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Chapter 14 Solutions

Chapter 14 Solutions - —'——fi Chapter 14 Chemical...

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Unformatted text preview: —'——fi Chapter 14: Chemical Kinetics 2. In each case, we draw the tangent line to the plotted curve. A[H202]_1.7 M —0.6 M _ = F9.2x10"‘ M s" 400 3—1600 5 (a) The slope of the line is A[i—12()2] Ar Reaction rate = - = 9.2x10" M s" 0)) Read the value where the horizontal line [HgOgl = 0.50 M, intersects the curve 2 21§0 s or = 36 minutes I I l l l l l MA] (0474 M—0.485 M) _ 3. Ratc=-— =— _1.0x10'3 Ms" ' Ar 32.45—71.55 A A — . 4 (a) Rate = —L = — 0'1498 0 1565M = 0.006”.Ir M min"1 1.00 mm — 0.00 mm Rate: _ A[A] = _0.1433 M—0.t493 M =0m65 M M4 . A: 2.00 min—1.00 mm (b) The rates are not equal because, in all except Zero-order reactions, the rate depends on the concentration of reactant. And, of course, as the reaction proceeds, reactant is consumed and its concentration decreases, the rate of the reaction decreases. ; (a) [A] = [AL +A[A] = 0.533 M —0.013 M 2 0.575 M (b) A[A] = 0.565 M —0.533 M = -0.023 M A; _—_ A[A]_§_ = fl: 1_[) mm A[A] ~2.2><10'2 Mjrnin time = t + A: = (4.40+1.0)n11n :54 min 6. Initial concentrations are [ch1.]:0.105 M and [Czof']=0.300 M. The initial rate of the reaction is 7.1x10‘5 M min”. Recall that the reaction is: 2 HgC12(aq) +C203‘(aq) 4, 2 Cl'(aq) + 2 C02(g) + Hg2C12(aq). The rate of reaction equals the rate of disappearance oszoi' . Then, after 1 hour, assuming that the rate is the same as the initial rate, (a) 3" | ' [chn]: 0.105 M— 7.1x10’5 waxl hxéomm = 0.096 M L- 5 Into! CEO‘t 1h (13) [clef]: 0.300 M—[Tlx 10'5 mi x1 h >< 60mm] =0.296 M L-min 1b 450 Chapter 14: Chemical Kinetics —A A A c - 7. (a) Rate: [ ]=-—-[—=l.76><10‘5 Ms" - - - A: 2A: A C ; % = 2><l.'l6><10*5 M s" = 3.52x10'5 Mfs _; Aw A[C] -. -. . . (b) ——~ = —— = —l.76x 10 M 3 Assume this rate IS constant. A: 2a: - [A]=0.3530 M+[—1.76x10“ M s"><1.001m'uix160.S ]=0.357 M min A A (c) [ ]= —1.76><10" M s" A: _ H At: A[A] =0.3500M—0.53580M=4‘5X102 s I —l.76x10‘5 W5 —1.'}'6x10 Ws 5.?XIO" mall-[202 x 1 mo] 02 l L soln- s 2 mol H202 An 0 s. (a) it 1] = 1.00 L solnx = 2.9x104 mol Ozfs o (h) M 2]=2.9x10“‘ “10102 x 60.5 =1.'l><10'1 molOzfmin .. g A: 3 1mm § 2 . (c) av[oz] z 13x10: 1110102 x 22,414 mL 02 at 3'1? = 3.8x10 02 at STP _ A: min 1 mo] 02 mm . ‘- ill“ a Notice that, for every 1000 mmHg drop in the pressure of A(g), there will be a corresponding 2000 mmHg rise in the pressure of B(g) plus a 1000 mmHg rise in the pressure of C(g). (a) We set up the calculation with three lines of information be10w the balanced equation: (I) the initial conditions, (2) the changes that occur, which are related to each other by reaction stoichiometry, and (3) the final conditions, which simply are initial conditions + changes. Ms) A 23(g) + C(g) Initial 1000. mmHg 0. mmHg 0. 