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# GEN COLLEGE CHEMISTRY II CHE 107

UK

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This 15 page Class Notes was uploaded by Michelle Gulgowski DVM on Friday October 23, 2015. The Class Notes belongs to CHE 107 at University of Kentucky taught by Harriet Ades in Fall. Since its upload, it has received 34 views. For similar materials see /class/228291/che-107-university-of-kentucky in Chemistry at University of Kentucky.

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Date Created: 10/23/15

Ch 13 Chemical Kinetics Or how fast reactions occur Chemical Kinetics We are going to look at 1 how fast chem rxns occur 2 how to express the rate of rxn as function of concs of reacting species 3 mathematics of different rate eqns 4 how rxns might occur on molecular level 5 how one can speed up rxn 132 The rate of a chemical rxn Rate change per unit time as miles per hour gallons per minute In chemistry we look at how concentrations of reactants and products change over time So in chemistry rate is frequently measured in terms of changes in concentrations as molLtime or VItime Mtime39lor mol L39l time391 and rate Rates are A means conc of A in Wm M mum lmumw mwwmwm N WWW Rate of Reaction A molecules 5 molecules Number al molecules 132 Look atA gt B Initially the com of A decreases mpidly and the com of B increases mpidly then changes slow down Why The mathematical expression for the rate of disappearance of A 1s rate 7 AA Where AA At A 1 there a minus sign for the mte of disappearance of A I And the expression for the rate of appearance of Bis I Iate AB At where AB B 39 B11 I Why is there no minus sign for this rate I Look at 2HZOZaq gt 2HZO1 02g by studying the rate of disappearance of H202 time s HZOZJM 232 0 300 1 86 600 1 49 1200 0 98 1800 0 62 3000 0 25 I time 5 At HZOZM AM 0 232 I gt300 gt 046 300 1 I 6 I gt300 gt037 I 600 149 I gt600 gt 051 I 1200 0 98 I gt gt 036 I 1800 62 I 1200 gt 037 I 3000 0 25 000123 000085 000060 000031 I What if study rate of appearance of 02g timelts 021M 0 0 300 0 23 600 0 42 1200 0 67 1800 0 85 3000 1 04 times m 021M AM AMAt 0 0 gt300 gt0 23 30 023 gt Ms 0 00077 0 00063 0 00042 0 00030 0 00016 I Compare Iates M s1 39 39 AinOz A02 At At I 153X10393 077X10393 I 123X10393 063X10393 I 085X10393 042X10393 I 060X10393 030X10393 I 031X10393 016X10393 I These are average rates over the indicated time interval To calc instantaneous rate rst plot COHC vs time and then draw a tangent to the line at that time The slope of the line at that point is the instantaneous rate nu l u zmm Reaction Rates and Stoichiometry How does one report the rate of a rxri ForaA bB cC dD we de ne rate So for ZHZOZ 4 ZHZO 02 39 39AinOz A02 At At 39 153X10393 077X10393 39 123X10393 063X10393 39 085X10393 042X10393 39 060X10393 030X10393 39 031X10393 016X10393 39 391 AinOz Aioz 39 2At At or A229 ZAioz At At For the oxidation of Br39 by BrO339 in acid solution 513p Br0339 6H a 3BrZ 3HZO If Br is reacting at a rate of020 Mmin at sometime what are the rates of appearance and disappearance of the other species at the same time What is the rate of appeamnce of H1 at the time when H2 is reacting at arate of 04 Mrnin HZ 12 9 2H1 133 The Rate Law I Object is to develop a mathematical relationship btn rate and concs of various species know as rate eqn law I for reactants gtproducts We say I or rate k AXBYCz Where k is the rate constant speci c rate constant and xy andz are nos to be determined from expt A B and C are species involved in the rxn Some information about the rate constant and rate law I 1 Numerical value of k corresponds to a speci c munder a speci ed set of conditions and at a speci ed temp I 2 kis a k does not change with time or with concentmtion changes However the mte of the rxn changes When the reactant concentrations change I 3 k does change if the temp changes I 4 The value of kmust be determined by expt I 5 The values of X y and z bear no relationship