1aa3-2002-classtest-solutions

1aa3-2002-classtest-solutions - Page 3 of 16 ENTER YOUR...

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Unformatted text preview: Page 3 of 16 ENTER YOUR ANSWER TO EACH QUESTION ON THE ANSWER SHEET, IN PENCIL. THERE IS NO ADDITIONAL PENALTY FOR INCORRECT ANSWERS. 1. Your version of this test is VERSION A. You need to identify that you are answering this version of the test by filling in the bubble for answer (A) for question 1 on the optical scan sheet. Your answer for question 1 is: (A) Questions 2 through 16 are each worth 2 marks. 2. For the following reaction at 25°C CaC03(s) H. CaO(s) + 002(g) Find the FALSE statements: i. Kp = Kc for this reaction. ii. Increasing the total pressure on the system will not change the value of KP. iii. If the concentration of C02 is doubled, the equilibrium constant will be halved. iv. Equilibrium is reached when the rates of forward and reverse reactions are equal. v. KC= [C02][CaO] . i, iii,v __ ® a .. . K? ‘ P002— (B) 1,11,1v 1A9) i, iii, iv,v (i3) ii, iii ,iv Roz (E) ii, iii,v l 0.5— Continued on nextpage Page 4 of 16 3. At a certain temperature, the equilibrium constant, Kc, for the reaction 3030;) + N0(g) "—“" N02(g) + 802(8) is found to be 0.500. If0.300 moles each of $03 and NO are placed in 2.00 L flask and allowed to react at that temperature, the equilibrium concentration of so2 , in mol L", will be 5W “ “52:20: “OM—“£6910 (B) 00500 3333’” Q53:ng 1: K TAG) 0.124 ohmaa ’ X '- X_ x 25” (D) 0100 eqmiwwc. O‘BOO‘X 03:3: K - 7 2. X X 2_ 2 (E) 00310 0500: Z X (0‘ :“02 XzOILZLf mo( 4. Consider the equilibrium : CC14 (g) C(s) + 2c12 (g) At 800 K pure CCl4 was placed in a closed container with an initial pressure of CC14 equal to 2.00 atm. Afier equilibrium has been reached, the total pressure is 2.45 atm. What is KP, in atm, for the equilibrium at 800 K? (114%) (,9 CCS) + 2 U26?) (A) 0.367 {mm 2.00 am :2“; (B) 0.818 wamfie « X 2% (C) 7 0.0343 840026, 200 " >< @ 0.523 ZX+ 0 g 0.581 X Zip'qg (Ah/2?le ,. L>1 K? :fp‘é'f” :L” 1 2.300% 200”Y Continued on next page (A) (B) (D) (E) Page 5 of 16 For the following reaction: 2 s03(g) 2 802(g) + 02(3) the equilibrium constant, Kc = 1.6 x 10'10. Calculate the value of K0 for SOa(g) SOz(g) + 1/2 02(g) 2 , 10 f 50 ¢ 8 o x 104 F 2— 2 ‘ KC ' " 2. l 10 “MO (so; - .L r p 143021 1.3x10'5 7 KC 2‘4“ 2' [5031 1.6 x 10 *5 r 6.25 x 109 K C ' C L For the following reaction, Kp = 1.1 x 10 “3 at 1000 K. 3 H2 (g) + F6203(s) W 2 Fe(s) + 3 H20(g) Which of the following statements are FALSE? i. Adding H2(g) to the equilibrium, at constant temperature, causes the partial pressure of H20(g) to increase. ii. Adding Fe203(s) to the equilibrium mixture, at constant temperature, does not change the partial pressure of hydrogen