Law and Activation Energy (2020/05/08 revised) Collect: 50 mL - PowerPoint PPT Presentation

Iodine Clock II – Integrated Rate Law and Activation Energy (2020/05/08 revised) Collect:  50 mL Erlenmeyer flask (10): wash clean, oven dry, and cool  10 mL graduated pipet (2), pipet filler (1)  Cork stopper (6)  Thermometer (1)  Stopwatch (1) and stir bar (1) (given by GTA)  Labels (label graduated pipets, beakers, and Erlenmeyer flasks) Prepare:  100 mL beaker (2):  Wash clean and oven dry  Label Na 2 S 2 O 3 and H 2 O, separately 1

Objective & Skills Objective  Determine the rate law of reaction 2- + 2I - → 2SO 4 2- + I 2 S 2 O 8 Rate = k [S 2 O 8 2- ] m [I - ] n 2- ) reacting  Add limiting amounts of thiosulfate ion (S 2 O 3 with iodine( I 2 ) as a measuring tool to determine the rate of the above reaction: 2- + I 2 → 2I - + S 4 O 6 2- (a fast reaction) 2S 2 O 3 Skills  Use of graduated pipet and magnetic stirrer  Graphic method - integrated rate law 2

Outline of Experiment I. Integrated rate law Rate = k’ [S 2 O 8 2- ] m II. E a , activation energy  E 1   a ln(k) = lnA R T III. Effect of catalyst 3

Theorem of Integrated Rate Law For a one reactant reaction, A  P Reaction Rate law Integrated rate law Plot Slope Intercept order   d[A]  Zero [A] = -kt + [A] o [A] vs. t -k [A] o rate k dt   d[A]  First ln[A] = -kt + ln[A] o ln[A] vs. t -k ln[A] o rate k[A] dt 1 1     d[A] kt Second  1/[A] vs. t k 1/[A] o k[A] 2 rate [A] [A] 0 dt * [A] o is the initial conc.; [A] is the conc. after reaction time, t. 4

Measurement of Concentration of 2- ] t with Time Reactant, [S 2 O 8 2- + 2I - → 2SO 4 2- + I 2 (1)  S 2 O 8 2- + I 2 → 2I - + S 4 O 6 2- (2) 2S 2 O 3 Limiting I - + I 2 → I 3 - + starch → dark blue,  t reagent  [S 2 O 3 2- ] ＝ 2  [S 2 O 8 2- ] 2- used up, I 2 reacts with  While the limiting reagent S 2 O 3 starch indicator, and the soln appears blue  Stop the timer and record  t  Calculate the  [S 2 O 8 2- ] left in reaction mixture at time  t 5

Trials of Integrated Rate Law V (mL) 1.0 M 0.20 M 0.15 M H 2 O 2% Starch Trial NaI Na 2 S 2 O 3 K 2 S 2 O 8 1 1.0 5.0 5.5 1.5 5.0 2 1.0 5.0 4.5 2.5 5.0 3 1.0 5.0 3.5 3.5 5.0 4 1.0 5.0 2.5 4.5 5.0 5 1.0 5.0 1.5 5.5 5.0 6 1.0 5.0 0.5 6.5 5.0  Keep the volume of solution same (18 mL) for each trial.  Keep concentration of reactants, [NaI] and [K 2 S 2 O 8 ], same at each trial  Vary the amount of limiting reagent, Na 2 S 2 O 3 , added  The concentration of NaI is much higher than that of K 2 S 2 O 8 and assume constant during the reaction 6  Investigate the reaction order of K 2 S 2 O 8 only.

Step 1 Integrated Rate Law  Wash clean 10 flasks, then oven dry, and cool to r.t.  Add reagents to each flasks * Add the last according to Table 2 reactant K 2 S 2 O 8  Add the last reactant K 2 S 2 O 8 from dispenser to the flask and start recording time  Stop the flask with cork and shake the flask to mix the reactants #4 #6 #5 #3 #2 #1  Since the solution turn to blue, stop recording time  You may add the last reatant Note: It may take 10 min. K 2 S 2 O 8 to each flask every 30 s. for flask #1 to change color. 7 Go figure.

Step 2 Determine the Activation Energy 2% Starch 1.0 M NaI 0.20 M Na 2 S 2 O 3 H 2 O 0.15 M K 2 S 2 O 8 No. 5 1.0 mL 5.0 mL 1.5 mL 5.5 mL 5.0 mL  Measure each reagents into (1) Ice-water bath the flask according to trial no. 5 Temp. ca. 2 o C  Balance the temp. of flasks at * Note: it needs (1) ice-water bath and (2) lots of ice to keep warm-water bath, respectively temp. constant  Add the last reactant K 2 S 2 O 8 and start timing (2) Warm-water bath  Record the time interval of Temp. ca. 40 o C color change,  t, for each trial * Note: measure the temp. of soln in the at various temp. flask 8

Step 3 The Effect of Catalyst on Reaction Rate  Prepare the reaction solution according to trial no. 5 and add 2 drops of Cu 2+ as catalyst  Compare the reaction rate with that without Cu 2+ added * Note: add the Cu 2+ just before the last reagent, K 2 S 2 O 8 2% 1.0 M 0.20 M H 2 O 0.020 M 0.15 M Starch CuSO 4 K 2 S 2 O 8 NaI Na 2 S 2 O 3 1.0 mL 5.0 mL 1.5 mL 5.5 mL 2 drops 5.0 mL 9