1111an Changes «4000. mmHg +2000. Iang +1000. mmI-Ig _ Final 0. l'l'll'an 2000. MmI-Ig 1000. mmHg Total final pressure = 0. mmI-Ig + 2000. mmHg + 1000. mmHg 2 3000. nuan 0:) Me) —> 213(3) + an Initial 1000. mian 0. mmHg 0. mmHg Changes —200. mmHg +400. mmHg +200. mint-1g Final 800 mmI-Ig 400. mmHg 200. mmHg Total pressure=800. mmHg+400. mmI-Ig +200. mmI-Ig: 1400. ang 451 flfi Chapter 14: Chemical Kinetics 10. (a) We will use the ideal gas law to determine NZO5 pressure O 1.00 gxl_m‘3$_t x0.08’206 L 3”“ x(273 + 65) K nRT 108.0g mol K 760 mmHg PENZOS} = —— = x V 15 L 1 atm (l1) After 2.38 min, one half—life passes. The initial pressure of NEO5 decreases by half to 6.5 mmHg. (c) From the balanced chemical equation, the reaction of 2 mol N105(g) produces 4 mol N02(g) and l molOAg) . That is, the consumption of 2 mo] of reactant gas produces 5 mol of product gas. When measured at the same temperature and confined to the same volume. pressures will behave as amounts: the reaction of 2 mmHg of reactant produces 5 111lan of product. = 13 mmHg 5 mmHgtpmducn pmml =13 N205 (initially) — 6.5 H‘IHIHg N105 (mt) mml‘lgtmcwnllxm] Method of Initial Rates u. (a) From Expt. 1 to Expt. 3, [A] is doubled, while [B] remains fixed. This causes the . 10“ " rate to increases by a factor of M = 2.01 = 2 3.35x 10 M s Thus, the reaction is first-order with respect to A. From Expt. 1 to Expt. 2, [B] doubles, while [A] remains fixed. This causes the rate —3 fl to increases by a factor of = 4.03 z 4 3.35 x 10 M s I Thus. the reaction is second-order with respect to B. (b) Overall reaction order : order with respect to A + order with respect to B = l + 2 = 3 The reaction is third—order overall. (c) Rate = 3.35x10" M s“ = k(0.185 M)(0.133 in)2 —4 —1 k4 3.35x10 Ms =0_102M_25_l " (0.185 mm. 133 M)2 i | l = (13—65 + 16) mmHg = 23 mmHg I 12. From Expt. 1 and Expt. 2 we see that [B] remains fixed while [A] triples. As a result. the initial rate increases from 4.2 ><10'3 Wmin to 1.3 ><10’2 Mfmin, that is, the initial reaction rate triples. Therefore, the reaction is first-order in [A]. Between Expt. 2 and Expt. 3, we see that [A] doubles, which would double the rate, and [B] doubles. As a consequence, the initial rate goes from 1.3)(10'2 Mfmin to 5.2 X 10‘1 Wmin, that is, the rate quadruples. Since an additional doubling of the rate is due to the change in [B], the reaction is first— order in [B]. Now we determine the value of the rate constant. Rate 52X 107: M; mi“ R :k A 1 B I ' = =—‘H—'—-—‘ ate [ ][ ] [A13] 3.00Mx3.00M The rate law is Rate = (5.8 x 10-3 L mol—1min" )[A]'[s]i = 5.8 ><10'3 L madman" 452 Chapter 14: Chemical Kinetics 13. From Experiment 1 to 2, [NO] remains constant while [C12] is doubled. At the same time the initial rate of reaction is found to double. Thus, the reaction is first-order with respect to [C12]. From experiment 1 to 3, [C12] remains constant, while [N0] is doubled, resulting in a quadrupling of the initial rate of reaction. Thus, the reaction must be second—order in [NO]. Overall the reaction is third-order: Rate = k [NO]2[C12]. The rate constant may be calculated from any one of the experiments. Using data from exp. 1, Rate 2.27x10‘5 M s" k = w=w [NO]2[C12] (0.0125 M)2(0.0255M) = 5.70 Mis"l 14. (a) From Expt. 1 to Expt. 2, [B] remains constant at 1.40 M and [C] remains constant at 1.00 M, but [A] is halved(x0.50). At the same time the rate is halved(x0.50) . Thus, the reaction is first-order with respect to A, since 0.50r = 0.50 when x = 1. From Expt. 2 to Expt_ 3, [A] remains constant at 0.70 M and [C] remains constant at 1.00 M, but [B] is halved(x0.50], from 1.40 M to 0.70 M. At the same time, the rate is quartered (x025) . Thus, the reaction is second-order with respect to B, since 0.50? = 0.25 when )2 = 2. From Expt. 1 to Expt. 4, [A] remains constant at 1.40 M and [B] remains constant at 1.40 M, but [C] is halved (x050), from 1.00 M to 0.50 M. At the same time, the rate is increased by a factor of 2.0. 