Answer Key. Chapter 1

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1 Aswer Key Chapter 1 Page 3, Quick Check a) Rate of decompositio of SO 2Cl 2. b) The slope becomes less ad less steep. c) The rate decreases as the reactio proceeds. d) mol/l/s (or +) e) Rate is greater at 300 s tha at 500 s. f) Fid the slope of a taget draw to the curve at that time. g) 500 s: 09.2 x 10-5 M/s < 300 s: = 2.1 x 10-4 M/s Page 6, Practice Problems a) mass of Cu or mass of Ag, [Cu(NO 3) 2] or [AgNO 3], blue colour of Cu(NO 3) 2 b) pressure of etire closed system will icrease or gree colour of chlorie gas c) volume or mass of carbo dioxide gas or icreased partial pressure of CO 2 (i a closed system). d) mass of BaSO 4 ppt formed, [either reactat], coductivity as ios are cosumed 2. Volume chage is geerally associated with the formatio of a gas. 3. Cocetratio chage is associated with the formatio or use of a aqueous species or a gas, ot a solid or a liquid (except i the rare case of solid or liquid mixtures). Page 8, Practice Problems a) g/mi b) Rate is greatest at the begiig c) Rate = 0 g/mi i the last icremet d) This is because the HCl is all goe (limitig). Excess Z left x 10-4 mol/l/s Page 9, Practice Problems x 1025 molecules/l mi s Page 11, 1.1 Review Questios 1. - The distace-time graph ivolves a vehicle chagig positio, while the rest of the data i this sectio ivolves a property of a chemical chagig durig a chemical reactio. - The vehicle accelerates ad the decelerates, while most reactios start at a maximum rate ad decrease rate the etire time. - The vehicle ca stop movig etirely ad the begi agai. Oce a reactio begis, it cotiues util it is fiished. 2. a) i) g/mi ii) g NO2/mi iii) mol HNO3/mi b) i) 840. ml ii) mol/l c) Δ Coc HNO 3 by Titratio usig Buret, Flasks, Magetic Mixer i M HNO 3/s to measure V f ad V i Δ Mass Cu by Balace to measure m f ad m i i uits of g Cu/s Δ Pressure NO 2 by Maometer i closed system to measure P f ad P i i uits of kpa/s NO 2 Δ Volume NO 2 by Eudiometer to measure V f ad V i i uits of ml/sno 2 Δ ph by ph meter to measure chage i ph of HNO 3 per secod Δ Mass of NO 2 gas by Balace to measure m i ad m f i uits of g/s Δ Coc CuNO 3 by Titratio usig buret, etc to fid V f ad V i i M/s 3.a)i) approx 32 ml/mi ii) approx 16 ml/mi iii) approx 3.3 ml/mi b) (a) > (b) > (c) As the reactio proceeds there are less reactat particles available to collide ad react. Cosequetly, the rate decreases with time. 4. a)1.29 x 10-4 mol/s b) The [HCl] was decreasig. This led to less successful collisios. c) Oe of the reactats (either the Sr disk or the HCl i solutio) was completely cosumed. Based o the reactio rate, the Sr disk was most likely the limitig reaget. The rate implies a reasoably large molarity of HCl. 5. Time (secods) Absorbaces (o uit) Cocetratio of Copper(II) Io (mol/l) 0 s 0 0 mol/l a) M HNO 3/s b) 0.67 g c) Cu(s) will decrease i size Blue Cu 2+ (aq) will appear Bubbles will be see as gas forms This is a very exothermic reactio so heat (ad potetially some steam) will be give off (Colourless NO(g) reacts immediately with oxyge i the air to form browish orage NO 2(g) or its dimer, N 2O 4(g) which is also browish orage.)

