CHEMISTRY 104 Practice Problems #1 Organic Chem and Ch. 11 Kinetics Do the topics appropriate for your lecture

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1 EMISTRY 104 Practice Problems #1 rganic hem and h. 11 Kinetics Do the topics appropriate for your lecture Prepared by Dr. Tony Jacob (Resource page) Suggestions on preparing for a chemistry exam: 1. rganize your materials (quizzes, notes, etc.) and find out exactly what material/chapters are on the exam. 2. Usually, a good method to prepare for a chem exam is by doing lots of problems. Re-reading sections of a chapter is fine, but rereading entire chapters takes up large amounts of time that is generally better spent doing problems. 3. ld exams and quizzes posted by your instructor should be completely worked though. ld exams give you a sense of how long the exam will be, the difficulty of the problems, the variability of the problems, and the style of your instructor. Good Luck! USEFUL INFRMATIN: R = J / mol K ; R = L atm / mol K rate = k[a] 0 = k; [A] t = -kt + [A] o ; t 1/2 = [A] o /2k rate = k[a] 1 ; ln[a] t = -kt + ln[a] o ; t 1/2 = 0.693/k rate = k[a] 2 ; 1 [A]t = kt + 1 [A]0 ; t 1/2 = 1/(k[A] o ) k = Ae -E a/rt E ; ln(k) = a R 1 T + lna ; ln k 1 = E a 1 1 k 2 R T 2 T 1 RGANI EMISTRY and EM 103 REVIEW 1. a. What s the hybridization of the N atom marked A? b. What s the angle formed by the N atoms marked B? c. What s the bond order for the bond labeled? What s the electron domain geometry around the atom labeled D? e. What orbitals overlap to generate the bonds labeled E f. What s the electron domain geometry around the atom labeled F? F A N B E D 2. Given the following molecule, answer the questions. a. What is the molecular geometry around the atom labeled A? b. What is the hybridization of the atom labeled B? D c. ow many π bonds are in this molecule? What is the hybridization of the atom labeled? e. What is the angle centered around the labeled D? F f. What is the molecular geometry around the atom labeled? g. What is the hybridization of the N atom? h. What is the electron domain geometry around the N atom? i. What orbitals overlap to form the bond labeled E between the and atoms? j. What orbitals overlap to form the bond labeled F between the and atoms? k. What is the molecular formula for the molecule (e.g., )? A E N B 3. Draw the resonance structure for this molecule. - (take a walk)

2 4. rder the molecules below from lowest to highest boiling point I. 3 3 II III. IV a. I < II < III < IV b. II < I < IV < III c. II < IV < I < III III < IV < I < II e. II < III < I < IV 5. Place the following chemicals in order of increasing boiling points, i.e., from lowest to highest boiling point. I. 3 F 3 F II. 3 3 III IV a. I < II < III < IV b. II < IV < I < III c. II < IV < III < I III < I < IV < II e. I < III < IV < II 6. In the molecule shown, F 2, choose the structure with the correct locations of the δ + and δ - symbols. a. δ - δ - δ 0 δ + δ - F δ - F δ - F b. δ - F δ - F c. δ - F δ - F δ - δ - δ + F δ - 7. Which molecule has the shortest bond length? - a. b. c. 8. Which molecule would be expected to be most soluble in water? e. All bonds are the same length a. b. c. e. 9. Which functional group is nonpolar? a. alcohol b. alkene c. carboxylic acid ketone e. all are polar 10. ircle and name all functional groups; where appropriate, include cis/trans, or 1, 2, or 3. (int: There are 9 functional groups.) N Which molecule could contain a triple bond? a. 2 6 b. 6 6 c e. None of these (have a sandwich)

3 12. The molecule shown has several functional groups. Which set of functional group names below matches the functional groups in the molecule? The answers are written in the order of A, B,, D. a. alkene, primary alcohol, aldehyde, ketone b. alkene, secondary alcohol, aldehyde, carboxylic acid c. alkene, tertiary alcohol, ketone, aldehyde alkene, secondary alcohol, ketone, carboxylic acid e. alkene, secondary alcohol, ketone, aldehyde A B D 13. Given the chemical formula 4 8, draw a molecule that contains the functional groups listed below. (Note: There are many different answers to each question.) a. A molecule that contains an aldehyde. b. A molecule that contains a ketone. c. A molecule that contains a primary alcohol and alkene. A molecule that contains a secondary alcohol and alkene. 14. Which name bests describes the molecule shown: a. trimethyl-2-hexanal b. 3,4,5-trimethyl-2-heptanoate c. 2,3,4-trimethyl-2-hexanone 2,3,4-trimethyl-5-hexanone e. 3,4,5-trimethyl-2-hexanone 15. (Some practice naming molecules.) Give the systematic name for each molecule. a. b. c. e. f. g. h. i. 16. (Some practice naming molecules.) Give the systematic name for each molecule a. 2 3 b ( 3 ) 3 2 c e. 3 2 ( 3 ) 2 f g (Some practice drawing molecules from their names.) Draw the line structure for each chemical name given. a. 2,2,3,3-tetramethylhexane b. trans-2-pentene c. 3-methyl-1-butyne cis-2,3-dichloro-2-pentene e. 3-ethyl-2-pentanol f. 3-bromo-2-butanone g. 3,4-dimethyl-2-pentanone h. 2,2-dichloro-3,3-difluoro-1-butanal i. 2-ethyl-1-pentanoic acid j. propylbutanoate (coffee time)

