Study guide for Exam 2
Study guide for Exam 2 CH102
Popular in Chemistry 102
Popular in Chemistry
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This 7 page Study Guide was uploaded by Malory Goetcheus on Saturday February 21, 2015. The Study Guide belongs to CH102 at University of Alabama - Tuscaloosa taught by Dr. Nikles in Fall. Since its upload, it has received 249 views. For similar materials see Chemistry 102 in Chemistry at University of Alabama - Tuscaloosa.
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Date Created: 02/21/15
This is a study guide for Dr Nikles chemistry 102 Exam 2 I have answered the questions given on the study guide CH 102 002 Name Exam 2 version 1 Spring 2014 CWID This is a multiple choice exam You may use a calculator Work the problem and circle the correct answer Enter your answer onto the machine readable answer sheet Make sure you have the correct number of signi cant gures Turn in your answer sheet showing your ACT card ID to the TA You may keep the exam questions as a record of your answers Gas constant R 83144621 Jmol K 1 Which is the following will not increase the rate of a first order chemical reaction a Increasing the concentration of the reactant b Increasing the reaction temperature c Adding a catalyst which creates a new reaction pathway with a lower activation energy barrier d Decreasing the concentration of the reactant e Adding more reactant In water persulfate ion is reduced by iodide ion to make sulfate ion and triiodide ion according to the following equation 8201300 3139 g a 28011200 1339aq Data on the effect of initial concentrations on the initial rates for this reaction are given in Table 1 Table 1 Effect of concentration on the initial rate for the reduction of persulfate ion with iodide ion 32082 M 139 M Initial rate Ms 022 021 167 030 042 454 044 021 333 044 042 665 Use the rate data to answer the following questions 2 What is the reaction order for S2084 a Zero b Third c First d Fourth e Second 3 What is the reaction order for I39 a Zero b First c Second d Third e Fourth 4 What is the value of the rate constant a 86 M39Zs39l b 196 M393s391 c 120 M39Zs39l d 43M391s391 e 36 M39ls391 In the gas phase chlorine monoxide decomposes to give chlorine and oxygen 2 C10 g gt C12g 02g In Table 2 is the data for ClO at different times during the reaction The temperature was 25 oC The data is plotted either as ClO as a function of time lnClO as a function of time or 1ClO as a function of time Use the table and the graphs to answer the following questions Table 2 Concentration of chlorine monoxide at different reaction times Time s ClO M 0 150 x 10398 10 719 x 10399 20 474 x 10399 30 352 x 10399 40 281 x 10399 100 127 x 10399 200 660 x 10391quot 160E08 0 140E08 120E08 2 100E08 14 800E09 0 g 600E09 400E09 200E09 000E00 1 1 1 f 1 0 50 100 150 200 250 Time s Figure 1 Plot of C10 as a function of time 5018 0 50 150 250 y 00135x 18759 R2 08355 lnClO Time s Figure 2 Plot of lnClO as a function of time Page 1 of 2 CH 102 002 1 60E09 1 40E09 1 20E09 1 00E09 800E08 600E08 400E08 200E08 00 E00 u 50 50 150 250 Time s H H H H H H 1 C10 H y 7E06x 7E07 R2 1 Figure 3 Plot of 1 C10 as a function of time 5 What is the reaction order for C10 a Zero b First c Second d Third 6 Fourth 6 What is the value of the rate constant a 135x10392s391 b 700x106M391s391 c 0751M391s391 d 00983 s391 e 23 M39Zs39l 7 How much time is required for the reaction to be 90 complete ie the concentration of chlorine oxide is onetenth its original concentration of 15 x 10398 M a 105 s d 48 s b 292 s c 479 s e 868 In the stratosphere bromine atoms react with ozone to make bromine monoxide and oxygen Brg 03g BI 0g 02 g The following table contains the rate constant measured at different temperatures The units of rate constant are a little unusual The concentration of bromine atoms was measured as molecules per cubic centimeter Use the data in table 2 and from the graph in Fig 4 to answer the following questions Exam 2 version 1 Spring 2014 Table 2 Temperature dependence of the second order rate constant for the gas phase reaction of bromine atoms with ozone Temperature K k cm3molecule s 238 59 x 103913 258 77 x 103913 278 96 x 103913 298 12 x 103912 274 H u 79203 0004 0005 276 y 83304x 24663 Q77 R2 0999 f 278 279 28 281 282 1T 1K Figure 4 Plot of lnk as a function of UT 8 10 What is the value of the activation energy for the reaction of bromine atoms with ozone a 391 kJrnol