2. For a one-point fixed time method, a pseudo-first order reaction obeys the equation 0.309

Transcription

1 Chaper 3. To derive an appropriae equaion we firs noe he following general relaionship beween he concenraion of A a ime, [A], he iniial concenraion of A, [A], and he concenraion of P a ime, [P] [ P] Subsiuing his relaionship ino equaion 3.8 for imes and 2, gives he desired resul 2 k l k l 2 e e ^ [ P] h^ [ P] 2h k l k l 2 e e [ P] + [ P] k l k l 2 e e 2 [ P] [ P] k l k l e e For a onepoin fixed ime mehod, a pseudofirs order reacion obeys he equaion k e K where A is phenylaceae and K is equal o e k. Using he sandard, we find ha K is 7. mm K(. 55 mm) K.7 mm.55 mm.39 Thus, for he sample, we have [phenylaceae].23 mm mm k You can, of course, use he equaion e and he resul for he sandard o calculae he rae consan, k, and hen use he same equaion and he resul for he sample o calculae he concenraion of phenylaceae. The rae consan has a value of.96 s. 3. Because we are following he change in concenraion for a produc, he kineics follow equaion 3.5 [2] [H2O] 2 k l e which we rearrange o solve for he produc s concenraion k [ 2] [H2O2] ^ e l h From Beer s law, we know ha he absorbance, A, is Chaper 3 Kineic Mehods 27

2 28 Soluions Manual for Analyical Chemisry 2. absorbance [H 2 O 2 ] (µm) Figure SM3. Calibraion daa (blue dos) and calibraion curve (blue line) for he daa in Problem 3. ( ) (s ) [glucose] (ppm) Figure SM3.2 Calibraion daa (blue dos) and calibraion curve (blue line) for he daa in Problem 5. A fb[ 2] Subsiuing his equaion back ino he previous equaion gives k l A ( fb) [ HO 2 2] ^ e h K[ HO 2 2] k l where K is equal o ( fb) ^ e h. Using he daa for he exernal sandards gives he calibraion curve shown in Figure SM3., he equaion for which is 3 A µm C H2O2 Subsiuing in he sample s absorbance of.669 gives he concenraion of H 2 O 2 as 286 µm. 4. For a wopoin fixedime mehod, we use equaion 3.8 [H2CrO 4] [H2CrO 4] 2 [H2CrO 4] k l k l 2 e e From Beer s law, we know ha A fb[ HCrO 2 4] A fb[ HCrO 2 4] 2 2 Solving hese wo equaions for he concenraion of chromic acid a imes and 2, and subsiuing back gives A A2 b b ( b) A A [H CrO ] f f f ^ 2h 2 4 k l k k k K A A l 2 l l 2 ^ 2h e e e e Using he daa for he exernal sandard, we find ha 4 [H2CrO 4] K 5. M 3 A A M ^ 2h The concenraion of chromic acid in he sample, herefore, is [H2CrO 4] K^A A2h M( ) 62. M 5. For a variable ime kineic mehod, here is an inverse relaionship beween he elapsed ime, D, and he concenraion of glucose. Figure SM3.2 shows he resuling calibraion daa and calibraion curve, for which he equaion is ( D ) 6.3 s +.5 s ppm 4 3 C glucose where D for each sandard is he average of he hree measuremens. Subsiuing in he sample s D of 34.6 s, or a (D) of.289 s, gives he concenraion of glucose as 9.7 ppm for he sample. The relaive error in he analysis is 9.7 ppm 2. ppm 2. ppm.5% error