gas. Kp = 19.1 x 102 is the equilibrium constant for the reverse reaction at the same temperature. Doubling the volume of an equilibrium mixture at constant temperature causes the partial pressure of H20(g) to increase. The only variable that can change the value of equilibrium constant is temperature, 3 H ii, iii K =fgzo'“ A iii, iv PH; g iv,V P {20V caveat/Se, mar/boa I ' 3 1,V K :‘Pgi—L‘ 1’2 P i, ii HLO Continued on next page 10 (A) (B) (C) (D) (E) [I Page 6 of 16 Consider the following system, which has already reached equilibrium: NH4HS(s) NHs(g) + H28(g) Which of the following changes will cause the equilibrium to shift to the RIGHT? F (i) Adding more NH3(g). F (ii) Adding NH4HS(s). T (iii) Increasing the volume of the container. (xiv) Adding a catalyst. T(v) Removal of NH3(g) from the system by reaction with HCl(aq). (ii) and (iii) "> eel” (“'me mm 3 fag (chat/id) (iii)and(v) Wm” WOJrCa‘a‘wie’ QQWMEWWGOJ (V) n n V P296 Heat O’t‘ 5mm (0 and (V) (ii) and (iv) K I: K {9 PM H3 ? H13 (i) and (iv) NOBr(g), initially at pressure Po, dissociates according to the equilibrium: 2 NOBr(g) 2 N0(g) + Br2(g) When equilibrium is established at 25°C, NOBr(g) is 34% dissociated and the equilibrium total pressure, P, is 0.25 atm. KP, in atm, for this equilibrium is 0.088 . ZNO/ir W) 2:2 QEOQ) + 0.21 3934 F0 0,34% 0. /7 P7 98x1“ Tom 7705590725 : 0.25am @-©,34/a}+034,6;0% 0019 012505544»: /§7fl9 0.0096 2%:0-2/3a25m- ,3 . :JGJELO—f— ’33:;— : €06 X/O KO (PA/0591 Continued on next page Page 7 of 16 9. In a certain experiment, 50.0 mL ofI-[Br react completely with 0.086 mol of NazO to produce sodium bromide and water. What was the original concentration of the HBr? balm/Lch Cq ucuH OM : (A) 3.40x10’aM 2 H quaq) + NCLZO(5)—72 Naggvf Hz/OCD (B) 4.30x10‘3M Opggmol N420 450.09: 447% O./72m0/eo H 81- 1 (C) 0.860M OR? Cl “ Aim) 1.72M MHszmyéf’gt @ 3.44 M 10. The order of decreasing aciditx of the oxo-acids of chlorine which decreases from left to right is: (strongest acid —> weakest acid). ~ Ma> HCIO4>HC103>HC102>H010 MA CHA (B) HClO > HClOz > ch3 > HClO4 1 (C) HClO > r10103 > Hc104 > HClOz (D) H0102 > HCIO > Hc103 > HClO4 (E) HCIO3 > HCIOZ > HClO > HClO4 Continued on next page Page 8 of 16 11. A 50.0 mL solution of 0.0500 M NaOH was treated with 700.0 mL of 0.00750 M HNO3 and diluted to 2.00 L with water. What is the pH of the 2.00 L solution? —3 ’5 (A) 2.46 #4140193 NQOH =2.5X IO A3 L03) 256 ,he woinNOB=525MO @ 2.86 gonna/(70m : NQOWMO 1, HNOBCM) = NqNO (D) 6'76 ( NLDDLQ.5X [0‘3 5 25"“? (E) “'1 2.?5-x lO—BMOl9) HMO3 m4 2757103 “400% L, / 382([0‘3 M [Wig]: 2L ' 12. Carbon dioxide dissolves in water and reacts to give HCO3— in the equilibrium shown below. Blood transpoxts C02 to the lungs and has a pH = 7.4. What is the ratio of [HCOg‘] to [C02] in blood at pH 7.4? co2 + 2H20 HCO3' + H3o+ K.