2 Page 22, 1.2 Review Questios 1. Factors affectig heterogeeous ad homogeeous reactios: Nature of reactats Temperature Cocetratio Presece of a catalyst 2. Surface Area. You ca oly icrease this area of cotact for solids ad rarely, liquids. 3. a) Platium i the catalytic coverter of a car. b) Iro is used i the Haber Process to make ammoia. c) Platium is used i cotact les disifectat disks to decompose hydroge peroxide producig oxyge gas. The bubblig process helps clea the surface of the leses. d) A Magesium Complex called Chlorophyll assists i photosythesis. 4. CaCO 3(s) + 2 HNO 3(aq) à Ca(NO 3) 2(aq) + H 2O(l) + CO 2(g) a) Icrease more frequet cotact betwee reactats b) Icrease more surface area leads to more cotact betwee reactats c) Decrease less eergetic collisios due to lower temperature d) o effect CO 2 is a product so icreased cotact has o effect o forward rate x 10-3 mol HCOOH/mi a) Less tha half as much time (rx most rapid a start) uder 30 secods. b) Cosiderably more time much > 1 miute. c) Approximately half as much time for each 10 o C icrease about 30 secods. 6. a) i) slow (due to solid C) ii) fast (ioic species) iii) slow (due to solid Mg) Fastest: (ii) > (iii) > (i) Slowest b) i) surface area of C(s), temperature of system, [O 2] (volume of cotaier) (partial pressure of O 2), a appropriate catalyst ii) [Pb 3+ ] ad [I - ], temperature of system, a appropriate catalyst iii) [CuCl 2], surface area of Mg(s), temperature of system, a appropriate catalyst 7. c > b > a The greatest opportuity for cotact betwee reactat particles is i system c. It appears to have the highest cocetratio/greatest partial pressure of each reactat gas. The ext would be system b ad the least opportuity would be system a. Page 25, Quick Check 1. Rate = k[c][d]. This meas that the rate of the reactio is directly related to the product of the cocetratios of reactats C ad D. Furthermore, it meas the reactio is 1 st order with respect to both reactats C ad D. (The overall reactio order = = 2.) 2. a) The rate would double. b) The rate would half. 3. If the volume of the cotaier were halved, the cocetratios of both A ad B would be doubled. Hece, the rate would be icreased by a factor of 2 x 2 = 4 times. 4. Icreased T à Icreased Rate. Icreased T must cause icreased rate costat, k. Page 27, Quick Check 1. Cocetratio (Volume of a Gas Sample), Surface Area, Temperature 2. Temperature, Presece of a Catalyst 3. Temperature Page 29, Practice Problems a) The added curve should have the same total area below it. The area below the curve ad to the right of the vertical lie represetig E a should be DOUBLED. b) The added curve should have the same total area below it. The area below the curve ad to the right of the vertical lie represetig E a should be SIGNIFICANTLY INCREASED. c) The added curve should have the same total area below it. The area below the curve ad to the right of the vertical lie represetig E a should be INCREASED BUT CLEARLY LESS THAN DOUBLE THAT IN CURVE (b).