3 Chaper 3 Kineic Mehods Subsiuing he sample s rae of µmol ml s ino he calibraion equaion gives he volume as µmol ml s 27. µmol ml s V µmol ml s V 95mL. This is he volume of he sandard enzyme ha has he same amoun of enzyme as is in he. ml sample; hus, he concenraion of enzyme in he sample is approximaely 5 more dilue han he concenraion of enzyme in he sandard. 7. For a firsorder reacion, a plo of [A] versus ime gives a sraigh line wih a slope equal o k and a yinercep equal o [A]. Figure SM3.3 shows he daa and he regression line, for which he equaion is. 469 ( s ) [A] ime (s) Figure SM3.3 Linearizaion of he daa from Problem 6 for a reacion ha is pseudofirs order in he analye. From he slope, we know ha he reacion s rae consan is.486 s. Using he yinercep, we know ha [A] is.469, which makes he iniial concenraion of A equal o.5 mm. 8. Under hese condiions a concenraion of aceylcholine ha is significanly smaller han he consan, K m we can wrie he MichaelisMenon equaion as k2[ E][] S R where [E] is he concenraion of enzyme and [S] is he concenraion of he subsrae aceylcholine; subsiuing in known values Ms (. 4 s )(6.6 M) 5 9 M and solving gives he concenraion of aceylcholine as.2 7 M. 9. Under hese condiions a concenraion of fumarae ha is significanly greaer han he consan, K m we can wrie he MichaelisMenon equaion as R k2[ E] where [E] is he concenraion of enzyme. Using he rae and concenraion of enzyme for he sandard, he value of k 2 is k [ E R 2. µm min 2 ]. 5 µm min Using his value for k 2 and he rae for he sample, we find ha he enzyme s concenraion in he sample is. [ E]. k R 5µMmin 863 µm min

4 22 Soluions Manual for Analyical Chemisry 2. /rae (µm s) /[urea] (µm) Figure SM3.4 LineweaverBurk plo of he daa from Problem. The blue dos are he reciprocals of he concenraion and rae daa provided in he problem, and he blue line is he resul of a regression analysis on he daa.. Figure SM3.4 shows a LineweaverBurk plo of /rae as a funcion of /[urea], for which a regression analysis gives an equaion of 6 (. 6 s) rae µm s + Curea From he yinercep we exrac he value for he maximum rae; hus V max y inercep µm s µm s or.46 M/s. From he slope, we deermine he value for K m, finding ha i is ( slope) Vmax 5 (. 6 s)(4.58 µm s ) 649 µm or M. Finally, we know ha Vmax k2[ E], which we use o calculae he value for k µ Ms 4 k2 5. µ M 8. s. f V max remains consan, hen he yinercep of a LineweaverBurk plo is independen of he inhibior s concenraion. f he value of K m increases and he value of V max remains consan for higher concenraions of he inhibior, hen he slope of a LineweaverBurk plo, which is equal o K m /V max, mus increase for higher concenraions of he inhibior. Figure 3.4 shows ha boh are consisen wih compeiive inhibiion. 2. For compeiive inhibiion, he iniial concenraion of enzyme is divided beween free enzyme, E, enzyme complexed wih he subsrae, ES, and enzyme complexed wih he inhibior, E; hus, a mass balance on he enzyme requires ha [ E] [ E] + [ ES] + [ E] f we assume ha k 2 is much smaller han k, hen we can simplify he equaion for K m o [ ][] K k k2 k E S m +. KES k k [ ES] where K ES is he equilibrium dissociaion consan for he enzymesubsrae complex. We also can wrie he equilibrium dissociaion consan for he enzymeinhibior complex, which is [ E][] KE [ E] Solving K m and K E for he concenraions of E and of E, respecively

5 Chaper 3 Kineic Mehods 22 [ ES] [ E][] [ E] [ E] KE and subsiuing back ino he mass balance equaion gives [ ES] [ E][] [ E] + [ ES] + KE [ ES] [ ES][] [ E] + [ ES] + KE Facoring ou [ES] from he righ side of he equaion [] [ E] [ ES] ' + + KE and hen solving for [ES] gives [ E] [ ES] [] ' + + KE Finally, he rae of he reacion, d[p]/d, is equal o k 2 [ES], or dp [ ] k2[ E] k2[ ES] d [] ' + + KE dp [ ] d dp [ ] d k2[ E][] S [] & + + KE K m Vmax[] S [] a + k K + E 3. For firsorder kineics, we know ha [ B] k A [ B] k B To obain. for [A] /[A] and.999 for [B] /[B], he raio of he rae consans mus be k A [ B] k B [ B] (. ) (. 999) k A 69 kb 4. Figure SM3.5 shows a plo of he daa, where we place [C] on he yaxis as he kineics are firsorder. Early in he reacion, he plo is curved because boh A and B are reacing. Because A reacs faser han B, evenually he reacion mixure consiss of B only, and he plo becomes linear. A linear regression analysis of he daa from 36 min o 7 min gives a regression equaion of [C] ime (min) Figure SM3.5 Plo of he daa for Problem 4. The blue dos are he original daa and he blue line is a regression analysis resriced daa from 36 min o 7 min when he reacion of A is complee.

