/9/03 This exercise explores the time depedece of the reactio of the dye: Kietics of Dye Bleachig Objective: To examie the kietics of a reactio i which a dye is bleached Last Update: /9/03 8:5 AM Malachite Gree, MG + H 3 C N CH 3 C Gree CH + 3 N Cl - CH 3 with OH - H 3 C N H 3 C To form the correspodig alcohol, MGOH which is colorless. OH N H 3 C CH 3 Colorless We write the reactio as k geerally MG + + OH - MGOH depeds o temperature but ot o cocetratios Rate law for the reactio is d[ MG + ] rate = = k [ OH - ] [ MG + ] m dt The rate of disappearace of MG + k = rate costat of the reactio m= order of the reactio with respect to MG + = order of the reactio with respect to OH - Our objective is to determie k, ad m 3 Sice the reactat, MG + is colored but either OH - or the product, MGOH, is (i.e., OH - bleaches the dye) We ca follow the cocetratio of MG + as the reactio proceeds usig the Absorbace of light ad Beer s Law A = s c Where, as usual, s will be the slope of the Beer s Law plot for the dye. (s = ε d) 4 d[ MG + ] rate = - = k [ OH - ] [ MG + ] m dt If [ OH - ] did t chage, the rate law would simplify to oe ivolvig oly MG oe + chagig + OH - cocetratio, MGOH [MG + ] [OH - ], [MG + ] d[ MG + ] ad [MGOH] rate = - = q [ MG + ] m all chage with ih time dt where q = k [ OH - ] We ca accomplish this experimetally by usig a large (excess), but, kow [OH - ] The, [ OH - ] will ot chage sigificatly durig the course of the reactio 5 To keep thigs simple, i this exercise, we limit the possible values of both m ad to either or For these two cases, calculus provides a expressio for the way i which [ MG + ] will vary with time, amely: m= -d[ MG + ] / dt = q [ MG + ] m= - d[ MG + ] / dt = q [ MG + ] Where [MG + ] 0 is the [ MG + ] at t = 0. [ MG + ] = [MG + ] 0 e -qt = ----- + qt [ MG + ] [MG + ] 0 6
/9/03 If we measure [ MG + ] at a sequece of times, we should be able to tell the value of m by plottig the cocetratio vs time ad seeig which of the above two fuctios it follows. But, we ca produce liear plots of [ MG + ] vs time for m = y = a + bt [ MG + ]= [ MG + ] 0 e -qt [MG + ] Cocetratio m= m = It will be very difficult to tell these two curves apart! takig the atural logarithm of each side produces: l [ MG + ] = l [ MG + ] 0 - q t Liear plots vs for m = time We have with slopes q or -q = + q t [ MG + ] [ MG + ] 0 Time, t 7 8 If the plot of -l [ MG + ] is liear with time, the reactio is first order with respect to MG + i.e., m = If the plot of ( / [ MG + ] ) is liear with time, the reactio is secod order with respect to MG + i.e., m = -l[mg + ] / [MG + ] slope = q slope = q d[ MG This is how + ] we determie m rate = - = k [ OH - ] [ MG + ] I either case, the dtslope of the lie is q = k [ OH - ] t t 9 For m = But we actually measure absorbace, ot cocetratio From Beer s law: Abs = s [MG + ] where s = εd So [MG + ] = Abs/s ad [MG + ] 0 = Abs 0 /s l [ MG + ] = l [ MG + ] 0 - q t l (Abs/s) = l (Abs 0 /s) q t l (Abs)-l (s) = l (Abs 0 ) l (s) q t l (Abs) = l (Abs 0 )- q t For this case, l (s) cacels out. We ca just plot l(abs) verses time to fid q (the slope). Is the same true for m =? 0 For m =: = + q t [ MG + ] [ MG + ] 0 Substitute i [ MG + ] = Abs / s ad divide each term by s = + (q /s) t Abs Abs 0 I.e., the slope of the secod order plot is q / s rather tha just q ( =k [ OH - ] )! Havig determied m, how do we determie, the order of the reactio with respect to OH -? Study reactio at two differet, kow, values of [ OH - ], c ad c, isurig that they are both i large excess over the cocetratio of MG +. This gives us two slopes q ad q, whose differet values deped o the two OH - cocetratios by q = k m = [ OH - ] q = k [ OH - ] / e = k c / e = = q = k [ OH - ] / e = k c / e log q / q = c / c = og (c / c )