3 2. Temperature 100 o C 300 o C Frequecy of 2.00 x s x s -1 Collisios Force of Collisios (% Possessig E a) 95.0% 97.0% Frequecy of Collisios Possessig Activatio Eergy 1.90 x s x s -1 a) 97.0% = 1.02X greater 95.0% b) 3.00 x s -1 = 1.50X greater 2.00 x s -1 c) 2.91 x s -1 = 1.53X greater 1.90 x s -1 d) Icreased frequecy of collisios. e) Extremely forceful collisios may be too forceful to trasfer KE of collisios ito PE of ew bods. These forceful collisios may be usuccessful resultig i reactats that simply bouce off oe aother. Page 33, 1.3 Review Questios 1. The frequecy of collisios ad the fractio of these collisios that succeed. 2. Sufficiet eergy (E a) ad proper collisio geometry. 3. a) Origial surface area + 6/6(origial surface area) sice six ew faces have bee exposed. Thus 6.00 cm 2 + 6/6(6.00 cm 2 ) = 12.0 cm 2. b) 6/6 + 6/6 = 12/6 or 2X as may collisios as before. c) 2 ( mol Z/mi) = mol Z/mi d) 242 ml H 2 4. a) 3 x 0.5 x (3.10 x 10-3 mol/l s) = 4.65 x 10-3 mol/l s b) 1 x 4 x (3.10 x 10-3 mol/l s) = 1.24 x 10-2 mol/l s c) The temperature was icreased, perhaps by approximately 10 o C. Or a catalyst was added. d) 0.5 x 0.5 (3.10 x 10-3 mol/l s) = 7.75 x 10-4 mol/l s (ote that both cocetratios were halved) 5. a) The greatest cocetratio of HCl ad the greatest surface area of CaCO 3 exists at the begiig of the reactio. Cosequetly, the reactio rate is greatest at the begiig as these circumstaces allow for more successful collisios. As the reactio proceeds, there are less particles ad surface available to collide ad so the reactio rate decreases. b) 6. III > I > II The powdered zic allows for the greatest cotact at the surface ad the 1.0M acid has the greatest umber of particles available for collisios. 7. The Cl atoms must be able to cotact oe aother to produce Cl 2 ad NO 2. The particles must be draw to allow this cotact to directly occur. 8. a) Low temperatures decrease the frequecy of collisios ad decrease the fractio of collisios reachig Ea. This keeps the food from spoilig. b) Fie bits of magesium i the shavigs (more surface area) allow more frequet collisios with the oxyge molecules. This allows a fire to start more readily. Its ature allows magesium to react readily with oxyge with a low Ea. c) Pt acts as a catalyst i the coversio of NO x s ito N 2. This meas the collisios required for a successful reactio ca occur with less eergy or at lower temperatures. Page 40, Practice Problems Rate = k[no][o 2] Trial 1 = X x 10 6 l/mol. sec 2. Rate = k[y] k = s!! 3. System 1: System 2: System 3: Rate = mol/l/s mol/l/s mol/l/s Page 42, 1.4 Review Questios 1. a) Rate = k[x] 2 [Y] b) k = 80 L! /mol! /s c) Rate = 0.4 mol/l/s 2. a) Rate = k[no 2][F 2] b) k = 40.0 L/mol/s 3. a) Rate = k[mo 4- ] 2 [H 2C 2O 4] k = ! L! /mol! / b) If all the cocetratios of all three reactats were doubled, the iitial reactio rate would be: (2) 2 (2) 1 (2) 0 = 8x. c) If the volume of the reactio system were doubled by addig distilled water, the reactio rate would be: (½) 2 (½) 1 (½) 0 = 1/8x. 4. a) Rate = k[a] 2 b) k = !! L/mol/s c) Trial 4: Rate = mol/l/s

4 5. a) Rate = k[mo 4- ] 2 b) k = L/mol/mi c) Rate = mol/l/mi d) MO!! = mol/l e) !!! mol/l/mi MO! 6. Rate = k[ch 3COCH 3] [H + ] k = L/mol/s Page 48, Practice Problems k = !! s!! 2. C = mol/l (0.98 mol/l) 3. C = !! mol/l (2.2 10!! mol/l) Page 49, Practice Problems t = 2900 s (48 mi) 2. Mass = 140 mg 3. t = 24 h Page 53, Practice Problems Pseudo itegrated rate aalysis Tim e (s) Cocetratio (mol/l) l Cocetratio ( ) (- 2.60) (- 2.90) (- 3.19) 1/Cocetrati o (L/mol) 10.0 The substitutio reactio is 1st order with respect to [(CH 3) 3CBr]. k = s!! 2. 