6 222 Soluions Manual for Analyical Chemisry 2. [A] ime (min) Figure SM3.6 Plo of he daa for Problem 4 afer we remove he conribuion from B. The blue dos are he recalculaed daa and he blue line is a regression analysis resriced daa from min o 3 min when he reacion of A is complee. 2 [ C]. [ B] ( min ) The yinercep of 2.82 is equivalen o [B] ; hus, [ B] e. 25 mm The slope of he regression line in Figure SM3.5 gives he rae consan k B, which is approximaely.332 min. To find values for [A] and for k A, we mus correc [C] for he conribuion of B. This is easy o do because we know ha and ha which means ha [ C] + [ B] [ B] [ B] e kb [ C] [ B] e For example, a ime, he concenraion of B is.29 mm and he concenraion of A is.33 mm.29 mm.92 mm. Figure SM3.6 shows a plo of [A] versus ime from min o 3 min. A regression analysis of he daa gives he following equaion kb (. 455 min ) from which he slope gives he value of k A as.46 min and he yinercep of.442 yields he iniial concenraion of A. 442 e. 236 mm 5. For radioacive decay, we know ha 2 /. 693/ m. Using he firs enry in Table 3. as an example, we find ha m 2.5 yr 5.54 yr 2 3 H The decay consans for he isoopes in Table 3. are provided here isoope halflife decay consan 3 H 2.5 yr 5.54 yr 2 4 C 573 yr.2 4 yr 32 P 4.3 d d 35 S 87. d d 45 Ca 52 d d 55 Fe 2.9 yr 2.38 yr 6 Co 5.3 yr.3 yr 3 8 d d

7 Chaper 3 Kineic Mehods Combining equaion 3.33 and equaion 3.37 allows us o calculae he number of aoms of 6 Co in a sample given he sample s aciviy, A, and he halflife for 6 Co N A / N 7 2. aoms s 36 s 24 h 365 d 53. h d yr yr N 56. aoms C The concenraion of 6 Co, herefore, is aoms Co 23 (6.22 aoms/mol)(.5 L) 6.7 M 7. Using he daa for he sandard, we know ha ( A) s 354 cpm 4 k w.593 gni 5.97 cpm/gni s. g g For he sample, herefore, we have ( A) x 2 cpm wx 4.79 gni k 5.97 cpm/gni Finally, he concenraion of Ni in he sample is.79 gni 34.2%w/w Ni.5 gsample 8. Using equaion 3.42, we find ha mass of viamin B 2 in he sample as analyzed is 572 cpm w 8.6 mg.5 mg mg 36 cpm This represens half of he original sample; hus, here are mg of viamin B 2 in he ables, or 5.79 mg/able. 9. For radioacive decay, we know ha A. A m / Subsiuing in /2 from Table 3. and leing 3 yr, gives A A yr 3 yr A A e. 266 The percenage of 4 C remaining, herefore, is 2.66%.

8 224 Soluions Manual for Analyical Chemisry Because we assume ha 4 Ar was no was presen in he original sample, we know ha he iniial moles of 4 K is he sum of he moles of 4 Ar and of 4 K presen when he sample is analyzed; hus 6 4 ( n 4 K) ( 4.63 mol K) (2.9 mol Ar) 6.72 mol Using he equaion for firsorder radioacive decay, we find ha ^n ^ n 6 6 h h 4 K 4 K mol mol k / (. 3725)( yr) 8 7. yr 9 yr 2. The relaionship beween he percen relaive sandard deviaion and he number of couns is ( v ) A rel M where M is he number of couns. To obain a percen relaive sandard deviaion of %, herefore, requires. M 2 M `. j couns To obain couns, we need a sample ha conains. gc couns 2 cpm 3.9 gc 6 min To obain a % relaive sandard deviaion when couning he radioacive decay from a.5 g sample of C, we mus coun for. gc couns 2 cpm.5 gc 333 min. 3 min 22. Sensiiviy in a flowinjecion analysis is direcly proporional o he heigh of an analye s peak in he fiagram, which, in urn, is proporional o he analye s concenraion. As he analye moves from he poin of injecion o he poin of deecion, i undergoes coninuous dispersion, as shown in Figure 3.9. Because dispersion reduces he analye s concenraion a he cener of is flow profile, anyhing ha limis dispersion will increase peak heigh and improve sensiiviy. ncreasing he flow rae or decreasing he lengh and diameer of he manifold allows less ime for dispersion, which improves sensiiviy. njecing a larger volume of sample means i will ake more ime for he analye s concenraion o decrease a he cener of is flow profile,