/9/03 m = This gives us two slopes q /s ad q /s, whose differet values deped o the two OH - cocetratios by q = k [ OH - ] q/s = k [ OH - ] /s q /s = k [ OH - ] / s = k c / s = = q /s = k [ OH - ] / s = k c / s log q / q = c / c = og (c / c ) 3 q ad q are kow from the slopes of our graphs log ( q / q ) = log ( c / c ) log ( q / q ) = log ( c / c ) c ad c from the way we made up the solutios. Whe we solve = log (q /q ) / log (c /c ) for, we will cosider oly the iteger solutios: 0, or This is how we determie 4 Havig determied both m ad, we ca ow go back ad calculate k, the specific rate. Agai, c ad c are kow from the way we made up the solutios. If m = q = k [ OH - ] = k c k = q /c q = k [ OH - ] = k c k = q /c If m = q /s = k [ OH - ] / s = kc /s k = s (q /s) / c q /s = k [ OH - ] / s = kc /s k = s (q /s) / c where s is the Beer s Law costat for the dye I either case, report the average of the two values PROCEDURE WORK IN PAIRS.) Prepare appropriate dilutios of MG + ad NaOH stock solutios.) Permit solutios to come to temperature of water bath Do ot measure absorbace of the dye solutio (procedure 3b) You will be give the Beer s Law slope i case you eed it! Blak should be NaOH use water istead 3.) Start timer, combie, ad mix solutios ALL AT THE SAME TIME (t=0) This is how we determie k 5 Do t stop the timer! 6 WAVELENGTH ADJUSTMENT: sets wavelegth (color) ZERO ADJUSTMENT: sets 0% trasmittace calibrates detector Needs oly Periodic checkig after warm-up LIGHT CONTROL: sets 00 % trasmittace Needs readjustmet before every measuremet of a sample. (Usig BLANK) 7 8 3
/9/03 PROCEDURE (cot d) 4.) Make a absorbace measuremet as soo as possible ad the, agai at appropriate time itervals as istructed (~60 sec, ~0 sec), Record absorbace & total elapsed time Do t stop the timer! 67 m 5.) Plot data o lab computer ad decide o m, the order of the reactio with respect to MG + Largest RSQ 6.) From the slope of the plot, determie the apparet specfic rate costat q This slope icludes the term [ OH - ] i.e., q = k [ OH - ] 9 Slope = q/s if m = 7.) Repeat Steps -6 usig differet [ OH - ] 8.) Plot ew data appropriately vs time, Use the same fuctio (i.e., l or /[MG + ] ) that produced liear plot i Ru 9.) From the sets of data, determie, ad the specific rate costat k Note: The cuvettes may become slightly staied by the dye. If so, they ca be cleaed by risig with -3 ml of 95% ethaol 0 Modificatios to the procedure: You will NOT do part 3b Determiatio of the Beer s Law costat for Malachite Gree. Should you eed it, the Beer s Law costat is 4.03 X 0 4 L/mol. It will be helpful to have a watch with a secod had for this exercise. You will ot do part 8 Determiatio of the rate costat for the bleachig of Crystal Violet. The Excel Workbook has two sheets scaled properly