=4.3 ><10_7 (0’89 : H 0* 55.98% 10.8 PH 1.4] écgggxg‘quo—Q (B) 4.3 LLB x (O : [C027 (C) 1.7 0.093 [HCOg] : g (S—(E) 5.8 x 10—8 E C027 ' Continued on next'page 13. 14. ;L (A) C (B) @ (D) (E) . Page 9 of16 A 0.300 M solution of hydrazine hydrochloride, NszJ’ 01', had a'pH of 4.50. Calculate the K, for hydrazine, N2H4. —4 + —— i: r. .4 1.05x10_5 N1H594QO‘) > NlHLtwg H (M:— + % 3.20x10 PH: tho... [Hljz'BJGX‘O‘ M 3.00><10'6 H 4 -9 + —--> N H «k ( x ‘ 3.33x10 NLH 6(m26— ‘ 2. wall 3672x105 3.13x10‘m £0.300M4J5A0) Blade 1 ~ 363 _ O S.) 3 Kai (0.300—3‘5“ ’ (0% :— 300x Ker,” (<3): MHZO Kl) A 50.00 mL sample of a weak acid HA was titrated to the equivalence point with 29.40 mL of 0.300 M NaOH. Ka for HA = 2.9 x 10—6. Calculate the pH at the equivalence point. HA + NQOH~> 'chJr/Pr H2059 4.63 (QC?) (960 @947 (“2 4.71 ft Arnolm NaOH = 8.82 >< (O 0 ~ m = . mL 9.29 Vplwmeg: 50.0304/14Li' 294-0 L Ezqzr , 8. 2x101ml fig 951 MG?) " M 10.8 "i’ H A 1— r” i “D X X aqua/t. 09Hle .. 3 ,233- z ’ O.lH—>< 34W“) xstH'] L MW 10'; POH¢t~¥i Page 10 of 16 15. “Concentrated” ammonia is a saturated solution of gaseous ammonia in water with a molarity of 17.0. What is the percentage ionization of ammonia in this i 5 solution? K, = 1.8 x 10‘5 for ammonia. /\ -gn CNHLCXCOW‘] (A) 0.0010 4.8 Y (O ‘ [NH31 (B) 0.0018 ' ‘ :_ NH+ + OH- (C) 10.018 NH‘ECa; H200) Lflqot) COL/“1) 010 max X K . 2 fig A (E) 0'18 mix ‘ {BRIO >5: 306% ‘0 X I: 000mm 4002: 00% 0000.16 (NmGL CONCENTE. l¥ 16. What is the pH of the solution at the half equivalence point of the titration of 50.0 mL of a 0.010 M solution of potassium propanoate with 0.010 M HBr (after V2 the required number of moles is added)? The Ka for propanoic acid (conjugate acid of potassium propanoate) is 1.3 x 10's. (A) 2.00 at haéfeqmmtmw/ [flow/i“ (B) 2.79 [HA] 2 LA} (C) 3.44 ( 0 00%61/ 50002000) /[A’l (E) _ 9.12 Continued on next page Questions 17 through 21 are worth 3 marks each. Page 11 of16 17. In a certain experiment, 0.750 atm H2 and 0.340 atm 12 are placed in a 5.00 L flask at 375 °C, and allowed to come to equilibrium according to: H2(g) + Mg) 2 PH(g) Kp for the reaction is 1.64 x 10-3. How many moles of H1 are present at 1 equilibrium? \ -00>< 'zmol A g? (A) 2 10 Help? Hf 1&2 W at) 1'88X10—3‘m’ mm ’1’ 075 on 0.3 0am + 2x (C) 2.63 x10-3mol WWW] — x _ x . v 2 X 03) 3.25 x 10-3 mol © FEOAX )4 (E) 3.76x10‘3mol N = Z ><=0.0/OO cdm ’ KP ‘(0.750-x)(o.31+0 ~><> 2 4x2 =-- [.64 X (0—3 (0255” (‘09 x + X > P H: (M eqmflbMW>v= @moom: #molen HI =55? = [88;«(0mo/ Continued on next page Page 12 of 16 18. This summer while at home you discover an old can in the basement that contains 50.0 g of lye (NaOH(s)). Rather than throw it in the garbage, to be “green”, you decide to