your complete working on each of problems 6—9 to receive full credit. 2. If necessary, please continue your solutions on the back of the preceding page (facing you). 3. YOU MUST Use black ink. 4. There are 9 problems on 11 pages. Please count them before you begin. Chemlstry 105 B M1dterm 2 Professor Bradforth Spring 2007 0 You must show your complete working on each of problems 6-9 to receive full credit. 0 State and justify any approximations you make. Useful Physical Constants and Equations TC =TK — 273.15 R = 8.314 J K'1 morl and R = 0.08206 L atm mor1 K“ Arrhenius Equation: k=Aexp (-EA/RT) ln(k) = ln(A) — %[ H Gaseous Equilibria: Kp = K (RT) A“ where An is the sum of the coefficients of gaseous products minus the sum of the coefficients of gaseous reactants. Aqueous Equilibria: Dissociation of an acid in water: HA + H20 = A' + H30+ Reaction of a base in water: B + H20 = HB+ + OH— Autoionization of water: Kw = [H30+] [OH_] = 1.0 x 10'14 @25°C pH = ~10g10[H3O+l pOH = - logm[0H ] Solution of the quadratic equation a x2 + b x + c = 0 is given by: x: —bi\lb2 —4ac 2a Name Formula Kn Values of Kn for Some Common MonoBrotic Acids N Hydrogen sulfate lon HSO4— 12 X 10.2 Chlorous acid HClOZ 1 2 X 10-2 Monochloroacetic acid HCZHZCIOZ 1 35 X 10—3 Hydrofluoric acid HF 7 2 X 10_4 Nitrous acid HNOZ 4 O X 10—4 Formic 1 8 X Acetic acid HC2H302 1 8 X Hydrated aluminum (III) lon [A1(H20)6]3+ 1'4 X 10.5 Hypochlorous acid HCIO 3 5 X 10—8 Hydrocyanic acid HCN 6 2 x 10—10 Ammonium ion NH] 5 6 x 10—10 Phenol HOCgHs 1 6 x 1 2 H He 1 00797 4.00260 3 4 5 5 7 i; 9 10 Li Be B C N O F Ne 6.941 9.01218 10.811 12.0111 14.0067 15.9994 18 9984 20179 H 12 13 14 15 16 17 is Na M g A1 Si P S C1 Ar 22.9897 24.305 26.9815 28.086 30.9737 32.064 35 453 39 948 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 390983 40.08 44 9559 47.90 50.941 51.996 54.9380 55.847 58.9332 58.69 63.546 65,377 69.72 72.59 74.9216 78.96 79.904 83.80 37 38 39 T 40 41 42 43 44 1 45 46 47 48 y 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe ( 85 467 87 62 88.9059 9122 929064 95.94 98.9062 101.07 102.905 106.4 107.868 112.40 114.82 118 69 121.75 127 60 126 904 131 30 l 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 C 5 Ba La Hf Ta W Re Os Ir Pt Au Hg T1 Pb Bi Po At Rn 132 905 137.34 138.905 178.49 180.948 183.85 186.2 190.2 192.2 195.09 196.966 200.59 204.38 207 19 208.980 (209) (210 (222) 87 88 89 104 105 r 106 107 108 109 Fr Ra Ac (223) 226.025 227.027 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140 12 140.907 144.24 150.35 151.96 157.25 158.925 162.50 164.930 167.26 4 168,934 173.04 174.97 90 91 92 93 94 95 4 96 97 98 99 100 101 102 103 4 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md N0 Lw 232.038 231.035 238.029 237.048 (244) (243) (247) (247) (251) (252) (257) (258) (259) Multiple Choice Questions 14 points each) Question 1) An equilibrium mixture of DNA and an enzyme called helicase is disturbed by increasing the temperature. The equilibrium is as follows: DNA(aq) +helicase(aq) <:> DNA'helicase(aq) AH = ~ 15 kJ/mol Which of the