4 18. Which functional group does not contain a carbonyl? a. aldehyde b. primary alcohol c. ketone carboxylic acid e. all contain a carbonyl 19. Which functional group can not exhibit -bonding? a. 1 alcohol b. 2 amine c. carboxylic acid ether e. all exhibit -bonding 20. Which one of the following molecules is not an isomer of the other ones? a. b. c. e. 21. Which pair of molecules is correctly assigned to the isomer name? 3 3 a. Enantiomer pair: l l b. Structural isomers: c. Geometric isomers: Structural isomers: 22. cis-2-butene and trans-2-butene exist as different isomers because it is not possible to rotate through the double bon Why is it so difficult to rotate through the double bond? 23. (Some practice drawing isomers.) a. Draw all eight structural isomers of 5 11 Br. b. Draw all four structural and geometric isomers of 3 5 l. Do not include cyclic compounds. c. Draw all seven structural and geometric isomers of 3 4 l 2. Do not include cyclic compounds. Draw all four ester isomers of Which molecule is chiral? Select answer e if none are chiral. N a. b. l c. Br e. none are chiral 25. ow many chiral carbon atoms does the molecule contain? a. 1 b. 2 c. 3 4 e. 6 (FB time)

5 26. Write the balanced combustion reaction of 1-heptene. (int: use the molecular formula for 1-heptene.) 27. Which molecule below when reacted with acetic acid yields? a. b. c. 28. For the reaction, choose the products forme + 2 S 4 a. + 2 b c. + 2 e When the chemical shown is oxidized, the product will be which chemical? a. 3-methyl-1-pentanal b. 3-methyl-1-pentanone c. 3-methyl-1-pentanoic acid 3-methyl-1-pentanol e. 3-methyl-1-pentane 30. Which chemicals do not react with an oxidizing agent? I II a. I b. II c. II and III I and IV e. II and IV (have a banana) III. 3 2 IV.

6 31. (Some practice doing organic reactions.) For each reaction below draw the products forme + 2 S 4 a. b. 2 N + 2 S S 4 c. + 2 S 4 N S 4 e. 32. (Some practice doing organic reactions.) For each reaction below draw the reactant(s) needed for the reaction. a. 2 S b. + 2 S c S N + 2 S 4 N + 2 (time for another walk)

7 33. (Some practice doing organic reactions.) For each reaction draw the product(s) forme a. [] [] [] b. c. [] e. [] [] f. g. (time for a nap)

8 34. (Some practice doing organic reactions.) For each reaction below draw the reactant(s) needed for the reaction. a. [] b. [] c. [] e. [] f. APTER 11 - KINETIS 1. For the reaction, S 2 (g) (g) 4 (g) S(g), if the rate of consumption of 2 (-Δ[ 2 ]/Δt) was 0.10M/s, what would be the rate of production of 2 S (Δ[ 2 S]/Δt)? a M/s b M/s c. 0.10M/s 0.20M/s e. 0.40M/s 2. What are the units for k in the rate law, rate = k [X]2 [Z] 2? [W] a. 1/sec b. 1/(M sec) c. 1/(M 2 sec) M/sec e. M 3. As the concentration of reactants is increased, the rate constant a. increases as the rate increases. b. increases because the number of collisions increases. c. decreases because the product concentration increases. decreases because the reverse reaction rate increases. e. is unchange (more coffee)

9 4. Given the reaction below which statement is correct? 2l 2 (g) (g) 2 (g) + 4l(g) a. The rate of consumption of l 2 is the same as the rate of production of 2. b. The rate of consumption of 2 is one half the rate of production of 2. c. The rate of production of l is one fourth the rate of production of 2. The rate of consumption of l 2 is twice the rate of production of l. e. None of the above statements are correct. 5. a. The hydrolysis of ethyl propanoate ester, , is performed: The concentration of the ester as it hydrolyzes is observed and its value is recorde time (s) [ester] x x x x a. What is the concentration of the ester at 600sec? b. What is the average rate of ester consumption at 600sec? c. What is the initial rate of ester consumption? 6. An experiment involving N and 2 was completed and the data collected is shown below. 2N(g) + 2 (g) 2N 2 (g) Experiment [N] o [ 2 ] o Initial Rate 2 (M/s) x x x x x x x x x x x 10-2? I. Given the above data, what is the rate law for the reaction. a. rate = k[n] b. rate = k[n][ 2 ] c. rate = k[ 2 ] rate = k[n] 2 [ 2 ] e. rate = k[n] 2 [ 2 ] 2 II. What is the overall reaction order? a. 1 b. 2 c. 3 4 e. none III. What is the value of the rate constant? a b c x 10 3 e x 10 5 IV. What is the initial rate for Experiment 4? a b c e V. ow does the rate change if the [N] was quadrupled and [ 2 ] was cut in half? a. rate cut in half b. rate unchanged c. rate doubles rate quadruples e. rate increases by 8 7. The reaction, A + 2B + 3 2B + A, has a rate law, rate = k [A]2[B]0[]1. If the concentration of all reactants was doubled, the rate will a. increase by a factor of 2 b. increase by a factor of 6 c. increase by a factor of 8 increase by a factor of 16 e. decrease by a factor of 8 (have some pizza)