2 NO 2 (g) à 2 NO (g) + O 2 (g) Tim e (s) Cocetratio (mol/l) l Cocetratio ( ) (- 4.84) (- 5.04) (- 5.34) (- 5.57) (- 5.75) a) Rate = k[no 2] 2 k = 0.54 L/mol/s /Cocetrati o (L/mol) b) i) If the iitial cocetratio of NO2 icreased, the reciprocal cocetratio would decrease, shiftig the straight lie graph dow. ii) If the temperature of the decompositio were elevated further, the slope (k) of the straight lie graph would become steeper (k icreases as temperature raises). Page 56, Practice Problems a) t!/! = ! s 770 mi (13 h) b) C = mol/l (0.09 mol/l) 2. a) k = !! s!! b) C = mol/l (0.04 mol/l) c) t = ! s Page 58, Practice Problems a) t!/! = 126 s b)! = 39.9 L/mol! C = !! mol/l 2. If the half-life of a reactio is idepedet of the iitial cocetratio of the reactats, the the reactio is first order. Page 60, 1.5 Review Questios 1. a) first order b) zero order c) secod order i) rate of Decreases Stays Decreases reactio costat ii) rate costat Stays costat Stays costat Stays costat iii) half life Stays costat Decreases (half the previous) Icreases (double the previous) 2. C = mol/l (0.22 mol/l) 3. C = 0.14 mol/l 4. C = 0.41 mol/l 5. a) k = h!! b) t = 56 h 6. k = s!! 7. t!/! = 7.30 s 8. Graphs for a 1st order equatio are show. a) decrease i the iitial cocetratio by a factor of ½ as evidet by examiatio of the y-axis b) decrease i the iitial rate: a taget draw to the curves i the first graph at time zero ow has about HALF the SLOPE c) rate costat, k (determied by fidig the slope of the lc vs t graph) remais the SAME d) temperature has icreased. Oly a chage i temperature ca chage k. e) the slope of the iitial taget lie (= the rate) has icreased ad the temperature could icrease the rate costat

5 9. Graphs for a 2d order equatio are show. a) decrease i the iitial cocetratio by a factor of ½ as evidet by examiatio of the y-axis b) decrease i the iitial rate: a taget draw to the curves i the first graph at time zero ow has about HALF the SLOPE c) rate costat, k (determied by fidig the slope of the 1/C vs t graph) remais the SAME d) temperature has decreased. Oly a chage i temperature ca chage k. e) the slope of the iitial taget lie (= the rate) has decreased ad the temperature could decrease the rate costat 10. Decompositio of aqueous sucrose to form the isomers glucose ad fructose: Time [C 12H 22O 11] l[c 12H 22O 11] 1/[C 12H 22O 11] (mi) (mol/l) (L/mol) ( ) ( ) ( ) ( ) ( ) 6.85 a) Rate = k[c 12H 22O 11] b) k = s!!. Approximate value from graph slope. c) t!/! ! s 189 mi (3.14 h). Approximate value from graph slope d) C = mol/l (0.17 mol/l). Approximate value from graph slope. 11. Time (s) [CH 3NC] (mol/l) l[ch 3NC] 1/[CH 3NC] (L/mol) ( ) (- 5.13) (- 5.78) (- 6.57) (- 7.3) 1000 a) Rate = k[ch 3NC] b) k == s!!. Approximate value from graph slope. c) t!/! = 3000 s 50 mi (0.9 h). Approximate value from graph slope. d) C = 2 10!! mol/l (10!! mol/l). Approximate value from graph slope. 12. a) k = L/mol/s b) t!/! = 316 s this is the 1 st half life 13. Assume 1.00 mol/l iitial sample: t = 34.2 years 14. a) t!/! = 20 C! s b) C = 0.13 mol/l 15. t = 75.4 s Page 66, Quick Check 2. 2 C 2H 6(g) + 7 O 2(g) à 4 CO 2(g) + 6 H 2O(l) kj/mol rx 2 C 2H 6(g) + 7 O 2(g) à 4 CO 2(g) + 6 H 2O(l) ΔH = kj/mol rx kj/mol rx + 2 HBr(g) à H 2(g) + Br 2(l) 2 HBr(g) à H 2(g) + Br 2(l) ΔH = kj/mol rx Page 70, Practice Problems a) edothermic b) i) 30 kj/mol ii) +20 kj/mol iii) 130 kj/mol iv) 140 kj/mol 2. I) Activatio eergy for the forward reactio. II) Potetial eergy for the reactats. III) Activatio eergy for the reverse reactio. IV) Ethalpy chage. Page 72, Quick Check A) Reverse ucatalyzed activatio eergy. E a(reverse Ucatalyzed) is #2 i Diagram 2. B) Forward catalyzed