first neutralize the lye and pour the resulting solution down the drain. How many litres of vinegar (5% CH3COOH by weight) would you need to treat the lye to reach the equivalence point? Assume no volume changes on addition of the solid lye, The Ka for acetic acid is 1.8 x 10'5. Assume that density of CHchOH is 1.00 g mL". (A) 0.75L 503r NQOH = [,25/molw (B) 1.0L N a OH + CH3COOH~7 NCLCH3COO * HAG 1K(C) 13L l-ngmolao CH3WOH “466099—09 1.5L 5% augment = Harcuaonm IOO% ofsolcdv‘ow (E) 3-” 0669mm cagin = 4 act w L ’ 5% wgwoaa Sarcwgom 4m IOOmL6ofafiOt/x ~/ MmngCOOM = 60am at 0.833 moi/L. 345 Wrote/3 : VUW 0333M = \/= 4.6L Continued on next page Page 13 of16 19. A solution of 0.01 M NH; is titrated with 0.01 M HCl. Which of the indicators below would you use? The K, for NH3 is 1.8 x 10"“. Answer Indicator pH range Colour Change BK. i Methyl orange 3.3 - 4.6 red - yellow 4.2 ii Methyl red 4.2 - 6.2 red - yellow 5.2 iii Bromthymol blue 6.0-7.8 Yellow - blue 7.2 iv Thymol blue 7.9 - 9.4 Yellow - blue 8.2 v Phenolphthalein 8.3 - 10.0 Colourless - red 9.5 (A) my Ollie/Claim 430ml: (QOOSM NHLTW?) iionly NHQOL‘Q ’“ H200)": NHS (00?) + EEO?L (C) iii only 0009* X 49‘ >< >< [IS—YD) ii or iv X2 .2 Ilyglelo _ S— gfi . (B) vonly 000 X ~£ X;[H30+] = [66 x (O Bibi-.31 ow MW (“4 aha mg a [9 H 20. What is the approximate pH of the endpoint of the titration of 100.0 mL of 0.050M HNO3 by 0.050 M KOH if phenolphthalein (see previous question) is used as an indicator. The pK;il for phenolphthalein is given in the table in Question 19, above. +n‘tmh‘om.' /D7L/l/O’Vtfi @0101, vs SW3 basal (A) 13 {3H 1?— ‘ ‘q 10 find“ lobeth’LPl/l/l‘aled/m Lg Sb WWW, 1+ 4‘s mace od'PH :40 /(C) 8.3 _ 25 (D) 7.0 (E) There is insufficient information provided to answer this question. Continued on next page Page 14 ofl6 21. What is the pH of the solution resulting from the addition of 7.80 grams of solid KOH to 750 mL of 0.350 M NH4Br? Assume no volume change occurs on addition of KOH solid. The K, for NH3 is 1.8 x 10'5. €45 1.8O¢KOH » 0. meme» KOH (A) 4.79 + _ (B) 4.85 NH 4% (cog #— @HCOLCD—W9 N Him) + H1003 ‘ . 0150an r1 @ 9.30 “WWW \I/ OJGBMMV) alagmolf/J (D) 11.5 t 03) 13.3 FWM 0.2650136) —— Ols-q mo 6/; , [9/3 a" W) 0W6 M m3 , + hum Comm—EV O.le M Nch I" ‘ NH3 (m) +H1©Q)<:Z NHWWL (0%) 0~\8§~>< 0.69, x _ (o. 169x) x 1.8% {O 5; Olga—x [Cl—r} +65% [05 1.02 FOH = #56? 13H: @305 Continued on next page ...
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This note was uploaded on 04/17/2008 for the course CHEMISTRY Chem 1AA3 taught by Professor Lock during the Fall '02 term at McMaster University.

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1aa3-2002-classtest-solutions - Page 3 of 16 ENTER YOUR...

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