to the stoichiometric coef cients Their values must be determined from expt I 6 Xy and z can be I For a rate law where rate k AKBYCZ I 7 say x is the order ofrxn with respect to A u u u u u u u u C Z Some sample rate laws 39 Hzg 12g H 2H1g rate kHz 12 I Hzg 12g gt 2Hlg high P Au catalyst rate k 2Nzo5 a 4N0 02 rate kNZOS 39 2N0g t 02 gt 2N02g rate kNOZOZ CHC13g Clzg gtCCl4g Hc1g rate kCHCl3ClZ Z 39 Hzg Brzg gt ZHBrg rate kHZBrZ Z l jHBrBrz Units of rate constant I Remember units of mte are Mtime I zero order mte I rst order rate I second order rate I third order mte I fourth order rate Effect of conc on rate Remember that conc changes do not affect the value of the rate constant but do affect rate For 2N0 BrZ gt 2NOBr experiments show that the rxn is second order inNO and rst order in Brz How does rate change if conc of Br2 is changed from 00022M to 00066M and the conc ofNO is unchanged Instead What ifNO is changed by from 00022M to 00066M BrZ doesn t change How do we nd the rate law expression from an experiment Method of initial rates I Study initial rate as a function of conc of reactants catalysts products etc I For reactants gtproducts I use initial rates because a as concs ofreactants dec rates can be dif cult to determine rxns can be reversible as products areactants I These complicate matters slogzr 217 2so4zr 1Z Expt Initial concI initial rate Ms 1 0 038 0060 14X10 5 2 0 076 0060 2 8X10 5 3 0076 0030 14 X 10395 Determine the rate law and the value of k What is the rate when 2082 0050M and 139 0025M I A B gt C I Expt A1I BMI A C AtMs initial I 1 010 020 500 I 2 020 020 1000 I 3 010 015 281 I Determine the rate law etc Determine the rate law for formation of phosgene C0 C1Z gt COClZ initial concM init rate ML Expt C0 C12 1 0053 023 37X10395 2 0106 023 74X10395 3 0106 046 10410395 I Consider the rxn A B gt products I From the following determine the rxn order I Expt initial conc1I initial rate Ms A B I 1 150 150 0320 I 2 150 250 0320 I 3 300 150 0640 134Facts about different order rate laws and integrated rate laws 1 zero order A rate kA0 k rate is a B units of k C rate AAAt k or ldA kldt and so A A kt or A kt A y mx b 39 Alf kt A10 I y mx b I plot A vs t get straight line whose slope is k and y intercept is AO slope k A time I halflife tiZ is time required for cone of reactant to decrease to half ofits original cone or A at tlZ AL2 I A kt AD and at tlz I AL2 ktlZ AD or 39 t12 2k I the tlZ depends on the initial cone ofA for zero order mte law and as the initial cone of A increases the t1 2 increases I 11 First order I rate I units of rate constant I rate AAAt kA I dAA kf dt I 1nAt1nAO kt or 1nAoA kt or1nAAo kt I 1nA 1nAO kt 1nA kt 1nAo y mx b I plot 1nA vs time get straight line whose e k slope k y intercept 1nA 39 1HAlAlo 39kT 39 at tlZ A Alo2 39 1HAo2Ao kt1Z or 39 1H2 0693 ktlZ Note that the halflife is indep of conc It is a constant for a given rxn under given conditions NalfLIfe in a FirstOrder Reanian Al I I 0 km mim m mm W my I halihle I m AlwrZ hall lMs Arm 3 ml hm Imm Second order rate units of rate constant integrated rate law lAt kt lAo y mx b plot 1 A vs time get straight line whose slope k MA slope k yintercept 1A 39 t12 17 kAo halflife depends on the initial conc for second order rate law The t1 2 increases as the initial conc of A decreases 1mm m husmuryhhl u mm mm mm um mu m m w u m kw m humm A y uw To determine rxn order l plot A vs time if straight line zero order 2 if not zero order plot lnA vs time if straight line rst order 3 if not zero or rst order plot 1 A vs time if straight line is second order n 111039 111w L A umzr r hm unisr luu 1 l A 1 H4 39 R Huo my Inn mdw u 004 mm 417er m k A39 i 13 5 THm m r 13 6 m Problems The rxn CZH5C1g gt CZH4g HClg is rst order in CZHSCI with a rate constant equal to 160 X 10 s 1 at 650 K Ifone starts with an initial conc of CZH5C1 of 0500 M what will the conc be a er 125 hours How long will it take the conc of CZHjCl to drop to 0150 M A rst order rxn is 