activatio eergy. #3 C) Decrease i E a due to catalysis. #4 D) ΔH. #7 E) Forward ucatalyzed activatio eergy. #6 F) Reverse catalyzed activatio eergy. #1 (Note that #5 is the PE of the activated complex forward ad reverse - ucatalyzed.) Page 74, 1.6 Review Questios 1. a) Fe 2O 3(s) + 2 Al(s) à Al 2O 3(s) + 2 Fe(s) ΔH = -852 kj/mol rx Fe 2O 3(s) + 2 Al(s) à Al 2O 3(s) + 2 Fe(s) kj/mol rx b) Ca(s) + O 2(g) + H 2(g) à Ca(OH) 2(s) ΔH = kj/mol rx Ca(s) + O 2(g) + H 2(g) à Ca(OH) 2(s) kj/mol rx c) 2 H 2O(l) à 2 H 2(g) + O 2(g) ΔH = +572 kj/mol rx 572 kj/mol rx + 2 H 2O(l) à 2 H 2(g) + O 2(g) 2. a&b)

6 c) The particles above the E a lie i the first graph represet the particles whose collisios are successful (have eergy > E a ad have good geometry). e) A edothermic reactio has a greater umber of successful collisios tha the correspodig exothermic reactio. This idicates that the edothermic reactio is much more sesitive to a icrease OR decrease i temperature s effect o its reactio rate. (Will fid i Chapter 2 that a icrease i T will always cause a shift i the ENDO directio.) 3. a) Edothermic sketch. b) 60 kj/mol c) -35 kj/mol 4. a) A = Activatio eergy for forward catalyzed reactio. B = ΔH C = Reverse ucatalyzed activatio eergy. D = Forward ucatalyzed activatio eergy. E = Reverse catalyzed activatio eergy. b) Exothermic c) No effect. 5. a) -50 kj/mol b) Q + S à T + 50 kj/mol Q + S à T ΔH = -50 kj/mol c) exothermic d) 150 kj/mol 6. The eergy released as more CO 2(g) ad H 2O(g) is formed provides the activatio eergy eeded to cotiue decomposig the hydrocarbo fuel. 7. Fill i the blaks for a EXOthermic reactio: The eergy released durig bod formig is greater tha the eergy required for bod breakig. As a result, there is a et release of eergy. The bods formed have _less_ potetial eergy tha the reactat bods did. 8. The diagrams would be the same. The potetial bod eergy would be the same. The reactio coordiate is NOT a time axis!! 9. As particles approach oe aother their KE DECREASES (bod breakig requires eergy KE à PE). Meawhile, their PE INCREASES (see above). Oce a activated complex has formed, the KE INCREASES (bod formig releases eergy) ad their PE decreases). The sum of KE ad PE remais costat the etire time. blue is PE, red (u-shape) is KE 10. a) exothermic b) 10 kj/mol c) 25 kj/mol d) -15 kj/mol - o chage at all whe catalyzed. e) 15 kj/mol lower with a catalyst. f) icreases the reactio rate. Page 80, Quick Check 2. a)bimolecular Elemetary Process b) Uimolecular Elemetary Process c) Termolecular Overall Reactio Page 83, Practice Problems O 3(g) à O 2(g) + O(g) (slow) O(g) + O 3(g) à 2 O 2(g) (ΔH value for etire reactio = kj/mol) O 3(g) à 3 O 2(g) There are o catalysts ad O is the itermediate. Overall E a is show o the y-axis. 2. H 2(g) + 2 Pd(s) 2 Pd-H(s) C 2H 4(g) + Pd-H(s) à C 2H 5-Pd(s) (slowest step) C 2H 5-Pd(s) + Pd-H(s) à C 2H 6(g) + 2 Pd(s) C 2H 4(g) + H 2(g) à C 2H 6(g) kj/mol

7 Catalyst is: Pd Itermediates are: Pd-H ad C 2H 5Pd Activatio eergy is betwee the dotted lies. 3. Cl 2(g) à Cl(g) + Cl(g) iitiatio CH 4(g) + Cl(g) à CH 3Cl(g) + H(g)) x 2 propagatio H(g) + H(g) à H 2(g) termiatio CH 4(g) + Cl 2(g) à 2 CH 3Cl(g) + H 2(g) chai mechaism There are o catalysts. LIGHT provides E afor the reactio. Itermediates iclude H s ad Cl s (free radicals). (The relative heights of the secod ad third step are ot actually kow from the iformatio provided. The reactio is oly slightly exothermic. Though the propagatio step occurs twice, the reactat ad product eergies are the same, so it is show oly oce.) Page 87, Practice Problems a) itermediates: HOCl, OH -, HIO; catalyst: H 2O b) Rate = k[i - ][OCl - ]/[OH - ] 2. a) Cl 2 2 Cl Cl + CHCl 3 à HCl + CCl 3 Cl + CCl 3 à CCl 4 Cl 2 + CHCl 3 à CCl 4 + HCl b) itermediates: Cl, CCl 3 c) Rate = k[cl 2] 1/2 [CHCl 3] Page 91, 1.7 Review Questios 1. A series of steps that may be added together to give a overall chemical reactio. 