10 8. Given the reaction below and the initial rate data, answer the following questions. 2 (g) + I 2 (g) 2I(g) Experiment [ 2 ] 0 [I 2 ] 0 Initial Rate [ 2 ] (M/s) x x x 10-2 I. What is the rate law? a. rate = k[ 2 ] b. rate = k[ 2 ][I 2 ] c. rate = k[i 2 ] rate = k[ 2 ] 2 [I 2 ] e. rate = k[ 2 ][I 2 ] 2 II. What is the reaction order? a. 1 b. 2 c. 3 none III. What is the value of the rate constant? a b c e IV. If [ 2 ] = 1.350M and [I 2 ] = 0.650M, what would be the initial rate? a b c e a. For the following reaction, A + AB + 2 A + B + A, and given the information below, write the rate law for the reaction. I. When the concentration of A is doubled the rate of reaction is double II. When the concentration of AB is cut in half the rate of reaction is cut in half. III. When the concentration of 2 is tripled the rate of reaction is increased by a factor of 9. b. If the initial concentrations are A = M, AB = 0.50 M, and 2 = M and the initial rate is 2.8 x 10-4 M/s, what is the rate constant for this reaction (include the proper units)? 10. Sulfuryl chloride, S 2 l 2, decomposes when heated: S 2 l 2 (g) S 2 (g) + l 2 (g) The order with respect to S 2 l 2 is first-order. If the initial concentration of S 2 l 2 was M, and the rate constant is 6.60 x 10-4 s -1, what is the concentration of S 2 l 2 after 1500s? a b c e Which of the following statements is incorrect for chemical reactions that follow first order kinetics? a. As the rate constant increases, the half-life decreases. b. When the reactant concentration increases, the rate constant also increases. c. If the temperature increases, the half-life decreases. As the reactant concentration increases, the half-life remains unchange e. All of the above statements are correct. 12. A certain first-order reaction is 45.0% complete in 65.0s. I. What is the value of the rate constant k? a x 10-2 s -1 b x 10-3 s -1 c x 10-3 s x 10-3 s -1 e x 10-3 s -1 II. What is the value of the half-life for this reaction? a. 56.3s b. 450s c. 75.3s 81.9s e. 100.s (watch some TV)

11 13. Given the first-order reaction, A 2 2A, if the initial concentration of A 2 was 0.150M and the rate constant is 2.50 x 10-3 s -1, how long will it take for the concentration to reach M? a b. 250 c e Which statement is incorrect? As the temperature of a reaction is increased a. the number of collisions increases. b. the number of molecules with energy greater than the activation energy increases. c. the rate of the reaction will increase. the value of the rate constant will increase. e. the activation energy will increase. 15. The reaction, A B, was analyzed with data plotted below. a. What is the rate law for this reaction? b. What is the value of the rate constant? c. What was the initial concentration of A? What is the concentration of A after 120s? 16. Answer the following statements either True (T) or False (F). a. The value of k is a function of temperature. b. The activation energy can be calculated if the rate constant is known at two different temperatures. c. An increase in the concentration of a reactant will always increase the rate of the reaction. A catalyst speeds up the forward and reverse reaction rates. e. An increase in temperature lowers the activation energy of the reaction. 17. The reaction, E F, was analyzed with data plotted below. a. What is the rate law for this reaction? b. What is the value of the rate constant (include units)? c. What was the initial concentration of E? What is the concentration of E after 1500s? (read a magazine)

12 18. A reaction, G, was studied and a plot of [G] versus time was prepared and a plot is shown below. a. What is the rate law? b. What is the value of the half-life? c. What is the rate constant k? Include units? 19. A reaction, D E, was studied and a plot of [D] versus time was prepared and is shown below. a. What is the rate law? b. What is the k value; include units? (eat an omelet)

13 20. The reaction: A + B 2 AB 2 was analyzed and is similar to the erioglaucine lab. In the first experiment, [A] o = M and [B 2 ] o = 0.30M (T = 298K). oncentration versus time of A was plotted (Graph 1). In the second experiment, [A] o = M and [B 2 ] o = 0.60M (T = 298K), and again the concentration versus time of A was plotted (Graph 2). Graph 1 Graph 2 a. What does the high concentration of B 2 relative to A do in terms of solving the rate law? Write the general rate law using k obs as you did in the erioglaucine lab. b. From Graph 1, determine the order with respect to A. c. What is the value of the observed rate constant, k obs1, from Graph 1? What is the value of the observed rate constant, k obs2, from Graph 2? e. What is the order with respect to B 2? f. What is the value of the rate constant k; include units? g. Write the rate law for this reaction with the orders include 21. The reaction: D + E DE was analyze In the first experiment, [E] o = 0.010M and [B] o = 1.0M; in the second experiment, [E] o = 0.010M and [D] o = 2.0M. Plots of 1/[E] t versus time were plotte a. Determine the order with respect to E. b. What is the order with respect to D? c. What is the value of the rate constant k; include units? (watch some more TV)

14 22. The reaction B + B was run and the kinetics were studie Two experiments were conducted during which chemical B was flooded in the reaction vessel. In the first experiment, [B] o = 1.0M and in the second experiment [B] o = 2.0M. The resulting data of the concentration of versus time is shown below in the graph. a. Write the general rate law. b. Since B was flooding the reaction, write a new simplified rate law. learly note any substitutions you made and what they represent. c. Determine the order of. Determine the order of B. e. Determine the rate constant. 23. Which statement is incorrect? In collisional theory, a. the molecules must collide. b. the molecules must have sufficient energy to overcome the activation energy of the reaction. c. no intermediate can be formed since these are unstable molecules. the proper orientation of the molecules colliding is require e. All of the above statements are correct.

