ENV 6015 Solutions to Kinetics Problem Set. Sucrose + H 2 O Glucose

Transcription

1 ENV 15 Slutins t inetics Prblem Set 1. The inversin f sucrse prceeded as shwn at 5 : The initial cncentratin f sucrse was 1. mles/liter. alculate the first rder rate cnstant and the half-life f the reactin. Why des this reactin fllw a first rder law despite the fact water enters int the stichimetric equatin? Hw lng wuld it take t invert 95% f a pund f sugar? 1 H O 11 + H O 6 H 1 O H 1 O 6, min Sucrse inverted-m/l Slutin: The slutin is diluted, [H O] 55.5 M/L vs [ 1 H 1 O 11 ] 1. M/L. The reactin is limited by the cncentratin f 1 H 1 O 11 and fllws first-rder kinetics. Sucrse + H O Glucse d dt k * after integratin: i k * t r -k*t + i Thus, if we plt [ ] vs time, the resulting slpe is equal t the rate cnstant k (min -1 )., min Sucrse inverted- M/L Sucrse remaining M/L Ln [ ]

2 Plt f vs time. (sucrse remaining) *time -.5 R (min) Frm the equatin: k.35 min -1 t half-life f the reactin: i.5* i i.35* t 1 t (.5) min T invert 95% f a pund f sugar:.5* i i.35* t 1 (.5) t 95% 856 min.35

3 . Find the reactin rder and rate cnstant fr the decmpsitin f diazbenzenechlride at 5 if the initial diazbenzenechlride cncentratin is 1 grams/liter. The reactin and nitrgen evlutin are shwn. Slutin: 6 H 5 N l 6 H 5 l + N, min N cm 3 evlved Fr t, V N 58.3 cm 3 ; using the ideal gas law: P * VN R *T 3 ( 1atm)( 58.3 cm ) 3 atm *cm ( 33 ) n N. mles 8 ml* by stichimetry,. mles f 6 H 5 N l prduce. mles f N. ssuming cmplete reactin: Initial mass f 6 H 5 N l (.mles)*(1 g/mle).39 grams Vlume f slutin 1 g L.39 g 1L * 1 cm cm 3 T determine the rder f the reactin and the rate cnstant, we need t graph: a) vs time fr zer-rder reactin b) Ln vs time fr first-rder reactin c) 1/ vs time fr secnd-rder reactin Nte: the graphs can als be made by using mass r mles f 6 H 5 N l in lieu f cncentratin (min) N cm 3 evlved N mles evlved (M/L f 6 H 5 N l remaining) 1/ (L/M)

4 a) Plt f vs time.8.7 ncentratin (M/L) *time +.55 R (min) b) Plt f vs time *time -.79 R (min) c) Plt f 1/ vs time 1 1 1/ncentratin (L/M) 8 1/.55*time R (min)

5 Frm the graphs we can see that the reactin fllws first-rder r secnd-rder kinetics. Firstrder kinetic is chsen because the data pints shw a nrmal distributin arund the trendline. The secnd-rder plt is nt apprpriate despite the higher R (.97) because the data pints shw a curvature. Frm the trendline equatin shwn in the first-rder graph (graph b), the rate cnstant is:.55 min -1 5

6 3. certain reactin is 1 percent cmplete in 15 minutes at and in 3 minutes at. Estimate its activatin energy. Slutin: ssuming first-rder kinetics: t k * t at.9 * k *15min (.9) 15 k k.7 min -1 at.9 * k *3min (.9) 3 k k.35 min -1 Using the rrhenius equatin: k E RT * e ( k) ( ) E - RT ( ) ( k) E + RT at : at E + RT ( ) ( k ) ( ) ( k ) then, E E ( k ) + ( k ) + RT RT E + RT k k E R * 1 T 1 T E k R * k 1 1 T T E J * O ml* O O E 69,8 J/ml 6

