Question

Chem 112 Lab Citrus Community College Spring 2020 Name: Date: Lab Section: Problem Session 2: Kineties General Instructions: Complete the following problems. Due Monday Tuesday, March 16/17 by 11:30 am (for the MW and T/Th Sections) in box stockroom window. Do work on separate sheets of paper. 1.) (Taken from General Chemistry & Ed, Ebbing and Gammon) You perform some experiments for the reaction A B +C and determine the rate has the form Rate = kar Calculate the value of x for each of the following (a) [A] is doubled and you observe no change in te (b) [A] is doubled and the rate goes up by a factor of 4 (c) [A] is doubled and the rate poes up by a factor of 9. 2.) (Modified Example Problem 14.53 from General Chemistry by Vining. Young Day, and Bosch) For the gas phase decomposition of ethyl chloroformate, the rate constant is 1.05 x 10s at 470 K and 111 x 10° s'at 508 K. Calculate the activation energy and the frequency factor for this reaction CICOOCH() - CH.CH) + CO.() 3.) (Taken from General Chemistry: Principles and Structure, 3 4, Brady) Suppose that the following sequence of reactions is proposed for a reaction: Step 1: A 2A Step 2: 2A B + 2D (a) What would be the overall reaction? (b) What would the rate law be if step I was slow and step 2 was fast? (c) What would the rate law be if step 2 was slow and step I reached a fast equilibrium? 4.) The integrated rate law for the uncatalyzed thermal decomposition of a chemical species, A. is given by the following for second and higher-order processes: -"-14, -") - nkt where, n is the order with respect to A. [AL. is the initial concentration of A, and Al, is the concentration of A at time, L. (b) What is the half-life for a third-order process in A? (c) What are the units of k for the third order process? 5.) At a particular temperature, Ke-6.5x10' for 2NO(g) + 2H:() N(e) + 2H2O(g) Calculate K, for each of the following reactions: (a) 3NO(g) + 3H (8) N (8) + 3H 0(g) (b) 2N (Ⓡ) + 4H2O(g) - 4NO(g) + 4H (B) 6.) The formation of NO, from N.O, is given by the following equation: N,048) 2NO(e) If the equilibrium constant (K.) for this reaction is 0.21 at 373 K and the initial concentration of N.O.is 0.045 M. what is the equilibrium concentration of NO,?

Answer 1

1 Toge- Reaction - A B+ c Rate = K [A]² 7 R EAJ is doubled but the rate becomes same. Therefore Rate does not depends on the concentration of A. reaction ". The rate is zero order with respect to A. . Rate = K [A] value of x = 0 (6) CAT is doubled and rate goes up by a factor 4. Therefore 4xRate = k{2AT - Dividing - 4x Rate = K [ A]" Rate = 4 = 2x KLAJK - 2- 2 " Therefore value of x=2 . of factor (c) g. кLAT ҳ EA] is doubed and rate goes by a exRate K [2AJ - w @ we have axRate - KERAJ = - Rare - 9 = 2* = ogg axloga » « = hogy = 3.17 e value of the x = 3.17 2 c ercoocahE (9) ensaca & concg). Jovki= 1.058 103 s',Ti Atok nants kon 1.11 x 102 s T2 = 508k consolle und According to Arrhenius equation, Ea (1 rog (22) = 2.303R ( 21 T 2 (508- 470 1.058 18 37 louro 2.303 R 508 X 470 1.1x102 WONEN 1508-470 A 12 log7.05x103) = 21303481314 508X 470 MS K E cu = 12 3207 I /mol on ER2 8.3 19 J. mort - 123.207K J Amor 111 X 102) M

page=2 - A è EA/RTA - roges = loga- Ea log (1.05 x 103) = LogA- rog A 228.551 A a 3.5568188 5 123. 207 8.319 x 15 3 x 470 Activation energy of the reaction is 123.207 KJ parol and frequency factor of the reaction is 3.556 8 128 51 Step-I = step - 1 - Az - zstaB zt c+20 overall reaction - A2+ B e + 2D (b) we know slowest step is the rate determining Step. As the step-I is slow, therefore it will be rate de termining step. . Rate = k [A ] [K= Rate constant] (c) As step-II is slow, therefore it will be rate determining step. i. Rate = K[A] [5] [k2 Rate constant] @. (Laoglan-[] in) ankt 0 n-3 so from the For a 3rd order reaction е ч ои - t 3 (EA] -3 - [A273) = 342 >> (4.J?- [] ?)= 3kt => [at? - Idegna -6kt 2 Kb = - + [alor Lega] 3 KEZ [ Lager ] 2) uz it I LAJE - LAJ This is the integrated rate law.

page-3 2. At za to, [A] = LATO ise, when the concentration becomes half of the initial value then it is called tvz (half life). So, from the integrated rate law equation, ks to lear-to- = sty ( ) K = 2t [Aar unit of for any order reaction, . General expression of a ka (L.mong time" n- 3 for 3rd order reaction, time! ka (t.moil) 3-2 = r. mot2 time (time can be in second, minutes or ha 2 No eq) + 2 H₂ (9) N₂Cq) + 2 H2O(g) 2 6.5 X 10 kez [Na] [H2o] x 102 [Nojor [H2J 2 3 Nocg) + 3H₂(g) 3 N₂Cq) + 3 H₂O (8) Kea [Na] [ he] 3 I NO J3 [H2J3 1 [Na] [H2J 2 3/2 [NO] [ELT = (6.5x12% 1.657 X10 1.

6 pageal 2N2 (8) + 4H2008 & No (8) + 4H₂(g) [NO] 4 [2] [N2Jr [H2OJA Ke- = (he) - ( 18 x 17) 6.5 = 1.538 X10 © N204 (9) 2NO2C9) initial 0.045 o conc. equilibrium (6.045-x) 2x ои с. 30:21 = (2x) 00095 x 42 - 0.045 0121 - 0.21x +0.212 - 0.045 0.21 = 0 0.21% - 9.45 x103 zo - 0121 I J 6,219 _ 4x4 (-3,45x103) (2x4) z x = -0.21 + .. x = -0.21 - 0.21 + 0.442 ( (-) ve value of x is not possible) 8 = 0.029 in equilibrium concentration of No2 = 2x - 2X0.029 M 2 0.058M