15 24. The reaction: A + B AB was analyzed and is similar to the erioglaucine experiment. In the first experiment, [A] o = M and [B] o = 0.50M; in the second experiment, [A] o = M and [B] o = 2.0M. In both cases B was flooding the reaction. Plots of ln[a] t versus time were plotte By flooding the reaction with excess B this yields: Rate = k obs [A] x where k obs = k[b] y a. Determine the order with respect to A. b. What is the value of the observed rate constant, k obs1, when [B] = 2.00M? c. What is the value of the observed rate constant, k obs2, when [B] = 0.50M? What is the order with respect to B? e. What is the value of the rate constant k; include units? 25. Using the mechanism and overall reaction shown below, draw a reaction coordinate diagram and label the axes, reactants, products, transition states, activation energy forward and reverse for each step, and the Δ rxn (also called ΔE rxn ). Rxn 1 Rxn 2 N 2 + N 2 N 3 + N (slow, endo) N 3 + N (fast, exo) verall N 2 + N + 2 (endo) 26. Below are two possible mechanisms for the reaction of N 2 to produce 2. Mechanism I: Rxn 1 N 2 + N 2 N 3 + N (slow) Rxn 2 N 3 + N (fast) Mechanism II: Rxn 1 N 2 N + (slow) Rxn (fast) a. For each mechanism determine the overall reaction. b. What is the molecularity of each step of each mechanism? c. Determine the rate law for each of the 2 possible mechanisms. It is found experimentally when the concentration of N 2 is doubled the reaction rate quadruples, and when the concentration of is doubled the reaction rate is unchange Which mechanism is valid and why? (text a friend)

16 27. a. For the mechanism below determine the overall reaction. b. List the reactants and products. c. List the catalysts and intermediates present. If there are none, write none. Determine a rate law from the mechanism. 2N 2 5 2N 2 + 2N 3 fast, equilibrium N 2 + N 3 N + N N 3 + N 2N 2 slow 28. a. For the mechanism below determine the overall reaction. b. List the reactants and products. c. List the catalysts and intermediates present. If there are none, write none. f. Determine a rate law from the mechanism. E + A 2 EA 2 fast, equilibrium EA EA + A 2 A 2 A + + EA A + E fast slow fast fast 29. For the reaction, B, the rate constant increases from 1100s -1 to 12,000s -1 when the temperature is increased from 25. to 60.. What is the activation energy for the reaction in kj/mol? a. 5.6 x 10 1 b. 5.6 x 10-1 c. 2.6 x x 10-4 e. 2.4 x The reaction 2N 2 2N + 2 has a rate law of rate = 1.4 x 10-2 [N 2 ] 2 at 500. K. What is the value of the rate constant at 298K if the activation energy is 80.kJ/mol? a. 3.0 x 10-8 b. 6.5 x 10 3 c. 1.3 x x 10-5 e. 4.0 x Which statement is incorrect for catalyzed versus uncatalyzed reactions? a. The relative energy of the reactants and products in a reaction coordinate diagram will be the same for the catalyzed and uncatalyzed reactions. b. The overall reaction is unchange c. The forward and reverse rates in the catalyzed reaction are faster. The activation energy for the uncatalyzed reaction is lower than for the catalyzed reaction. e. The reaction coordinate diagram for the catalyzed and uncatalyzed reactions are different. 32. If a reaction has an activation energy of 52.5kJ how much will the rate increase by when the temperature is increased from 65.0 to 96.9? (almost done)

17 33. Using the plot of ln(k) versus 1/T, answer the following questions. Assume the graph values have 3 sig figs. I. What is the value of the frequency factor, A? a b c e. None of these II. What is the value of the activation energy in J/mol? a. 500 b c e III. What is the value of the rate constant at 400K? a x 10-3 b c e. none of these (yea done) ANSWERS hem 103 Review and rganic 1. a. sp 3 b. 180 c. 1.5 trigonal planar e. σ: sp 2 () sp 2 (); π: p p f. tetrahedral 2. a. tetrahedral b. sp 2 c. 3 sp 3 e. 120 f. bent g. sp 3 h. tetrahedral i. (sp 2 ) (sp 3 ) j. (sp 3 ) (sp 3 ) k N areas of resonance: benzene ring and - group (carboxylate group) - 4. b {determine IMFs for each molecule; III has -bonding while none of the others do greatest IMFs highest bp; other 3 are hydrocarbons LDFs; II smallest #e - smallest LDFs lowest bp; I and IV have same number of e - but I is more branched smaller LDFs lower bp than IV; II < I < IV < III} 5. d {I has dipole-dipole IMFs; II has -bonding IMFs; III has LDF IMFs; IV has -bonding IMFs; IMFs Boiling Point ; since III only has LDFs III has lowest IMFs of the other choices; since I has dipole-dipole it has greater IMFs than III; since IV has one group for -bonding while II has three groups for -bonding IV has lower IMFs than II; IMFs increase as: III < I < IV < II and the boiling points follow this trend in IMFs: III < I < IV < II; note that the size of the molecules are similar and the molecules with the stronger IMFs also generally have more electrons; if molecule III was significantly larger than the other molecules then the larger number of electrons in molecule III may increase its LDF above the dipole-dipole interactions of molecule I or even the -bonding interactions of molecule IV} 6. b {compare each pair of atoms; the more EN atom gets a δ - while the less EN atom receives a δ + ; EN > EN ; EN F > EN } 7. c {greater B smaller bond length; c has B = 2} 8. c { a and e are hydrocarbons and are nonpolar and not soluble; d is an ether and is polar but less polar than molecules with groups; between b and c c has more groups which makes it more polar and the short carbon chain also increases the solubility and as a carboxylic acid it will also in this case} 9. b {other groups have or = groups both of which are polar}