for the time scales of the two rus Iput OH - cocetratio i reactio mix [OH - ] =.00E-07 Scheider Time(sec) Abs -l(abs) /Abs Akhtar Iput Names 33.340-0.9 7.46E-0 00 sec itervals 00 sec itervals 04.60-0.5 8.6E-0 Iput: elapsed Output: time 80 ad 0.960 0.04.04E+00 Slope, Itercept, & Graphs absorbace RSQ 48 0.80 0.0.E+00 350 0.636 0.45.57E+00 LOG PLOT RECIPROCAL PLOT 450 0.50 0.69.99E+00 slope.386e-03 slope 4.36E-03 575 0.380 0.97.63E+00 itercept -3.867E-0 itercept 3.033E-0 7 0.60.35 3.85E+00 Rsq 0.9996 Rsq 0.944 3 4 4
/9/03 Time (sec) Abs -l(abs) /Abs 0 0.500 0.693.000 75 0.450 0.799. 377 0.400 0.96.500 55 0.350.050.857 733 0.300.04 3.333 888 0.50.386 4.000 070 0.00.609 5.000 303 0.50.897 6.667 Spreadsheets require exactly 8 etries I order of icreasig time LOG PLOT RECIPROCAL PLOT slope 9.90E-04 slope 3.408E-03 itercept 6.090E-0 itercept.45e+00 Rsq 0.97 Rsq 0.8875 The relevat plot is the oe that produces the largest value of R Note that times are the total elapsed secods 5 6 You must check the lab bulleti boards for ay chages i the exercise before proceedig. Do ot assume that ay cocetratios of reagets are as described i the lab maual. 7 A example - the logic ad calculatios - Bleachig of Cogo Red (CR) Ru: 0 ml of CR stock solutio (.9x0-4 M) diluted to 00 ml 0 ml of NaOH stock solutio (.x0 - M) diluted to 00 ml Let solutios come to bath temperaturet at t=0, two above solutios are mixed (V=00 ml) CR Absorbace is measured at 90 sec itervals (Beer s Law costat =. x 0 5 ) [CR] 0 = (0 ml)(.9x0-4 mmol/ml)/ 00 ml =.9x0-5 M [OH - ] = (0 ml)(.x0 - mmol/ml)/ 00 ml =.x0 - M 8 Kietics Data Time (sec) Abs -L(Abs) / Abs Scheider Akhtar 8 0.500 0.693.000 9 0.450 0.799. Cogo Red - 90 sec 78 0.400 0.96.500 [OH-] =. X 0-80 0.350.050.857 359 0.300 04.04 3.333333 445 0.50.386 4.000 LOG PLOT RECIPROCAL 549 0.00.609 5.000 slope 7.90 E-04 slope 7.30E-0 640 0.50.897 6.667 itrcpt 6.090 E-0 itrcpt.45e+00 Rsq 0.833 Rsq 0.96 9 R = 0.833 Slope =.x0-3 s - R = 0.96 Slope = 7.3x0 - mol/ls Coclusio: rate is secod order i CR 30 5
/9/03 Ru: 0 ml of CR stock solutio diluted to 00 ml 0 ml of NaOH stock solutio diluted to 00 ml at t=0, two above solutios are mixed CR Absorbace is measured at 60 sec itervals c =. X 0 - M c = [ OH - ] = (0mL)(.x0 - mmol/ml )/ 00 ml =.x0 - M Ru 0 ml Ru 90 sec Log Plot Slope = 30.6 X 0 - s - Reciprocal Plot Slope = 30.6x0 - mol/ls 3 3 Ru: 0 ml of CR stock solutio diluted to 00 ml 0 ml of NaOH stock solutio diluted to 00 ml at t=0, two above solutios are mixed CR Absorbace is measured at 60 sec itervals Reciprocal plot of CR absorbace gives q /s = 30.6x0 - mol/ls q /s = 7.3x0 - mol/ls log ( q / q ) = --------------------- log ( c / c ) c =. X 0 - M c =. X 0 - M log ( 7.3 X 0 - / 30.6 X 0 - ) = --------------------------------------- =.06 log (. X 0 - /. X 0 - ) 33 k = (q /s) s / c = 7.3 X 0 - X. X 0 5 / (. X 0 - ) = 7. X 0 7 L 3 / mol 3 -sec From st ru k = (q /s) s / c = 30.6 X 0 - X X0. 0 5 /(X0 (. - ) = 7.6 X 0 7 L 3 / mol 3 -sec Our rate law ca ow be writte as: From d ru rate = 7.4 X 0 7 [ CR ] [ OH - ] M / sec 34 FOR NEXT WEEK SYNTHESIS OF ASPIRIN READ SUSB - 08 DO PRE-LAB You will eed your 5 ml Erlemeyer to be clea ad dry for the aspiri sythesis. You might wish to wash it at the ed of the kietics exercise. 35 6