CHEM 162 2001 HOURLY EXAM I ANSWERS CHAPTER 12 KINETICS RATESRATE CONSTANTS

Chem 162 2001 hourly exam i answers chapter 12

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6. CHEM 162-2001 HOURLY EXAM I + ANSWERSCHAPTER 12 - KINETICSRATES,RATE CONSTANTS, REACTION ORDERS, HALF-LIFES CALCULATIONSINTEGRATED RATE LAWConsider the reaction 2H2O2(l) 2H2O(l) + O2(g)41
Assume that this reaction follows first-order rate kinetics and that its half-life is 10.0 minutes. How long would it take for 80.0% of an original H2O2sample to decompose? A.8.05 minutesB.15.0 minutesC. 23.2 minutesD.40.5 minutesE.55.2 minutest1/2= 10 t1/2= 0.693/k k = 0.693/10 = 0.0693ln(C/Co) = -ktln(0.2/1) = -0.0693 x t-1.609 = 0.0693 x tt = 23.22 minutes 8.CHEM 162-2001 HOURLY EXAM I + ANSWERSCHAPTER 12 - KINETICSRATES,RATE CONSTANTS, REACTION ORDERS, HALF-LIFES CALCULATIONSREACTION RATESConsider the reactionH2(g) + I2(g) 2 HI(g)Assume that 0.100 mol H2and 0.100 mol I2are placed in a 5.0 L container. After 30.0 minutes, 0.080 mol of H2remains in the container. What is the average rate of disappearance of H2over this 30.0 minute interval? H2I2VTConditions 1 0.100 mol = 0.020 M 0.100 mol 5.0 L 0 minConditions 2 0.080 mol = 0.016 M 30 minRate = (0.016 - 0.020)/(30 - 0) = 0.004/30 = 1.33 x 10-4Mmin-142
12.CHEM 162-2001 HOURLY EXAM I + ANSWERSCHAPTER 12 - KINETICSRATES,RATE CONSTANTS, REACTION ORDERS, HALF-LIFES CALCULATIONSREACTION ORDERGiven the following data for a reaction between A and B:Initial Rate (M s-1)Initial Concentrations (M)[A] [B] 0.0030 0.040 0.0300.0060 0.040 0.0600.0270 0.120 0.030What is the rate law for this reaction?A.Rate = k[A][B]B.Rate = k[A][B]2C. Rate = k[A]2[B]D.Rate = k[A]2[B]2E.Rate = k[A]2Rate = k[A]m[B]nRate = k[A]m[B]n2 = 2n[B]n= [B]1n = 19 = [3]mm = 2rate = [A]2[B]1 17.CHEM 162-2001 HOURLY EXAM I + ANSWERSCHAPTER 12 - KINETICSRATES,RATE CONSTANTS, REACTION ORDERS, HALF-LIFES CALCULATIONSINTEGRATED RATE LAWThe reaction 2NO(g) →Ν2(g) + O2(g)follows the rate law: Rate = k[NO]2In an experiment, it is found that an initial NO concentration of 0.0028 M falls to a concentration of 0.0020 M in 33 minutes. What is the rate constant for this reaction? 43
D.4.3 M-1min-1(1/C - 1/Co) = kt((1/0.0020) - (1/0.0028)) = k x 334.33 = k 20.CHEM 162-2001 HOURLY EXAM I + ANSWERSCHAPTER 12 - KINETICSRATES,RATE CONSTANTS, REACTION ORDERS, HALF-LIFES CALCULATIONSREACTION RATEFor the gas phase decomposition of acetaldehyde, CH3CHO, the experimental rate law is:Rate = k[CH3CHO]3/2If [CH3CHO] is doubled, the rate will

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