18 10. A = carboxylic acid; B = trans-alkene; = ketone; D = alkyne; E = benzene/aromatic; F = secondary alcohol; G = trans-alkene; = aldehyde; I = primary amine A F B G D N I 2 E 11. b { a and d are alkanes n 2n+2 ; c is either a ring or an alkene n 2n ; b could contain a triple bond; e.g., 12. e ) 13. a. or b. c. or or or or or or 14. e 15. a. 2,3,4-trimethylhexane b. 3-methyl-1-butene c. trans-4-methyl-2-pentene 3,3-dimethyl-1-butanol e. 2,2,3-trimethylbutanal f. 3,4-dimethyl-2-pentanone g. 1-propanoic acid h. 2,3,4-trimethyl-1-pentanoic acid i. butylethanoate 16. a. 3-ethylhexane b. 2-methyl-1-pentanal c. 2-butanol 3-methyl-1-pentyne e. 2,2-dimethyl-1-butanoic acid f. ethylpropanoate 17. a. b. c. Br F F l l e. f. g. h. l l i. l. 18. b {primary alcohol does not contain a carbonyl; alcohol: R } 19. d {for -bonding there must be a bonded to N,, or F; ethers do not have this} 20. d { a = 6 12 ; b = 6 12 ; c = 6 12 ; d = 6 14 ; e = 6 12 ; a, b, and c are structural isomers of one another; e is a geometric isomer of a } 21. d { a same molecule rotated; b formulas are different, 5 12 and 5 10 ; hence they re not isomers; c structural isomers not geometric isomers} 22. The π bond needs to be broken to allow rotation through the double bon

19 23. Br Br Br a. Br l Br Br Br Br b. l l l l l l l l l c. l l l l l l l l 24. c {need to find a atom with 4 different groups attached to it} * * 25. b c {the product is an ester; the reactant given is a carboxylic acid so an alcohol is needed; only answer c is an alcohol} 28. d + [] 2 S c 30. e {xidizing agents react with 1 and 2 alcohols, and aldehydes. They do react with carboxylic acids (II) and tertiary alcohols (IV).} S a. N b. 2 N + 2 S S 4 + c. + 2 S N S 4 2 N + e.

20 32. 2 S a. b. + 2 S c. N S N 33. N S 4 N + 2 a. c. [] b. [] [] no reaction [] e. (no reaction) [] [] [] f. g. 34. a. [] b. c. [] e. [] f. [] hapter 11 Kinetics 1. b {relative rates: 1 Δ[ 2 S] = 1 $ 2 Δt 4 Δ[ 2 ] ' & ) = Δ[ 2 S] = 2 $ % Δt ( Δt 4 Δ[ 2 ] ' & ) = ( 1 / % Δt ( 2 )(0.1) = 0.05M/s} 2. c {substitute for rate: M/sec and for the chemicals: M; M/sec = k(m) 2 (M) 2 /(M); M/sec = km 3 ; k = 1/(M 2 sec)} 3. e {k is a function of T only} 4. e {the rate relationships are: 1 2 (rate l 2 ) = 1 2 (rate 2 ) = rate 2 = 1 4 (rate l) and none of the answers correctly relate these rates; a : rate l 2 = 2(rate 2 ); b : rate 2 = 2(rate 2 ); c : rate l = 4(rate 2 ); d : rate l 2 = 1/2(rate l)}

