Question

Exam CSUS Department of Chemistry Experiment 11 Chem. 1A EXPERIMENT 11: Chemical Kinetics Name: PRE-LABORATORY ASSIGNMENT Section: Score: 10 (Turn this in with you lab] 1. Three runs were used to study the reaction: A+B 3C+D The following data were obtained initial concentrations of Cand D = 0): Initial Concentration (M) Concentration after 150 $ Run (A) IAL [B] [C] 1 3.000 2.000 2.970M 2 6.000 2.000 5.940M 3 3.000 4.000 2.880M (a) Fill in the missing concentrations in the table above based on the reaction stoichiometry [B] (b) Calculate the rate of the reaction in M/s for each of the three runs. Rate x a At (c) Using the method of initial rates, determine the order with respect to A and B. (d) Calculate the rate constant for the reaction, including units. Rev. F14 Page 1 of 8

chemical kinetics

Answer 1

Answer 01. given reaction A +8 → 30 +D. (a). Initialeones concentration after. 1505. Run. [A] [B] [A] TB1 [C]. 1. 3.00 2.00 2.97 M. 1.97 M 3X0-03 = 0.09.M 2. 6.00 .2.00 5.94M 1.94M3x0.06 = 0.18M 3. 3.00..4.00 2.88'M 3.88M 3 x 0.12 = (-36M Note: - From stoichiometry i mole g A Heact with Imole 9B to give, : 3 mores of C o dels solve first run to understand. Initial concentration AB, [A]:- 3.00m. concentration of A after 150 second, [A]:50 = 2097M now, concentration of A get reacted - 3.00M- 2.97M = 0.03 M. . Hences. 0.03 M i B will also be reacted. And, production of G will be 3 times. .; 3* 0.83 M = 0.09. mo produced, (b). calculating the reale, M5-1.. Lgiren Rale=- 1 x <[A]. a = coeficient of A. st: ingiven to Trun-1} Rates - 1 x TA]. – A31500? a. = . 50 see . = -1 3000M – 2.97M. Ata] - change in conen = -1x 0.03M Ato change in time 150 sec. - išo second. = 2.0 x 10-4 MS-! 150 sec

Run-2) Rates-1x [47o – [A]jso Isos - 1x 600M - 5.94 M 1503 . : 0.06M 7505 0.0004 [Run-3). =40810-4 MS- Rate = -1% [a]. – [13150 : 1500 2 -1* [3.00] - [ 2.88 m] 150 0.12.M 150 s :a 810-4 MS-! :: initial conen. [A] [B] rose rake .-1 Coro Oo NOTE:-|M= moll-17 Run. 3.00 2.00 2x10-4 mollis 4X10-4 moltt's- 6.00 *3:00 2.00 4.00 3. 6 x 10-4 mol l-Is-1 (e) - finding order, het, nate = k [A] [B] B & and ß are order of the reaction with respect to From, Run, - 1. 4 and 3 raut 2x10-4 = k [3.00] [2.0078 tis. From-Run-2 4810*4= K.[6.00]"q 2.007 ---(ij) From Run-3. 6x10-4= K[3.00]*[4.00] P. (iii)

2x10-4 2x10-4 ondividing (i) by is we get, -4 X10-4 = K[6.00]“ [2000] p [ 3.003° [2.00] P 4. .: 2 = [27°. on, a=1. : On dividing Gee) by ci) - we get 6x10-4 K[3.00] [ 4.00] P 'k [3.00]a [ 2.00] ... 3 = [2] P. or $3808 (2)' = [23.8.; Baş se rate = * [A]”[8] %. d). det, ros.consider rutin 1, we get, rate = K[A]?[B75/3. 2x10-42. K [3112] 5/3 He 2x10-4 mot L-les-/= [3]137502 190101) .:K = 2.0x10=4 cx (moll-1) s-/ x[mol L-4) 4+5. i (molt-1) 8/3 2.0 x10-4 3*3 * (mot k-1)=5435-1. :.; K = 0.22 x10=4 pol mot 2-1)-575-, . . . 3x (2)543