255 complete in 490 rnjn at 25 C What is the rate constant and halflife The rxn 4PH3 gt P4 6HZ is rst order in PH3 What percent is le a er 151 s ifthe halflife is 45 s at 680 C The rate constant for the secondorder rxn 2NOBrg gt 2NOg Brzg is 080M39lS39l at 10 C Starting with a conc of086M calc the conc of NOBr a er 22s How much NO and BrZ have been formed at this point in time What are the halflives when the initial conc ofNOBr is 086 M 043 M The decomposition of NZO5 at 335 K follows rstorder kinetics Starting with 250 g ofNZO5 at 335 Kit is found that 150 g remains a er 109 s What mass of NZO5 will remain after 300s 135 Activation Energy and Temperature Dependence of Rate Constants Collision Model or how might these rxns occur on arnolecular level How might Hzgg 12g gt 2Hlg on a molecular level To react molecules must collide with each other amp with walls of the container Molecules that move have KE Energy transfer can occur on collision More reactant molecules presentmore collisions can occur So rate on collisions When molecules collide some of the KB is changed into vibrational energy of the bonds Sometimes there is enough energy gained through collision to break a bond and initiate an If the energy is not enough to break the bond the molecules bounce of one another with no DCH occurring Some reactant molecules gain enough enough through collisions to break chemical bonds and start the rxn Not all collisions lead to rxn just those that have enough energy to initiate rxn The molecules must also collide in a correct orientation to react NOCllglANKXIlu zNolgidTUw V Ineffective collision Ineffective collision J Tective collision J There is some minimum collision energy below which no rxn occurs Minimum energy above room temperature energy needed for rxn to start occuran Generally the smaller the value of the activation energy the faster the rxn Ezlsa Activation Energy 11H i Ujg zngoiw mvmimnmiu u g it iv V rm m n m i dnnunn n IIKI Pmuuu its Minn pmgn Energy oi veanants Energy at pmdum s dCimn mm I Increase temperature IXH faster Why mm WWHWWWWW mm quotquotquot 39 Potential Energy Profiles for Exothermic a d Endolhermic Reactions I More molecules VVlll have mama saggy nmmmiun Mum my complex complex H a a E 5 E gt g A a 39g E n quot Azwanrm o S my t I C 4 D A A B amazon wows Reaciiun pogltss a b i exothermic endothennic I k Aexp EzRT I at constant temp as Ea T eXpE2 RTL k i I at constant E2 as T T expE2RT TkT The Arrhenius equation I experimental eqn k AexpE2RT I Larger the value of k the faster the rxn I A collision frequencyinumber of times reactants approach activation barrierunit time I expEaRT fmction of approaches that lead to reaction get over activation barrier and form products I The frequency factor A pz in collision theory I p orientation factor iread about on p 587 I F collision frequency more moleculesvolume more collisions Arrhenius equation k AexpEzRT lnk EaR1T 1nA mx b plot ln k vs 1T get straight line whose slope Ea unitskJmol R 8314x10393kJKmol Ynmpnmnneml Rwle unsmnHM Ms m w 7 ramp me an CnnslnHNr39s n m 7 h m r Slope yrinxercepl s 39 LIZXIO A l we n I l I 0 00005 quot00 00m in AK m u Comparative form of the Arrhenius eqn lnkZk1 EaR 1T11TZ If the rxn rate doubles when the temp is raised from 25 C to 45 C what is the activation energy for the rxn I When heated to a high temp cyclopropane is convened to propene The activation energy is 270kJmol Lfthe rate constant is 110 x10quot s39l at 470 C What s the mte constant at 510 C The rxn CZHSBr gt CZH4 HBr in the gas phase is rst order in CZHSBr The rate constant is 20 X 10395 s 1 at 377 C and the activation energy is 226 klmol At what temp 1s the rate constant 60 10395 S4 15 X10 s 1Calc rate constant at 25 C Ea568 kJmol and the frequency factor is 136 Reaction Mechanisms and 137 Catalysis Experimental rate law for CH33CC1 OH39 gtCH33COH C139 is rate kCH33CCl and for 39 Hzg 1265 2H1 is rate lez 12 Why does the rate law sometimes agree with the stoichiometry