following statements about raising the temperature is true: As the binding reaction is exothermic, the new equilibrium constant is smaller. The reaction moves to the left to achieve a new equilibrium (b) As the binding reaction is exothermic, the new equilibrium constant is smaller. The reaction moves to the right to achieve a new equilibrium (0) As the binding reaction is exothermic, the new equilibrium constant is larger. The reaction moves to the left to achieve a new equilibrium ((1) As the binding reaction is exothermic, the new equilibrium constant is larger. The reaction moves to the right to achieve a new equilibrium (e) The equilibrium constant does not change on” changing temperature. The reaction moves to the left to achieve a new equilibrium (1) The equilibrium constant does not change on changing temperature. The reaction moves to the right to achieve a new equilibrium Question 2) Gaseous hydrogen iodide is placed in a closed container at 425°C, where it partially decomposes to hydrogen and iodine: 2Hl(g) = H2(g) + 12(g) At equilibrium, it is found that [HI] = 3.53 x 10'3 M; [H2] = 4.79 x 10'4 M and [12] = 4.79 x 10' 4 M. What is the value of K at this temperature? (a) 6.50 x 10‘5 .84 x 10'2 (c) 2.71 ><10‘1 (d) 1.54 x 104 (e) 5.43 x 101 Short Answer Question gno working reguired) Question 3) What is the pH of a solution of 0.001 M HBr? H3, 4443} «'9 #30+ + 3V, (4 points) Question 4) (4 points) Using equilibria expressions, explain why the ammonium (NH4+) ion forms acidic solutions in water and the acetate (CH3COO') ion forms basic solutions in water. NH¢+ ’L #30 :1“ N”; 1" 7430+ am! a) but (1) [‘41, (I) acid/(3) CH 607) " k H, 0 : Ci/sdool/ +— 0H ’ 9 5w (I) any!» def/(l) ,6“; (a) Question 5) (4 points) Do you expect acidic, neutral or basic solutions from the following dissolved in water? (circle) Na20 acidic neutral KF acidic neutral F -,L/{z 0 fi ,L a” "‘ AlCl3 neutral basic CsBr acidic @ basic Long Answer Questions. Full credit will require providing all your working. Question 6) (15 points) Ammonium carbamate (NHZCOONH4) is found in the blood and urine of mammals. At 250°C, K = 1.58 x 10‘8 for the following equilibrium: NHZCOONI-L; (S) <=> 2NH3 (g) + C02 (g) (a) Using the Law of Mass Action, write the equilibrium constant for the reaction. H 2' Co z: t» ,7 i] (b) Convert K into Kp for the reaction at 250°C. KP: (QM/=33 3 K —. [flax/5‘9 (o.o€wé)[$‘2a,t)] P- K = 0. 00(25' (5 Pts) (0) If 7.80 g of solid NHzCOONH4 is put into an evacuated container at 25 0°C, what is the total pressure at equilibrium? KP = PIN/9 P002. Nigmnw ::‘ 21W, + 00,, 1: ~ 0 ° mm C —x +2.>< H‘ E _. 