1 [mtg =16 rate3 =16><Zrate2 = 4 rate2 =4x31Eratel = 2Xrate1] Thus, the order of the reaction with respect to C is —l , since 0.52 = 2.0 when z = *1. l 2 WI ('3) rates = k(0.70 M)1 (0.70 M)2 (0.50 M)" = [{1'4: M] (M: M] [1‘02 M] 11-2-1 2 =k§ (1.40 M)‘ g (1.40 Mn" (1.00 M)‘l = meg] = ratel[%] = fi-ratel This is based on rate, = k (1.40 M)‘ (1.40 M)2 (1.00 M)‘l First-Order Reactions g (a) TRUE The rate of the reaction does decrease as more and more of B and C are formed. but not because more and more of B and C are formed. Rather the rate decreases because the concentration of A must decrease to form more and more of B and C. (b) FALSE The time required for one half of substance A to reactmthe half-life—is independent of the quantity of “A” present. 453 a Chapter 14: Chemical Kinetics 16. (a) FALSE For first-order reactions, a plot of In [A] or log [A] vs. time yields a straight line. A graph of [A] vs. time yields a curved line. (b) TRUE The rate of formation of “C” is related to the rate of disappearance of “A” by the stoichiometry of the reaction. LL. (3) Since the half-life is 180 s, after 900 3 five half-lives have elapsed, and the original quantity of A has been cut in half five times. final quantity of A = (0.5)5 x initial quantity of A = 0.03125x initial quantity of A About 3.13% of the original quantity of A remains unreacted after 900 s. or More generally, we would calculate the value of the rate constant, k. using k = .192 _ 0593 = 000335 5—1 Now ln(% unreacted) : -10 = 0.00385 s‘*x(900s) = —3.46§ rm _ 130 s (% unreacted) = 0.0313 % (b) Rate = k[A] = 0.00335 5" x050 M = 0.00193 Mrs 18. (a) We note that the final concentration is one-eighth of the initial concentration. Thus, three half-lives have elapsed, the first to reduce [A] from 0.800 M to 0.400 M. the second to reduce [A] from 0.400 M to 0.200 M, and the third half—life to reduce [A] from 0.200 M to 0.100 M. From this we see 3mm = 54 min, therefore, tug = 18 min. To reduce the concentration even further requires two additional half-lives: the first of these to lower [A] from 0.100 M to 0.050 M and the second of them to reduce [A] from 0.050 M to 0.025 M. These two half-lives equal 2x18 min = 36 rnin . Thus, at 54 min +36 min = 90 min from the start of the reaction, [A] = 0.025 M. (b) We determine the rate constant for this first-order reaction: k _ 0.693 _ 0.693 in, ' 18 rnin Rate = k[A]l = 0.039 min“ x 0.025 M = 9.3 x10"1 Mmun = 0.039 rnin'1 Then we determine the rate: 19. (a) The mass of A has decreased to one fourth of its original value, from 1.60 g to . l 1 . 0.40 g. Since 1 = ~2— x i , we see that two half—lives have elapsed. 