b 470 kJrnol c 833 Jrnol d 69 kJrnol e 831 Jrnol The ozone layer is in the lower stratosphere The temperature is 270 K What is the value of the rate constant in cm3molecule at this temperature a 127 x 1019 b 99 x 1013 c 89 X 1013 d 255 e 35200 A catalyst a inhibits a reaction by increasing the activation energy thereby slowing the reaction rate b increases the rate of a reaction by providing a new reaction pathway with a lower activation energy c increases the reaction rate by causing a large change in the enthalpy d has no effect on the rate of a chemical reaction Page 2 of 2 QWUOM gmduz Mew o I 3 9x C 0amp2qu N wwwmjim Mum3 Wl vMe 3 P wonWM W ovuemxwh39m LouWA N we 39 L We 1 013 13 c M vm vow CWWJC mm WWWV WVMCW K M mom 34 K 1amp3quot JVOM39S39 E bazanl g o 301C098 S C 8amp0de WLW Uuu 650Lm0 1563 BMW 0 Iwmo 7 Y5 01 5 W L 39 44 7397 1quot1 59 39 6 oxo I mgk lue C1030 T7 C 1010 K C 3 610020 cachet ITOO SX Oqu 0102 quot r97 YM mo g M K37 K 7 MD b Hi 3 0 xlob vr39S t I S bmm 3096 LL 0 39SKDPa Q SLOW J g39our CamPI a 13 melc gm 6330LL3 13 Tme 039 kImm ID 1 UJov um NH M oomvxg m N ohm1 110 F 13 m WWW 26 OWLZW W1 WWW WWW6 WWW Mm vaw 77m b w M0 is V cM39W m of O mwa 31c 5 W WM we MM W mm m WMWW we WNW Wm WM MM WW WW MW W W A mm m mm mm Ruiz 2 B k 2 J Hz Wm uwmm m E I MAmm wvwmm m om mm 13 39 Q j 39 mum WWV mmm Jaw 39 wmw MWMMW Wm w W Q T WWWNWUX WM 39M DADlb M13 9 I 816 um mAHoE39PcCX dLD g9 K 39 l gz y j g 5 3 3 n v o w owUY 939thme 0 wwww m WV KU J W vawwwmm Tm WWMMNA nsz Mow mammal n WMBNVWWWW O zmmwr mme m WWW WWW W W waw 06 WW m M L WN mm m wwwmmx W WWW 39 WWM WW WWW Mwww 0W mm 1 mst WM 1x MWWM mum a a cvw cam oM W mm WWW L m h ML h 0 0L0 42quot KM 0 as 039 Z hzm 1 1m m MkooWS HMOX Wat 39 m m Avm wmw mm 5W1 ovdw va LCMWUID S n 9 V5 mamax W qum vmmm WWW ohL mmmmw 6 mammm m with walV EN wst MWM WWW MW umeva we pg WNW 7 quot9 ikX Mot S vev E fllo v t WEP XO WNW Swdommxu M39xkcwwmmm xx mmm m wme are A 3 993 v y l amp WK 39 N0 Jc UQ U o 3 M 1 A first order reaction looks like this rkAn when integrated it looks like this lnAtktlnAo Things that affect a 1St order rate include Concentration of reactant increasing concentration increases rate decreasing concentration decreases rate Temperature increased temperature increases rate decreasing temperature decreases rate Catalysts catalysts speed up first order reactions Amount of reactant the rate increases with increasing amounts of reactant A second order reaction looks like this rkA2 when integrated it looks like this 1Atkt 1Ao Things that affect a second order rate include Concentration of reactant increasing concentration increases rate decreasing concentration decreases rate Temperature increasing temperature increases rate decreasing temperature decreases rate Catalysts catalysts speed up second order reactions Amount of reactant the rate increases with increasing amounts of reactant A zero order reaction looks like this rkA0 the integrated form looks like A tk The rate of a zero order reaction only depends on the rate constant However it will speed up if there is a catalyst present or is heated up The concentration does not affect the rate constant for zero order 2 In order to figure out the order of a product or reactant you need to look at the chart and notice trends To find the order for 8208 look at the middle column and see where the concentration for I is constant On the first and third row the concentration OH is 021 so this is where it is constant Now look at the rate on the third column On row one the rate is 167 and the concentration of 8208 is 022 On row three the rate is 333 and the concentration is 044 So the concentration doubled and the rate doubled This means that 8203 is first order 3 Do the same procedure as in number 2 This will give you first order 8 You will need to remember the equation slopeEaRT Slope will come from the graph Eaactivation energy and it is measured in kImol R is the gas law constant which is equal to 83144 Imol K T is the temperature given and must be in Kelvin For this problem the R is in Ioules so make sure you convert your answer to kImol Wt mm 393 WWW m WW gtua L M MW KM mm m xmemmm M W w m n m WM mm mmv 32m Wig W JAR mamas WW 3mm 39 11 XL 39 W L WWW WMQ WWWM W MMMVWWW mm 00 mem mm N MM M wsz 39 NM NM W 6 WM w MW WWW W W Wwwm xg m w awemu w um me t 12 LCA39 WM UR wmmmw M ummmmmx Wu mm mm WMWW I9R W 00 mm M kWM we WY WWW W 9 11 quot iii 3
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