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# General College Chemistry II CHE 152

Monroe Community College

GPA 3.9

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This 7 page Class Notes was uploaded by Dr. Golda Donnelly on Thursday October 15, 2015. The Class Notes belongs to CHE 152 at Monroe Community College taught by Sherman Henzel in Fall. Since its upload, it has received 24 views. For similar materials see /class/223550/che-152-monroe-community-college in Chemistry at Monroe Community College.

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Date Created: 10/15/15

CHE152 Instructor Sherman Henzel CHEMICAL KINETICS REVIEW There have been several concepts presented in our study of chemical kinetics It is all too easy to miss how these concepts are related Often concepts are compartmentalized and the interrelationships are missed In other words the solo multibranched members tend to produce a condition of myopia with regard to viewing the veritable plethora of coniferous and deciduous growth With the above in mind we will consider the following reaction CH220 g H CH4 g CO g with regard to the interrelationships of the following 1 The instantaneous rate of the reaction at a given time t 2 The rate equation for the reaction 3 The order of the reaction 4 The half life of the reaction 1 THE INSTANTANEOUS RATE OF A REACTION The instantaneous rate of a reaction at a given time t is determined by plotting either the concentration or pressure of a reactant or product vs time an drawing a tangent line at the point on the resulting curve at time t Graph 0 The slope of this tangent line will give the instantaneous rate of the reaction at time t with respect to wrt the reactant or product being graphed The kinetic data for the reaction is presented in Table I We will calculate the instantaneous rate at t 800 min Table I Time min PCH2ZO mmHg PCH4 mmHg 00 12494 00 100 11074 1420 200 9821 2373 400 7723 4771 600 6073 6421 1000 3754 8740 2000 1122 11372 NOTE all graphs start on page 6 CHE152 Kinetic Review page 2 The instantaneous rate with respect to CH220 is calculated from the tangent drawn to the CH220 curve at 80 minutes Using the points at the extreme ends of line 0 90 and 152 0 APCHZZO 9O 0 mmHg 05921mmHg At 0 152 min 39 rate 2 mm rate 05921 mmHg min391 at t 80 min wrt CH220 The instantaneous rate with respect to CH4 can be calculated from tangent line drawn to the CH4 curve at 80 minutes Using the points at the extreme ends of line calculate the rate wrt CH4 Answer based on these two points 0 33 and 175 140 rate 06114 mmHgmin at t 80 min wrt CH4 Note that the two rates are very close and differ by only 33 How close these two rates are to each other is a function of how the tangent lines are drawn Ideally the rates should be the same if the tangents are drawn correctly One word of caution is needed The two rates are the same or very nearly so because for each mole of CH220 that decomposes one mole of CH4 is formed This would not be true for reactions where the mole amounts are different 2 THE RATE EQUATION The rate equation rate kPESHZhO is readily calculated from initial rate data Here the initial pressure of CH220 is measured from a tangent line drawn at t 0 The initial rate is equal to the slope of this tangent line The graphs for this series of experiments will not be shown The data is presented in Table II Table II Experimen PCH2ZOi Ratei mmHg min391 t mm g 1 12494 165 2 30000 389 3 50000 658 CHE152 Kinetic Review page 3 The exponent m is calculated as follows ratel k 111211 165 mmHg min l k12494 mmHgm rate2 k PE 1 9 389 mmHg min 1 k30000 mmHgm CHZZO 2 042416 041647quot ln042416 m ln041647 O85763 m 087595 m 0979 1 Using Experiments 2 and 3 verify that m l 7 1 Smce m 7 1 then rate lltPltCHZgtZO We will use this equation to nd k rate kP gHZZO from Experiment 1 165 mmHg min 1 k12494 mmHg 165 mmHg min 1 00132 min 1 note the unit for k 12494 mmHg Use the data from Experiments 2 and 3 to nd k from Experiment 2 k 00130 min391 from experiment 3 k 00132 min39l k 0013l3min391 average deviation A 001313 min391 00132min391 00001min391 A2 49 x 10399 min392 001313 min391 00130min391 000013 min391A2 169 x 10398 min392 001313 min391 00132min391 00001min391 A2 49 x 10399 min392 2A2 2 4 x 109 min392 16 x 108 min392 4 x 109 min392 267 x 10398 min392 2 78 72 s m 23967X10 mm 2000012min 1 N l 3 1 The standard deviation s is small indicating that k is constant CHE152 Kinetic Review page 4 3 REACTION ORDER From the rate equation it is apparent that the reaction is rst order with respect to CH220 and first order overall The order of the reaction could also be determined by graphing the original experimental data presented in Table I We need to add two new columns to the original data set ln13CHZZO and P CHzzOl Table 111 time min PCH2ZO mmHg 1nPCH2ZO 1PCH2ZO mmHg391 0 12494 482783 80038 10 11074 470718 90302 20 9821 45871 1018 400 7723 43468 1295 600 6073 41064 1647 1000 3754 36254 2664 2000 1122 24177 8913 By graphing each function of pressure versus time the order of the reaction can be determined A plot of P vs time was made in Graph 1 Since it did not yield a straight line the reaction is not zero order The only plot that will yield a straight line is that of lnP vs time Graph 11 thus indicating a rst order reaction as expected The plot of UP vs time does not yield a straight line Graph 111 Thus the reaction is not a second order reaction Since the reaction is rst order the equation for this reaction can be written as lnPKCanoit k 39 t lnPKCanoit From the initial rate data in Table II we determined the average value of the rate constant k to be 001313 min39l Determine the slope of the line for Graph 11 From the slope determine k Does this value agree with the value calculated from the initial rate data The points I chose were 0 485 and 200 220 this gives a slope of 001325 min391 and k 001325 min391 which agrees quite well with our initial rate calculation Using the point 200 220 we can solve for the intercept b 485 Our equation 71 1s now ln13CHZZOt 00132 mm t 485 CHE152 Kinetic Review page 5 4 HALF LIFE OF A REACTION 0693 0693 For a first order reaction tlz k 523 min 5 mln Another method for the determination of half life is as follows The half life of this reaction occurs when the pressure of CH220 is equal to one half of the initial pressure This occurs when P l2starting pressure From Table I our starting pressure was 12494 mmHg On half of 12494 mmHg 6247 mmHg We can use our rst order rate law to solve for t 1nPltCHZgtZot k 39t 1nPltCHZgtZot ln6247 001325 min l tln12494 4135 OOl325 min 1 t 4828 O693 OOl325 min 1 t t 523 min 5 SUMMARY As we have seen all of the following expressions can be used to describe the reaction CH220 g H CH4 g CO g APCHZZO APCsz rate rate 2 AP rate 001313 min l PJCHMO lnP 71 CHZW 1 001325 mm t 1nPCHZZO0 0693 0693 U2 2 71 523 mm k 001325 mm Which expression we use depends upon what we want to know about the reaction rate nal or initial concentrations half life or time We choose the expression that has the value we wish to solve for based on the given values CHE152 Kinetic Review page 6 Graph 0 P mmHg 0 20 40 60 80 100 120 140 160 180 200 Time min Graph I 100 150 P CH220 mmHg 200 CHE152 Kinetic Review page 7 Graph 11 y 390013lx 48278 R2 1 4 4gtUIUI l11PCH22O N u N 0 50 100 150 200 Time min Graph III 012 0 008 006 004 002 1PCH22O 0 50 100 150 200 Time min

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