2 Thus, 2mm = 38 min, or rm :19 min. . A (b) k : 0.6931“:m = 0 69,3 = 0.036 min"l ln[—-]’— s —kt = —0.036 rnin'l x60 min = #22 19 min [AL A iA—{f e‘” =01; or [A1=[AL e‘“ =l.60gAXO.ll:0.l§gA A 20. (a) in i 1 = —kr : 1n 0'6” M = 4.256 k = —_0—'25§—= 0.0160 min" [A10 0.816 M 16.0 min {b} 1m 2 fl — mm = 43.3 min I: _ 0.0160 min—l 454 Chapter 14: Chemical Kinetics (c) We need to solve the integrated rate equation to find the elapsed time. A mu=uhfln023SM =—1.245=u0,0160min-Ixt t_ —l.24S [AL 0.816 M _W = (d) In [A] =-—kt becomes [A] =e'“ which in turn becomes [Aio [Ale [A]: [n]D e‘" = 0.816 M e:t[:u[—0.0160min_I x25 bx 60’1"“ 1h = 0.816x0.0907 = 0.074 M 21. We determine the value of the first-order rate constant and from that we can calculate the half-life. If the reactant is 99% decomposed in 137 min, then only 1% (0.010) of the initial concentration remains. A [ ]‘ = —kr = In 0010 = --4.61 = —k x137min k = 4-6]: = 0.0336 min—1 [A]o 1.000 137 mm 0.0693 _ 0.693 In t = 20.6 min 22. If 99% of the radioactivity of 32F is lost, 1% (0.010) of that radioactivity remains. First we compute the value of the rate constant from the half-life. k = 0:593 = = 0.0435 11" [1‘2 ' Then we use the integrated rate equation to determine the elapsed time. A A ln[ 1‘ = —kr t = —lln[ 1‘ = —m———L———_!—ln (1010 = 95 days [A]o 1: [AL 0.0435 d 1.000 35 W g; (a) In [A] = ln(0.35) = —kr = (—4.31 x 10“3 min“): t: 218 min. 0 Note: We did not need to know the initial concentration of acetoacetic acid to answer the question. (b) Let’s assume that the reaction takes placed in a 1.00 L container. 1 mo! acetoacetic acid 10.0 g acetoacetic acid X = 0.09795 mol acetoacetic acid. 102.090 g aoetoacetic acid After 575 min. (-— 4 half lives, hence, we expect ~ 6.25% remains), use integrated form of the rate law to find [AL = 575 min. ln[ :2} J = —kr = (4.81 x 10‘3 mm“‘)(575 min) = —2.76§ [A]! : e'm‘j = 0.06293 (~ 6.3% remains) —[fl~—— [A]. 0.09795M 455 ' = 0.063 [A]l : 6.2 XIO'3 moles. Lilith" an...“ In Chapter 14: Chemical Kinetics (a) (b) (a) (b) (c) [Mmcled = [A]o — [A][ = (0.098 — 6.2 x 10'3) moles = 0.092 moles acetoacetic acid. The stoichiometry is such that for every mole of acetoacetic acid consumed. one mole of C03 forms. Hence, we need to determine the volume of 0.0913 moles CO; at 24.5 °C (297.6; K) and 748 ton- (0.984 atm). Latm RT 0091311101 0.08206 K 1 297.65 K v = " -_- "1° = 2.3 L C02 P 0.934 atrn A lngz— [:11] g =—3.47=“6.2X10_43-II, IZflrj= S [A]0 80.0 g 6.2 x10 5 We substituted masses for concentrations, because the same substance (with the same molar mass) is present initially at time t, and because it is a closed system. 1 mol N205 1 mol o2 0 t0 =77.5 N o x————x———=0.359 r0010 am "n 2 g 2 5 103.0 g N205 2 mol N205 2 RT 0.359 mo] 02 x0_08206 L x(45+273) K v: " : n“: m =956LO2 P 745 mmHgX——— 760 1111an If the reaction is first-order, we will obtain the same value of the rate constant from several sets of data. A ln[ 1 = —t: = mm: —k><100 s =—0.1ss, k = 0‘188 =1.33x10—3 5" [AL 0.600 M 100 s A lug: —k1 = In 0344 M = - x300 s=—0.556, k = 0556 :1.85><10'3 5'1 [AL 0.600 M 300 s A ln[ 1 = —k1 = In 0285 M = —kx400 s = 41744, I: = 0344 =1.36><10‘3 s“1 [AL 0.600M 400 s A ln£—= —k: = ln 0'198 M = w1005.00 s = —1.109, I: = L109 =1.85x10‘3 5" [AL 0.600 M 600 s The virtual constancy of the rate constant throughout the time of the reaction confirms that the reaction is first-order. For this part, we assume that the rate constant equals the average of the values obtained in part (a). k =1.88+l.85:l.8.6+1.85><10_3 54:1.86x104 3-] We use the integrated first-order rate equation: [A1750 = [ ALepoz): 0.600 M exp(—1.86x10'3 5“ X750 3) .