Notice Full version report  Erlenmeyer flasks should be oven-dried and cooled down to r.t.; do not wipe dry with paper towel  Use the same set of reagents for whole series of trials  Start recording time since the last reactant, K 2 S 2 O 8 , is added to the solution (remove pipet filler to drain all liquid); the way of recording time should keep constant throughout experiment  Use enough ice in the ice-water bath to keep the temp. constant during the reaction  Take the temp. of reaction solution; not the temp. of the water bath  After class, wash clean the Erlenmeyer flasks and put them into oven to dry 10  Hand the stir bar and timer to TA

2- ] t vs. Data Analysis of [S 2 O 8 Time • Tabulate your calculation in details 2- ) added 2- ) used [S 2 O 8 2- ] left (S 2 O 3 (S 2 O 8 2- 2- n of Na 2 S 2 O 3 n of S 2 O 8 [S 2 O 8 ]left  t 2- ] t / M 2- ] 2- ] No. added / mmol used up / mmol i.e. [S 2 O 8 ln[S 2 O 8 1/[S 2 O 8 (0.15*5-0.55)/18 #1 5.5*0.2=1.1 1.1/2 = 0.55 = 0.0111 -4.50 90.0 t 1 (0.15*5-0.45)/18 #2 4.5*0.2=0.9 0.9/2 = 0.45 = 0.0166 -4.09 60.0 t 2 (0.15*5-0.35)/18 #3 3.5*0.2=0.7 0.7/2 = 0.35 = 0.0222 -3.81 45.0 t 3 (0.15*5-0.25)/18 #4 2.5*0.2=0.5 0.5/2 = 0.25 = 0.0277 -3.58 36.0 t 4 (0.15*5-0.15)/18 #5 1.5*0.2=0.3 0.3/2 = 0.15 = 0.0333 -3.40 30.0 t 5 (0.15*5-0.05)/18 #6 0.5*0.2=0.1 0.1/2 = 0.05 = 0.0388 -3.25 25.7 t 6 11

Example of Integrated Rate Law 2- ] vs. time ln [S 2 O 8 2- ] vs. time [S 2 O 8 -3 0.050 0 200 400 600 800 0.040 -3.5 0.030 [A] ln[A] -4 0.020 0.010 -4.5 0.000 -5 0 100 200 300 400 500 600 700 y = -4E-05x + 0.0385 y = -0.002x - 3.1581 Time（s） Time（s） R 2 = 0.965 R 2 = 0.9983 2- ] vs. time 1/[S 2 O 8 100 2- ] [A]: [S 2 O 8 80 According to the plots 60 1/[A]  ln[A] vs. time 40  It’s a first order reaction, m = 1 20  According the slope, k = 2.0 x 10 -3 0 0 200 400 600 800 y = 0.1039x + 16.575 12 Time(s) R 2 = 0.9698

Data Analysis of the Activation Energy  Determine the rate constants, k, of reaction at various temperatures  Use Arrhenius eqn. to calculate the activation energy  E 1   a ln(k) = lnA R T  t (s) ln(1/  t) Temp. ( o C) 1/T (K -1 ) Trial  t 1 ln(1/  t 1 ) 1) Room temp. T 1 1/T 1  t 2 ln(1/  t 2 ) T 2 2) Ice-water bath 1/T 2  t 3 ln(1/  t 3 ) 3) Warm-water bath (ca. 40 o C) T 3 1/T 3 13

Plot ln(1/  t) vs. 1/T to Obtain E a Derived from Arrhenius eqn.:  1 E 1    a ln( ) = lnA lnc Δt R T y = -5824.7x + 14.597 R 2 = 0.9994 0.00 0.00320 0.00330 0.00340 0.00350 0.00360 0.00370 -2.00 ln(1/ t) -4.00 -6.00 -8.00 1/T E a = 5824.7 × 8.314 = 48 kJ/mol 14

Notice of Report Full version report  List the calculation in details  Print and attach the tables and figures with report  Determine the reaction order, rate constant, and activation energy with appropriate significant figures and units  The effect of CuSO 4 T ( ℃ ) △ t (s) No. 5 T (K) Without Cu 2+ 270 14.0 287.2 With CuSO 4 92 14.5 287.7 * Conclusion: add Cu 2+ could increase reaction rate 15

T12 - Graduated Pipet Ex.1 Deliver 5.00 mL solution: Partially deliver Wash a 10 mL pipet thoroughly.  Pipette 安全吸球 A Rinse twice with small portion of sample filler  Aspirate solution. S A 排氣閥 E valve Press valve A of pipet filler and squeeze Bulb 安全吸球  10 ml S Ex E bulb to expel the air inside and create a Suction 0 吸液閥 valve 1 vacuum. 25 2 排液閥 Empty ml 3 Insert the top of pipet into pipet filler, press  valve 4 valve S to draw liquid to equal to the mark 5 of 0.00 mL. Hold pipet vertically and transfer liquid into  container. (One hand hold pipet and the other hand hold container to operate.) Transfer Pipet Press valve E to drain liquid to the mark of  pipet 5.00 mL. Wash thoroughly after use.  16