2 x K Question 7) (20 points) Consider the following reaction: 2N20(g) == 2N2(g) + 02(g) AH = ~164 kJ moi1 The reaction is catalyzed by a gold surface. (a) What type of catalyst is gold in this reaction? (2 ms) \AUWyeneoM (b) Sketch the energy diagram for the catalyzed and uncatalyzed reaction. Mark on the relevant energies to the forward and backward reaction rate as well as the overall thermochemistry. (5 pts) (c) Does addition of the catalyst increase the equilibrium yield of nitrogen and oxygen? Explain your answer. (2 pts) No dag/SQ” TR GLde mack («pie/1k ((1) Would doubling the overall pressure increase or decrease the equilibrium reaction yield? (Hint: imagine the same amounts of reactants placed in a container half the volume at the same temperature as earlier in the question). MW 63mm WW6 m RHS (3 pts) W V‘M) Saw. k) “a (e) What happens to the reaction yield and the rates of the forward and backward reactions if the reaction temperature is increased? may (3 pts) AT W “Mic, LNU‘CQAM Wfiwymxer pW’k/I ream“ R Haxd JmLSL. \/' dim,m-s ‘1‘de at (f) It is found that the forward reaction is accelerated by a factor of 101% adding the copper. Calculate by how much the activation energy is decreased for the forward reaction on addition of the gold catalyst. . ( i41st A \M\®l W224i 1*) (5 pts) \RW‘W -: a 0N” ("ng/ \ E cat— 5 {LA Cc} k, <_ CL + a C T «— “- (jL WA ‘ q RT R’T ‘QUflCD’.’ (Em: wot) : (E‘~%\H)C7;Cb8) (on to"): “74; arr/Wt N Question 8) (20 points) Acetylsalicylic acid (aspirin) is the most widely used pain reliever and fever reducer. It is a weak acid with molecular formula HC9H7O4 (Ka for HC9H7O4 at 37°C is 3.6 x 104). What is the pH of 0.018 M aqueous aspirin at body temperature (37°C). For full credit you must carefully show your method and working. Detail and check all approximations you make. HCSH?0+ a— Hmo :F’ (gnaoqr' + H30? , q." I 0.018 " O '0 'V c ‘7K ~+x 4- x I, t (o OLE'1\ 7‘ 9" \/ * 3. 6 ‘7L lo"H : [Cal—h9g1 (“30% 3 3(2' (omxzx) Ecg W} 0 H1 \.Yx\o“‘ 1, 7..<§ Xlo" OM" WNNK NW ems'mac x, 2.7;! x to”; , «3 \O N\ 4 Question 8) Additional space for your working. §uanvl€ W who (A'NLdfinf “ho .. ‘g A - .MEL .. 5.7m" A 239ij (0, 018- Z..‘-\ Xlo‘z)’ :5. pH-y—Lw(cum)=z.c7— h» 'zmwxplmfl: (.15: O‘DX‘ 15' <<. out” Mex CM wdk rt;qu mo,“ my (TLuS‘ Omqur WW xrzS'w ARV—3 M PH: Question 9) (25 points) A solution of the sodium hypochlorite salt (NaOCl) is sold as “chlorine bleach”. (a) Is the solution acidic or basic? Why? NA 0 0L“, ‘3 Naif (MO "‘K‘ UCLMLOq') 00‘- + Hue ..—_‘—__——‘» Hook 4» a CM- OOL— (1:11” av: a Lag/52., (5 pts) (b) Use the data provided at the front of the exam to come up with the K3 for NaOCl (if it is an acid) or the Kb for NaOCl (if it is a base). Ma (“0%) 3 3.3- X IO"x .' .J “5(0Q’): Kw w '.OOXIO (f KACMDOQ 33x [0“‘3’ . - 71- : 2 . S x \O (5 pts) 10 (15 pts) Question 9 continued) (c) Using your result to part (b), calculate the pH of a 0.01 M solution of “chlorine bleach”. Carefully show your working and detail and check any approximations you make. oer A H10 :4 hou + (DH‘ ,1 O .0 l O K342 O C 1 3L, 6 (o owe) x 7» % Goajfouj It :_ 2-3 'X‘Dhl b, I: 005‘] f (00‘ ‘DO x<<O on to ot~30£t 00‘ z a 9 1" o ,0»? O 01) 2-. . 'X ’0 x ( 2. 3K ) ( ‘- _. t - 5 A— ,; 5.9 xlo = ‘Q‘Ts 0'5“ . O” N 5;;7‘ x 100 -; mi) l: O I: q X ‘ O M O I / P O H : L+~ 7-?" ll ...
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Exam2_Spring06 - Chem 105b Exam 2 Print Name Last Thursday...

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