[ ALSO = 0.600 M a” = 0.143 M 456 \ Chapter 14: Chemical Kinetics (CH3)EO(g) -> CHAg) + H2(g) + C0(g) Initial 312 mmI-Ig 0 l'l'll'an 0 mmI-Ig 0 mnng Changes —109 mmHg +109 rang +109 mmHg +l09 mmHg Final 203 mmHg 109 mmHg 109 1711an 109 mmHg Plow] : PDME + Pmlham +Rlydmgen +Pco = 203 mmHg+109 mmHg-t-IOQ I‘ang +109 mlan = 530. mmHg Reactions of Various Orders E (a) Set 11 is data from a zero-order reaction. We know this because the rate of set II is constant. 0.25 W25 5 = 0.010 M s‘l . Zero—order reactions have constant rates of reaction. (1)) A firstworder reaction has a constant half-life. In set I, the first half-life is slightly less than 75 sec, since the concentration decreases by slightly more than half (from 1.00 M to 0.47 M) in 75 5. Again, from 75 s to 150 s the concentration decreases from 0.47 M to 0.22 M, again by slightly more than half, in a time of 75 3. Finally, two halfilives should see the concentration decrease to one-fourth of its initial value. This, in fact, is what we see. From 100 s to 250 sec, 150 s of elapsed time, the concentration decreases from 0.37 M to 0.08 M. i.e.. to slightly less than one-fourth of its initial value- Notice that we cannot make the same statement of constancy of half-life for set III. The first half-life is 100 s. but it takes more than 150 s (from 100 s to 250 s) for [A] to again decrease by half. (e) For a second-order reaction, 1! [A]l — I! [A]0 = kt . For the initial 100 s in set 111, we have l — l =1.0Lmol“=k100s, lc=0.010Lmol"s'1 0.50 M 1.00 M For the initial 200 s, we have I - 1 = 2.0 L moi" = k 200 s, k = 0.010 L mol" 3'1 0.33 M 1.00 M Since we obtain the same value of the rate constant using the equation for second- order kinetics. set [11 must be second—order. 28. For a zero-order reaction (set II), the slope equals the rate constant: k = —a[A]lot=1.00 W100 s =0.0100 W3 29. Set I is the data for a first-order reaction; we can analyze those items of data to determine the half-life. In the first 75 s, the concentration decreases by a bit more than half. This implies a half- life slightly less than 75 s, perhaps 70 s. This is consistent with the other time periods noted in the answer to Review Question 18 (b) and also to the fact that in the 150 s interval from 50 s to 200 s, the concentration decreases from 0.61 M to 0.14 M, which is a bit more than a factor-of-four decrease. The factor—of-four decrease. to one-fourth of the initial value, is what we would expect for two successive half-lives. We can determine the half-life more accurately, by obtaining a value of it from the relation ln([A1f[AL): —kt followed by rm = 0.693; k For instance, ln(0.78l1.00)= —k (25 s); k = 9.93 x103 s'l. Thus. to; = 0.693l9.9gl_ x103 5" = 70 s. 458 Chapter 14: Chemical Kinetics 30. We can determine an approximate initial rate by using data from the first 25 s. A A _ Rate = -—£-I—] = —w = 0.0080 M s“ 25 5—05 3. The approximate rate at 75 s can be taken as the rate over the time period from 50 s to 100 s. A[A] _ _ 0.00 M — 0.50 M Rt =— =0.010M ‘1 (a) 36" At 100 5—503 5 AA . fl . _ (b) Ratel =__[_]=_w=0_0043 M 5‘ AI IOOs-SOs A -- . (.3) Rate!