21 5. a. 0.69M {read right off the table, 0.69M} b. 4.3 x 10-4 M/s {rate = -Δ[]/Δt = -( )/( ) = 4.25 x 10-4 M/s} c. 6.0 x 10-4 M/s {rate = -Δ[]/Δt = -( )/(0 200) = 6.0 x 10-4 M/s} 6. I. d {for N, use Exp 1 & 2: rate 2 = k[n] 2 x y [ 2 ] 2 rate 1 k[n] x y 1 [ 2 ] ; = [0.0250]x [0.0250] y [0.0125] x [0.0250] y ; 4 =! $ x! $ # & # & " % " % y ; 4 = 2 x ; x = 2 for 2, use Exp 1 & 3: rate3 = k[n] 3 x y [ 2 ] 3 rate 1 k[n] x y 1 [ 2 ] ; = [0.0125]1 [0.0500] y [0.0125] 1 [0.0250] y ; 2 =! $ 1! $ y # & # & ; 2 = 2 y ; y = 1 " % " % II. c {add the orders of: rate = k[n] 2 [ 2 ] = = 3} III. d {use any experiment; I used Experiment 1: 2.8 x 10-2 = k(1.25 x 10-2 ) 2 (2.5 x 10-2 ); solve for k; k = 7168M -2 s -1 } IV. d {rate = k[n] 2 [ 2 ] = (7170)(5 x 10-2 ) 2 (5 x 10-2 ) = 8.96 x 10-1 M/s} V. e {rate = k[n] 2 [ 2 ]; use k = 1 and all concentrations = 1; rate = (1)(1) 2 (1) = 1; now change concentrations: rate = (1)[4] 2 [ 1 / 2 ] = 8; rate increases by a factor of 8} 7. c { use k = 1 and all concentrations = 1; rate = (1)(1) 2 (1) 0 (1) = 1; now double the concentrations: rate = (1)(2) 2 (2) 0 (2) = 8; rate increases by a factor of 8} 8. I. d {for I 2, use Exp 1 & 2: rate 1 = k[ 2 ] 1 x y [I 2 ] 1 rate 2 x y k[ 2 ] 2[I2 ] ; = [0.500]x [0.450] y [0.500] x [0.150] y ; 3 =! $ x y! $ # & # & ; 3 = 3 y ; y = 1 " % " % for 2, use Exp 2 & 3: rate 3 = k[ 2 ] 3 x y [I 2 ] 3 rate x y 2 k[ 2 ] 2[I2 ] ; = [1.000]x [0.300] 1 [0.500] x [0.150] 1 ; 8 =! $ x! $ # & # & " % " % II. c {add the orders of the individual chemicals: rate = k[ 2 ] 2 [I 2 ] = = 3} III. d 1 ; 8 = 2 x 2 1 ; 4 = 2 x ; x = 2 {use any experiment; I chose experiment 3; Exp 3: rate = k[ 2 ] 2 [I 2 ]; 1.40 x 10-2 = k(1.00) 2 (0.300); k = 4.67 x /(M 2 s)} IV. d {rate = k[ 2 ] 2 [I 2 ]; rate = (0.0467)[1.350] 2 [0.650] = M/s} 9. a. rate = k[a] 1 [AB] 1 [ 2 ] 2 {rate = k[a] a [AB] b [ 2 ] c ; rate 1 rate 2 = ([A] 1 [A] 2 )a ; 2 1 = (2 1 ) a ; a = 1; rate1 rate 2 = ([AB] 1 [AB] 2 )b ; 1/2 1 = (1/2 1 ) b ; b = 1; rate 1 rate 2 = ([ 2]1 [ 2]2 )c ; 9 1 = (3 1 ) c ; c = 2} b. 1.0 x 10 2 M -3 s -1 {rate = k[a] 1 [AB] 1 [ 2 ] 2 ; 2.8 x 10-4 = k(0.025)(0.50)(0.015) 2 ; 2.8 x 10-4 = k(2.81 x 10-6 ); k = 99.56(1/(M 3 s))} 10. a {time integrated rate laws; ln[s 2 l 2 ] = -(6.6 x 10-4 s -1 )(1500s) + ln(0.545); ln[s 2 l 2 ] = ; [S 2 l 2 ] = M} 11. b {The rate can increase by increasing reactant concentrations but k remains constant and is only a function of T.} 12. I. c {45% complete means 55% remains use ln([a] t /[A] o ) = -kt; ln(0.55) = -k(65); k = 9.20 x 10-3 s -1 } II. c {t 1/2 = 0.693/k = 0.693/9.20 x 10-3 = 75.35s} 13. d {time integrated rate laws; ln[a] t = -kt + ln[a] o ; ln(0.05) = -(2.5 x 10-3 )(t) + ln(0.15); = -2.5 x 10-3 t; t = 439.4s} 14. e {E a is not a function of T; the activation energy will only change if a catalyst is added} 15. a. rate = k[a] {since the plot of ln[a] vs time yields a line 1st order} b s -1 {slope = -k; slope was given in the equation of the line; k = 0.01; can also calculate the slope if the equation of the line was not given; slope = Δy/Δx = [-3 (-5)]/[0 200] = 0.01; k = 0.010} c M {the y-intercept of the line equation is ln[a] 0 ; ln[a] 0 = ; [A] 0 = e = M = 0.050M} 0.015M {can be calculated: ln[a] t = -kt + ln[a] 0 = -(0.01)(120) = ; [A] t = e = M; can also be read from the graph easiest graph to use is the [A] versus time; this yields ~0.0158M} 16. a. T b. T c. F T e. F