7 In English units: E al * O ml* O O E 16,68 al/ml 7

8 . The fllwing data were btained thrugh a treatment plant treating a pulp and paper mill waste. The vlume f the aeratin tanks is,, gallns divided int fur bays. Determine the first rder rate cnstant using a retardant kinetic mdel. Slutin: The retardant reactin equatin is k X 1 ( 1+ α r * t) α r Test 1 Test Flw, MGD (m 3 ) 5. (9,5).5 (77,5) Recycle, MGD (m 3 ) 7.85 (9,7) 7.9 (9,9) BOD Influent, mg/l BOD Recycle, mg/l 67 BOD, Bay 1, mg/l BOD, Bay, mg/l 7 5 BOD, Bay 3, mg/l 39 3 BOD, Bay, mg/l 3 6 VLSS, mg/l 185 The BOD cncentratin entering Bay 1 during test 1 was calculated as fllws 6 gallns mg 6 gallns mg 5*1 * *1 * 67 day L day L BOD Bay1 6 gallns 6 gallns 5* *1 day day BOD Bay mg/l X B1 BOD Bay1 1, then X B mg/l X B BOD Bay1, then X B mg/l X B3 BOD Bay1 3, then X B mg/l X B BOD Bay1, then X B mg/l Vlume Ttal time ( + ) Q Recycle Q IN 5*1 6.*1 gallns gallns *1 day 6 6 gallns day.7 days 1.77 hr 1.77hr 1.77hr Bay 1.hr Bay *.88hr 1.77hr Bay 3 3* 1.33hr Bay 1.77hr 8

9 The BOD cncentratin entering Bay 1 during test was calculated as fllws 6 gallns mg 6 gallns mg.5*1 * *1 * day L day L BOD Bay1 6 gallns 6 gallns.5* *1 day day BOD Bay1 131 mg/l X B1 BOD Bay1 1, then X B mg/l X B BOD Bay1, then X B mg/l X B3 BOD Bay1 3, then X B mg/l X B BOD Bay1, then X B mg/l 6 Vlume.*1 gallns.85 days.5 hr Q Recycle Q IN 6 gallns 6 gallns.5* *1 day day.5 hr.5 hr Bay 1.51 hr Bay * 1. hr.5 hr Bay 3 3* 1.53 hr Bay.5 hr Ttal time ( + ) Using nn-linear regressin with the equatin fr retardant reactin and the data fr X Bi and Bay i fr test 1 and test (shwn in next table) we can btain the values fr α and k. X Bay 1, Test Bay, Test Bay 3, Test Bay, Test Bay 1, Test Bay, Test Bay 3, Test Bay, Test.5 11 Frm the print ut f the nnlinear regressin presented n the next page: α r 1.93 hr -1 and k 1.9 hr -1 r 9

10 The fllwing plts shw the predicting capabilities f the mdel btained: Retardant inetic Mdel Regressin X, mg/l , hurs Initial BOD, mg/l ctual vs Predicted BOD 8 7 Predicted BOD, mg/l ctual BOD, mg/l 1

11 Print ut f the nn-linear regressin (Sigmaplt 9.) Retardant inetic Mdel Regressin Equatin X *(1-(1+a r *time) -k/ar ) Number f Iteratins Perfrmed 13 R Rsqr dj Rsqr Standard Errr f Estimate efficient Std. Errr t P VIF a r < k < nalysis f Variance: Uncrrected fr the mean f the bservatins: DF SS MS Regressin Residual Ttal rrected fr the mean f the bservatins: DF SS MS F P Regressin <.1 Ttal Statistical Tests: PRESS Durbin-Watsn Statistic.985 Failed Nrmality Test Passed (P.983) -S Statistic.1537 Significance Level.983 nstant Variance Test Passed (P.979) Pwer f perfrmed test with alpha.5:.999 Regressin Diagnstics: Rw Predicted Residual Std. Res. Stud. Res. Stud. Del. Res <

12 5. The data presented in the table belw were btained frm a batch reactr experiment with an initial reactant cncentratin f 1 mg/l. Determine the rder f reactin. The value f activatin energy and the rate f cefficients fr each temperature. Slutin: Reactant ncentratins (mg/l) (days) The rder f reactin is determined using plts f vs time, vs time, and 1/ vs time. a) vs time 1 Plt f vs time 1 3 Reactant cncentratin (mg/l) (days) 1

13 b) vs time Plt f vs time *time *time *time (days) c) 1/ vs time Plt f 1/ vs time Reactant cncentratin -1 (L/mg) (days) Frm the plts we can bserve that the reactin fllws first-rder kinetics. 13

14 The rate cnstants are: 1 1. day day day -1 The activatin energy: E a 1 * R *T T T 1 *T 1.98 J *8.31 * 93 * 83. ml* E a 33, J/mle In English units: E a 1 * R *T T T 1 *T 1.98 al *1.987 * 93 * 83. ml* E a 8,3 al/mle 1

15 6. Temperature-rate relatinships are ften represented by an expressin shwn belw: Using temperature in degrees elsius, determine the value f θ using the data f Prblem 5. ( T ) * θ Slutin: Fr 3 : ( 3 ) * θ θ 1 θ 1 θ Fr 1 : ( 1).98 1 * θ θ 1 θ 1 θ 1.5. Then: θ T 15