2. The slowest elemetary process i a reactio mechaism determies the overall reactio rate ad is called the RATE-DETERMINING STEP. 3. Chemists determie these models by alterig the cocetratio of species ivolved i the mechaism ad examiig the effects these alteratios produce. Computer models are also very useful i the pursuit of mechaisms. 4. a) Hetero b) Homo c) Hetero 5. 1 st : H 2O 2(aq) + I - (aq) à H 2O(l) + IO - (l) slow 2 d H 2O 2(aq) + IO - (aq) à H 2O(l) + O 2(g) + I - (aq) fast a) Overall Rx: 2 H 2O 2(aq) à 2 H 2O(l) + O 2(g) b) Itermediates: IO - c) catalyst is I - 6. NO 2 + F 2 à NO 2F + F (RDS) NO 2 + F à NO 2F NO 2 + F 2 à 2 NO 2F 7. a) 3 b) 2 d step (middle step or step C) c) C d) B e) F f) edothermic 8. a) Mechaism Oe: Step 1: 2 NO 2(g) à NO 3(g) + NO(g) slow Step 2: CO + NO 3 à CO 2 + NO 2 fast Overall Rx: CO(g) + NO 2(g) à CO 2(g) + NO(g) (Note that this is a example of authocatalysis. The reactat, NO 2, also acts as a catalyst for the reactio. Some autocatalyzed reactios ivolve the formatio of the catalyst from the reactats i a early step.) Mechaism Two: Step 1: 2 NO 2(g) à N 2O 4(g) fast Step 2: CO + N so 4 à NO 2 + CO 2 + NO slow Overall Rx: CO(g) + NO 2(g) à CO 2(g) + NO(g) b) Mechaism 1 is correct as the CO is NOT i the RDS! 9. a) No too may reactat particles (five) for a sigle step. b) HBr + O 2 à HOOBr

8 10. HBr + HOOBr à 2 HOBr HBr + HOBr à H 2O + Br 2 HBr + HOBr à H 2O + Br 2 4 HBr + O 2 à 2 H 2O + 2 Br 2 c) Step Oe (both HBr ad O 2 ifluece reactio rate). d) K!" = [M] A [B] [M] = K eq[a][b] Rate = k K eq[a] 2 [B] Rate = k[a] 2 [B] 13. a) Step 1: NO + O2 NO 3 Step 2: NO 3 + NO à 2 NO 2 2 NO + O 2 à 2 NO 2 b) Rate = k[no] 2 [O 2] 14. a) NH 3 + OBr - NH 2Br + OH - NH 2Br + NH 3 à N 2H Br - N 2H OH - à N 2H 4 + H 2O 2 NH 3 + OBr - à N 2H 4 + Br - + H 2O (basic) b) itermediates: NH 2Br, OH -, N 2H + 5 c) Rate = k[nh 3] 2 [OBr - ]/[OH - ] (Note the relative heights of steps oe ad two do t matter, but there must be AT LEAST two steps.) 11. a) itermediate: N 2O 2 b) Rate = k[no] 2 [O 2] 12. Startig with each mechaism s temporary rate law based upo the slow step: Mechaism Oe: icosistet with the provided rate law, rate = k[a][b] Rate = k [M][A] C [M] K!" = A [B] [M] = K eq[a][b]/[c] Rate = k K eq[a] 2 [B]/[C] Rate = k[a] 2 [B] /[C] N/A: if [M] i the rate law was substituted with the statemet above, o-reactat species are also icluded Mechaism Two: cosistet with the provided rate law, rate = k[a][b] Rate = k [M][A] K!" = [M] B [M] = K eq[b] Rate = k K eq[a][b] Rate = k[a][b] Mechaism Three: icosistet with the provided rate law, rate = k[a][b] Rate = k [M][A]

1. Collision Theory 2. Activation Energy 3. Potential Energy Diagrams

1. Collision Theory 2. Activation Energy 3. Potential Energy Diagrams Chemistry 12 Reactio Kietics II Name: Date: Block: 1. Collisio Theory 2. Activatio Eergy 3. Potetial Eergy Diagrams Collisio Theory (Kietic Molecular Theory) I order for two molecules to react, they must