22 17. a. rate = k[e] 2 {since the plot of 1/[E] vs time yields a line 2nd order} b M -1 s -1 {slope = k; slope = ; k = M -1 s -1 ; can also calculate slope if the equation of the line was not given} c. 0.90M {from the 1st graph [E] o = 0.90M; from the equation y-intercept = 1/[E] o : 1/[E] o = ; [E] o = 1/ = M = 0.90M} 0.092M {time integrated rate laws; 1/[E] t = kt + 1/[E] o = (0.0065)(1500) + (1.1114) = ; [E] t = 1/ = M} 18. a. Rate = k[g] 1 {not 0 order since 0 order would yield a line when [G] versus time was plotted; so it is either first or second order; to determine which, see what happens to the half-life; the half life is constant 1st order since t 1/2 = 0.693/k} b. 38sec {this value is approximate; find the time that is required to drop from 2.0M to 1.0M and from 1.0M to 0.5M, etc.} c. k = s -1 {from first half life: 38sec = 0.693/k; 38k = 0.693; k = s -1 } 19. a. Rate = k[d] 2 {not 0 order since 0 order would yield a line when [G] versus time was plotted; so it is either first or second order; to determine which, see what happens to the half-life; the half life is increasing 2nd order since t 1/2 = 1/k[D] o } b. k = M -1 s -1 {from first half life: 28sec = 1/k(2.0) where 2.0 = the initial concentration of D; 56k = 1; k = M -1 s -1 ; if the second half life was used then the same result would be obtained: 56sec = 1/k(1.0); 56k = 1; k = M -1 s -1 ; 1.0M is the new [D] 0 for the second half-life}

23 20. a. Since the concentration of B 2 is much larger than the concentration of A, the concentration of B 2 can be considered constant. This therefore simplifies the rate law from: rate = k[a] x [B] y to: rate = k obs [A] x where k obs = k[b 2 ] y b. order for A = 0 {since a plot of [A] versus time yielded a line 0 order; x = 0} c. k obs1 = {k obs1 = -slope from Graph 1; k obs1 = 3.5 x 10-6 } k obs2 = {k obs2 = -slope from Graph 2; k obs2 = 1.4 x 10-5 } e. order for B 2 = 2 {When [B 2 ] doubled going from 0.30M to 0.60M, k obs increased by a factor of 4 ( / = 4). Since k obs = k[b 2 ] y and k is constant, then if [B 2 ] doubles and k obs increases by 4 y = 2} using y = ln k obs1 ln [B 2 ] 1 ; y = ln 3.5x10 6 k obs2 [B 2 ] 2 1.4x10 5 ln 0.30 ; y = ln(0.25) 0.60 ln(0.5) ; y = 2 } f. k = 3.89 x /(Ms) {use k obs1 = k[b 2 ] 1 2 ; 3.5 x 10-6 = k(0.3) 2 ; k = 3.89 x 10-5 ; can also use kobs2 = k[b 2 ] 2 2 ; 1.4 x 10-5 = k(0.6) 2 ; k = 3.89 x 10-5 ; if this was experimental data you would do both and average the k values} g. rate = k[a] 0 [B 2 ] 2 = k[b 2 ] a. order for E = 2 {if 1/[E] t versus time yields a line 2nd order; rate = k[d] x [E] y ; rate = k obs [E] y where k obs = k[d] x } order for D = 1 {When [D] was doubled going from 1.0M to 2.0M, k obs doubled as well (0.1/0.05 = 2). Since k obs = k[d] x and k is constant, then if [D] doubles and k obs doubles x = 1; note k obs is the slope of the line given in the equation of the line shown on the plot} e. k = M -2 s -1 {use k obs = k[d] x ; can use either experiment: 0.1 = k(2) 1 ; k = 0.05M -2 s -1 ; for the other experiment: 0.05 = k(1) 2 ; k = 0.05M -2 s -1 } 22. a. Rate = k[b] x [] y b. The new rate law is now: rate = k obs [] y. Since [B] is approximately constant since it is flooding the reaction, then let k obs = k[b] x. c. order with respect to = 1 {Using the B = 1.0M graph (you could also have used the [B] = 2.0M graph), determine the half life for two intervals. t 1/2 (1) = (9 0) = 9s; t 1/2 (2) = (18-9) = 9s; this is constant so the order of is 1. If the half life decreased by a factor of 2 it would be zero order; if it increased by a factor of 2 it would be second order.} order with respect to B = 0 {The half life of the other plot, [B] = 2.0M, was determined to be approximately the same as in the first graph. t 1/2 (1) = (9 0) = 9s; t 1/2 (2) = (18-9) = 9s; Since half lives for the two graphs are unchanged, then changing [B] has no effect on the reaction; hence, the order of B = 0. Likewise, you could use x = {ln(k obs(1) /k obs(2) }/{ln([b] 1 /[B] 2 )} this expression was derived in the lab. Since half lives are inverse to the rate, the inverse of the times can be used for the k obs. x = {ln((1/9)/(1/9))}/{ln(1.0/2.0)}; x = ln(1)/ln(0.5); x = 0. Alternatively, k obs can be calculated as was done in the lab and then used rather than the inverse of the t 1/2. k obs = 0.693/t 1/2 ; k obs = 0.693/9 = 0.077; x = {ln(k obs(1) /k obs(2) }/{ln([b] 1 /[B] 2 )}; x = {ln(0.077/0.077)}/{ln(1.0/2.0)}; x = ln(1)/ln(0.5); x = 0} e. k = 0.077s -1 {k obs(1) = k([b] 1 ) x ; (0.077) = k(1.0) 0 ; k = 0.077} 23. c {intermediates are not part of collisional theory} 24. a. order for A = 1 {a plot of ln[a] t versus time that yields a line 1st order} b. k obs1 = {k obs1 is the slope of the line given in the equation of the line shown on the plot} c. k obs2 = {k obs2 is the slope of the line given in the equation of the line shown on the plot} order for B = 1 {When [B] quadrupled going from 0.50M to 2.0M, k obs quadrupled as well ( / = 4). Since k obs = k[b] y and k is constant, then if [B] quadrupled and k obs quadrupled y = 1} y = ln k obs1 ln [B 2] 1 y = ln k using obs2 [B 2 ] ln 2.00 y = ln(4) 2 ; ; ln(4) ; y =1 } e. k = M -1 s -1 {use k obs = k[b] y ; can use either experiment: = k(2) 1 ; k = M -1 s -1 ; for the other experiment: = k(0.5) 1 ; k = M -1 s -1 }

