Question

From the data table determine the rate law for reactions 1-4 and calculate the value of k for each. Thanks

Reactants 8 ml 30% HO, 2 mL 05MK 8 mL 30% H0, 2 mL 0 25 MKO 8 ml 15 HO, 2 mL 0 5 MI 8mL 30% H2O2 2 mL 05MK Temperature Initial rate CC) (kPa's) 250 0.1803 250 0.03642 25°C 0.06140 35°C 046451 DATA ANALYSIS Rate Law 1. From the data collected during this experiment we will be determining the rate law for the reaction. The rate law has the form: Rate = k[H2O2) 11-1 To determine the reaction order we must complete the following table: 0.00M Hoog 0.5M KI | Part Initial rate (mol/L-S) [H2O2) after mixing after mixing 101 To convert the initial rate from kPa/s to mol/L-s we use the equation mol/L - P/RT. We can use R = 8.314 L kPa/K mol. Rate Constant 1. To calculate the value of k, the rate constant, you simply plug into the rate law the values of the concentrations, the orders, and the rate of the reaction and solve fork. Do this for for Parts I, II and Ill and then determine an average value for the rate constant. Also calculate the rate constant for Part IV to use to calculate the activation energy. Part Initial rate (molL-) THOal after moing r after micing Rate constant Average value of k from parts 1. Il and I (Be sure to include the units)

I-IV

BACKGROUND INFORMATION

Given: 3% H2O2 Concentration= 0.88M & 1.5% H2O2 Concentration= 0.44 M

•Calculation of H2O2 after mixing for parts I,II, IV= 0.704 M

•Calculation of I^- after mixing for parts I,II, IV= 0.10 M

•Calculation of Initial (mol/L-s)

Filled in Table

Now the solutions for Rate Order and Rate Constant are needed

Answer 1

Solution Melanity of 3.1. H2O2 = 0.88 m Molarity 81574202 = 0.44 m Totul volume of schalim = 10 mL Table : obsevation initial fake Purt kf pufos 4 | neleg (изод 1 кг] H202 kr 1 a mixing! I after I at mixing kPa/sec H202 0.1803 0.1 &mL 12 ml 0.88 m. 105M 17.04x16" 1-0 X 16P 10.704 0.08642 | 2me &mL 17:04X163 10.5x103 10.704 | 0.05 0.25M 17 (0.88m ame. 3.52x16"-oxió? 10.352 0.1 0.06140 8 me 10.44m 0.5m 2mL 0-6451 -0X10 TV 7.04X16 0.704 qmi 10.8qM Calculation is moles of H2O2 = volume of M20 2.X molarity For Part (1) 8 mL X IL x 0.88 med i looom = 7.04X163 med. . ) moles of kl = volume of KI X melarity Scanned with CamScanner

.. ..... . RLON HON D E R WERDE for part (1) males of KI = 2ml x 12 x 0.5 mol ) 1ooom - 1.ox 103 mol. (15) Concentration of H2O2 after mixing = miles of H2O2 .. ties . ..total volume. for part (I)- = 7.04x10's met .. lo me X IL = 7.00 Xoo"? x 100 = 0.704 mel . 1ooomL - Concentration of I After mixing = moles of kI .. . totul volume. (1) = 10 X 10-3 mol For part 10.0 mL X IL locomt = 0.1 med i Scanned with CamScanner

AT ! ... . . ....... ... . . . . Conversion of initial rate from ke fals to mol/l-s PV = nRT :p= ( RT p= CRT, where (= concentration e = 2 c= 1 for Past (I) where p = pressure (inkPa) ... R = 8.314 L- kPa / kmal T: absolute temperature c = 0.1803 kPa 8.314x L-k ja x 298 k e-mg c = 7.277 x16" (mol) do for other parts. Similarly art Initial rahe (mol/L) 111) after mixing Rate constant (mooth bgt Ilho, aftermixing 1 1.232 X10-3 1 7.177 X1050.704 0.05 1.181 X10 2-7992 3.488810 5 10.704 1.186 X103 T. 12.478x1050-352 V 12.519810 4 1 o Jou Skanned with 1 426 X10 3 = 4.264 x 10-3 CamScanner

order of Reaction T + 2H202 - H20 + O2(g) . Rate = k [1] 4:08] from Table - 2 Vis * [0.70] [ 0.13 Y, = k [ 344 ( ver] . Se est = (4) zb = 7.407 = 2.0862 b bagca) log (1-086) 6 = log (2.086) = 1:06o7 . lig(2) . b. = Lofot (b=1 Therefore order of rexh with respect to i is one Cs Scanned with CamScanner

Again from Table 1 V = K[ 0.352)" [oi]" . - r - 10.704)a (2) 2.478x105 Mens = oja 7.27X 105 or 24 = 7,277 = 2.9319 2.478 a (loga) = = log (7.9369) log (3.9369) = 1:55 going (3) Therefore order with respect to H2O2 is 3 Overall order of reaction = 1+ - 2.5 Therefore, I rake = k [102] [1] cs Scanned with CamScanner

tale of reaction and Rare constant (k) From Table ① For Part (1) n=k (0.764) 10.1) 7.277X1015 - k X0.05906 k = 7.277x165 = 123.195 X105 - 0.05906 k = 1.232 X16? (moh) Ik -3/ 23/2-1 = 1.232 X163 mol is Similarly do for part I, I and IV. avererge, value of k foom pass (1) () 4 () = (1232+ 1.181+ 1.186) Xio 312, 312_- Kuva = 3 mol (224 1.199 X103 mol h 3/2017 . tsat 25°C 37. k= 4.264 x 103 mol stat 35°c Scanned with CamScanner