“ = -9U = _w = 00034 M s" A: lOOs-—50s Alternatively we can use {A} at 75 s (the values given in the table) in the relationship Rate = k[A]"' , where m=0 , 1, or 2. (a) Raten = 0.010 M 3" x(0.25 mollL)D = 0.010 M s" (b) Since rm = 705, l: 2 0.693 I 705 '= 0.00993" RateI = 0.0099 s" x (0.47 rnolt'L)I = 0.0047 Ms" (c) Ratem = 0.010 L mol'1 3" ><(0.57 rims/L)2 = 00032 M 5‘1 32. We can combine the approximate rates from Question 31, with the fact that 10 s have elapsed, and the concentration at 100 s. (3) [AL 2 0.00 M There is no reactant left after 100 s. (b) [a]l = [A]m —(10 sx rate) = 0.37 M —(10 sx0.0047 M s“) = 0.32 M (c) [a]In = [ALm ~(10 sx rate): 0.50 M —(10 sx0.0032 M s") = 0.47 M 33. Substitute the given values into the rate equation to obtain the rate of reaction. Rate = k[A]2[B]" = (0.0103 M'lmin")(0.116 M)” (3.33 M)D = 1.39 x 10“ Mi min 34. (a) A first-order reaction has a constant half-life. Thus, half of the initial concentration remains after 30.0 minutes. and at the end of another half-life—60.0 minutes total— half of the concentration present at 30.0 minutes will have reacted: the concentration has decreased to one-quarter of its initial value. Or, we could say that the reaction is 75% complete after two half-lives—60.0 minutes. {h\ A 79rn_nrdar Martini-u nmnaar‘n nt- 4-- nnnnan... —nl— 411...... :c a... _.-..-..: .. ,. 1. :nm _ _ _-_,1 _-_ Chapter l4: Chemical Kinetics (d) We use the same equation as in part (b), but solve for: , rather than k. 1 1 1 1 1 1 . t: _ fi_ = M— = 219 min I: [111 [A]o 0.137 Lind—{mind 00250 M 0.100 M (e) We use the same equation as in part (b), but solve for: , rather than k. 1 1 1 1 1 1 . I:__ ___M =_M_l ‘ _! _—#——— =136m1n k [A [A 013?me min 0.0350M 0.100M 37. (a) In the first 22 s, [A] decreases from 0.715 M to 0.605 M. that is, A[A] = —0.1 10 M. The average rate over these 22 s is then determined as follows. —A A . Rate: [ 1=M=5.0X10'3Ms Ar 22 s In the first 74 s, A[A] = 0.345 M ~0.'115 M, and the rate is determined. #A A . Rate: [ 1: W=50x104MIs T4 s Finally, in the first 132 s, A[A] = 0.055 M —0.715M. and the rate is determined as —A A follows. Rate = if ] = M = 5.0x10'3Mt's. Since the rate is constant for this 3 reaction, it must be zero-order in [A]. (b) The half-life of this reaction is the time needed for one half of the initial [A] to react. Thus, MA]: 0715 M+2=0.358M and 1m =—°'—35§—M——=72 s. 5.0)<10_3 MIS 38. (a) We can either graph ll[C “1116] vs. time and obtain a straight line, or we can determine the second-order rate constant from several data points. Then, if k indeed is a constant, the reaction is demonstrated to be second-order. We shall use the second technique in this case. First we do a bit of algebra. 1 1 _, les 1 = FETAL—k I[[A], [AL] k h 1 1 '12.13 min [0.0144 M _ 0.0169 M = 0.843 L mol'lmjn" k 1 1 k=—~—_— “Law—i— =0.a'151.1111;114:1111:-1 2455 mm 0.0124 M 0.0169 M l k = ——-—] = 0.892 L mol"min"' 42.50 rmn 0.0103 M 0.0169 M l k = = 0.870 1.. moi—Imm—l 68.05 nun 0.00845 M 0.0169 M The fact that each calculation generates similar values for the rate constant indicates that the reaction is second-order. Chapter 14: Chemical Kinetics _. The half-life of the reaction depends on the concentration of “A” and, thus, this reaction cannot be first—order. For a second—order reaction, the half‘life varies inversely with the reaction rate: rm = lt[k[ AL) ork = If (rl,2[ AL). Let us attempt to verify the second- order nature of this reaction by seeing if the rate constant is fixed. k = ———-1——— = 0.020 L mol'lmin" 1.00 MXSO min I: = = 0...
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