24 a. N 2 + N + 2 b. Mechanism I, Rxn 1: bimolecular, Rxn 2: bimolecular; Mechanism II, Rxn 1: unimolecular, Rxn 2: bimolecular c. Mechanism I: rate = k 1 [N 2 ] 2 and Mechanism II: rate = k 1 [N 2 ] (use the slow step to determine the rate law; check to make sure no intermediates are present in the rate law; they re not) The general rate law: rate = k[n 2 ] 2. Therefore, Mechanism I is consistent with this experimental evidence. 27. a. 2N 2 5 4N b. reactant: N 2 5 ; products: N 2, 2 c. no catalysts; intermediates: N 3, N rate = k[n 2 5 ] where k = k 2 (k 1 /k -1 ) 1/2 {use slow step to determine rate law: rate = k 2 [N 2 ][N 3 ]; [N 3 ] is an intermediate so need to make a substitution for [N 3 ]; from step 1 which is equilibrium step: set rate forward = rate reverse: k 1 [N 2 5 ] 2 = k -1 [N 2 ] 2 [N 3 ] 2 ; rearrange to solve for [N 3 ]: [N 3 ] 2 = k 1 [N 2 5 ] 2 /(k -1 [N 2 ] 2 ); [N 3 ] = (k 1 /k -1 ) 1/2 [N 2 5 ]/[N 2 ]; substitute into the original rate law: rate = k 2 [N 2 ]{(k 1 /k -1 ) 1/2 [N 2 5 ]/[N 2 ]}; collect terms: rate = k 2 (k 1 /k -1 ) 1/2 [N 2 5 ]; simplify: rate = k[n 2 5 ] where k = k 2 (k 1 /k -1 ) 1/2 } 28. a. 2 + A 2 2A b. reactants: 2, A 2 ; products: A c. catalyst: E; intermediates: EA 2, EA, A 2, rate = k[e][a 2 ][ 2 ] where k = k 2 k 1 /k -1 {use slow step to determine rate law: rate = k 2 [EA 2 ][ 2 ]; [EA 2 ] is an intermediate so need to make a substitution for [EA 2 ]; from step 1 which is equilibrium step: set rate forward = rate reverse: k 1 [E][A 2 ] = k -1 [EA 2 ]; rearrange to solve for [EA 2 ]: [EA 2 ] = k 1 [E][A 2 ]/(k -1 ); substitute [EA 2 ] into the original rate law: rate = k 2 {k 1 [E][A 2 ]/k -1 }[ 2 ]; collect terms: rate = (k 2 k 1 /k -1 )[E][A 2 ][ 2 ]; simplify: rate = k[e][a 2 ][ 2 ] where k = k 2 k 1 /k -1 } 29. a {ln(k 1 /k 2 ) = E a /R(1/T 2 1/T 1 ); ln(1100/12000) = [E a /(8.314)][1/333 1/298]; = [E a /8.314][ x 10-4 ]; E a = 56,338 = 5.6 x 10 4 J/mol x 1kJ/1000J = 5.6 x 10 1 kj/mol} 30. a {ln(k 1 /k 2 ) = (E a /R)( 1 /T 2-1 /T 1 ) with R = 8.314; k 1 = 1.4 x 10-2 ; T 1 = 500; T 2 = 298; E a = 80,000J/mol; solve for k 2 ; ln(1.4 x 10-2 /k 2 ) = (80,000)/(8.314)[1/298 1/500]; ln(1.4 x 10-2 /k 2 ) = 13.04; 1.4 x 10-2 /k 2 = x 10 5 ; k 2 = 1.4 x 10-2 /4.605 x 10 5 = x 10-8 } 31. d {catalyzed reaction has lower activation energy.} times greater {solve for the rate constant ratio, k 1 /k 2 since the rate constant ratio reflects the rate ratio; let k 2 correspond to 65.0 and k 1 correspond to 96.9 ; ln(k 1 /k 2 ) = (E a /R)[1/T 2 1/T 1 ]; ln(k 1 /k 2 ) = (52500J/8.314J/molK)[(1/( ) 1/( )]; ln(k 1 /k 2 ) = (6314.6)[ ]; ln(k 1 /k 2 ) = (6314.6)[ ]; ln(k 1 /k 2 ) = ; k 1 /k 2 = e ; k 1 /k 2 = 5.00; since the rate constant at the higher temperature is 5 times larger than the rate constant at 65.0, the rate will increase by a factor of 5} 33. I. d {from y = mx + b = ln(k) = (-E a /R)(1/T) + ln(a), b = ln(a) and hence, ln(a) = 9.25; A = e 9.25 = 10405} II. c {slope = -500 = -E a /R = -E a /8.314; E a = (500)(8.314) = 4157J = 4160J} III. d {400K = T; 1/T = ; lnk = -500(0.0025) = 8.00; k = e 8 = 2981}