9 Chaper 3 Kineic Mehods 225 pump loop injecor DPKH NaOH HCl channel juncion mixing/reacion coil mixing/reacion coil mixing/reacion coil which also improves sensiiviy. Finally, injecing he analye ino a channel resuls in is diluion and a loss of sensiiviy. f we merge his channel wih anoher channel, hen we dilue furher he analye; whenever possible, we wan o dilue he analye jus once, when we injec i ino he manifold. 23. Depending on your measuremens, your answers may vary slighly from hose given here: he ravel ime, a, is 4. s; he residence ime, T, is 5.8 s; he baselineobaseline ime, D, is 5.2 s; he reurn ime, T l, is 3.5 s; and he difference beween he residence ime and he ravel ime, l, is.7 s. The peak heigh is.762 absorbance unis; hus, he sensiiviy is k C A.762. ppm ppm We can make injecions a a rae of one per uni reurn ime, which for his sysem is every 3.5 s; hus, in one hour we can analyze hr 36 s sample hr 3.5 s o 27 samples/hr 24. Figure SM3.7 shows one possible manifold. Separae reagen channels of DPKH and NaOH are merged ogeher and mixed, and he sample injeced ino heir combined channel. Afer allowing sufficien ime for he reacion o occur, he carrier sream is merged wih a reagen channel ha conains HCl, he concenraion of which is sufficien o neuralize he NaOH and o make he carrier sream acidic. 25. Figure SM3.8 shows he calibraion daa and he calibraion curve, he equaion for which is A ( ppm ) C deecor Subsiuing in he sample s absorbance of.37 gives he concenraion of Cl as ppm in he sample as analyzed, which means he concenraion of Cl in he original sample of seawaer is ppmcl 5. ml. ml 37 ppm Cl ppm wase Figure SM3.7 One possible FA manifold for he analysis described in Problem 24. absorbance [Cl ] (ppm) Figure SM3.8 Calibraion daa (blue dos) and calibraion curve (blue line) for he daa in Problem 25.

10 226 Soluions Manual for Analyical Chemisry 2. (s) Figure SM3.9 Calibraion daa (blue dos) and calibraion curve (blue line) for he daa in Problem 26. signal (arbirary unis) [cocaine] (µm) Figure SM3. Calibraion daa (blue dos) and calibraion curve (blue line) for he daa in Problem 28. volume (ml) log[hcl] [H 2 SO 4 ] (mm ) Figure SM3. Calibraion daa (blue dos) and calibraion curve (blue line) for he daa in Problem Figure SM3.9 shows he calibraion daa and he calibraion curve, which for an FA iraion is a plo of D as a funcion of log[hcl]. The equaion for his calibraion curve is D s + (4.33 s) log[hcl] The average D for he five rials is s. Subsiuing his back ino he calibraion equaion gives log[ HCl] s s s.5 [HCl].76 M 27. Using he daa for he single exernal sandard, we know ha k C S 7.3 na.29 na mm glucose 6.93 mm Using his value for k, he concenraion of glucose in he sample is C.5 nm glucose.2 mm.29 na mm 28. (a) The mean and he sandard deviaion for he 2 replicae samples are and.65, respecively. The relaive sandard deviaion, herefore, is % (b) Figure SM3. shows he calibraion daa and he calibraion curve, for which he equaion is S C cocaine Subsiuing he sample s signal of 2.4 ino he calibraion equaion gives he concenraion of cocaine as µm as analyzed. The concenraion of cocaine in he original sample, herefore, is Z 6 _ mol L 25. ml.25 ml ] b [ g mg ` ].25 L mol g b \ a. mg 94.% 8 w/w 29. Figure SM3. shows he calibraion daa and he calibraion curve, for which he equaion is mm ml V ml + C H2SO4 Subsiuing in a volume of.57 ml for he sample, gives he concenraion of H 2 SO 4 as.627 mm.