and sometimes it doesn t on molecular level Balanced eqn doesn t tell us how rxn occurs A mechanism has been proposed for the previous rxn lCH33CCl gtCH33C Cl39 slow 2CH33COH39 gt CH33COH fast These steps are madeup and represent how someone thinks the ECU might happen Mechanism made up of a series of elementary steps that represent what might occur in the course of a rxn The mechanism is a theoretical construct 2 rules for mechanisms 1 sum of steps of mechanism must agree with the overall stoichiometry 2 rate law predicted by mechanism must agree with experimental rate law For elementary step can write rate law from rxn Also talk about molecularity of steps A gt products rate kA A B gt products rate kAB or A A gt products rate kAZ A B C 4 products ratekABCt Order is experimental concept molecularity is theoretical concept Back to lCH33CCl gtCH33C Cl39 slow Ratel k1CH33CC1 2CH33C OH39 gt CH33COH fast Ratez kzrltCH33cOH39 CH33CC1 OH39 gtCH33COH or overall In a mechanism some steps are slow steps Slow step is known as the rate determining step When guring out rate law from mechanism forget about all steps a er slow step The faster steps don t affect the rate Slow step lCH33CCl gtCH33C Cl39 slow Rate1 k1 CH33CC1 2CH33C OH39 gt CH33COH fast RateZ kzCH33COH Step 1 is the rate determining step and agrees with the experimental rate law rate kCH33CCl Energy Diagram lor a TwoStep Mamanism l Hrrgv may 1 immunity mm Mum r mm lcy l ummmu m amva um i Jumnww Mail i H naanams 34 Pmdum m iv v New 1 imum warm Catalyst speeds up rate of rxn by giving a new path for rxn to occur This new path has a lower activation energy A catalyst is not consumed in ECU and does not appear in the balanced eqn Homogeneous and heterogeneous catalysis Enzymes are biological catalysts In a mechanism catalyst rst appears as Comparison oft he Activation Energy Barriers of an Uncatalyzed Reaction Same Reaction with 3 Catalyst Potential 1an s o nncllnn woman a Pmemial enemy a M Human mama M Reaction intermediates species that you create to explain Ixn as activated complex It may not exist in real life for measurable time Rxn intermediate ri does not appear in balance rxn lCH33CCl gtCH33CJr Cl slow 2CH33COH39 a CH33COH fast CH33CC1 overall Identify rxn i OH gtCH33COH Cl ntermediates and catalysts N20 is relatively unreactive at room temp but at 600 C decomposes 2Nzoe 2NZ 02 Mechanism 1 N20 gt N2 o 20NZO gt NZOZ Ea for rxn 24016 A catalyst lowers the activation energy by giving a new path for the rxn as In presence small amts of C12 1 C12 gt 2C1 2 N20 c1 gt NZ C10 2a N20 c1 gt N2 C10 3 2c1o gt c1Z 02 18 140kJmol ZNZO gt 2NZ 02 Rxn 34 X 1017 at 298K times faster with lower Ea The activation energy for the rxn B 12C is 250 kJmol The enthalpy change is 388ldmol What is the activation energy for the reverse rxn Mechanism 1 c12 gt 2C1 2 C1 CO gtCOC1 3 COC1C1 gt COClZ What s the balanced rxn ri catalyst In recent years ozone in the stratosphere has been depleted at an alarming rate by CFC s A CFC molecule as CFC13 is first decomposed by uv rad n CFC13 gt CFC1Z c1 The C1 radical reacts with 03 a C103 gt CIOOZ bCIOOe C1OZ What are roles of Cl and C10 and overall rxn oflast two steps Decomposition of H202 ZHZOZ gt ZHZO 02 Add KI catalyst H202 139 gt H20 01 H20Z or gt H20 02 17 Overall ZHZOZ gt ZHZO 02 The rate law for the rxn 2HZ 2N0 gt N2 ZHZO is rate kHZNOZ Which ofthe following mechanisms can be ruled out 1 H2 NO 9 H20 N slow N NO 9 N2 0 fast 0 H2 gt HZO fast 11 H2 2N0 gt N20 H20 slow N20 H2 a N2 H20 fast What if have a fast initial step as the following gas phase reaction C12 3 2C1 fast equilibrium c1CHc13 gt Hc1cc13 slow C1 CCI3 gt CCI4 fast TO increase fXIl rates 1 Increase conc of reactants that appear in rate aw 2 Increase temp 3 Increase surface area of reactants stir etc 4 Add a catalyst The halflife of tritium is 123yr How much time must pass before the mass of a sample of tritium is reduced to 5 of its original value Radioactive decay follows a rstorder mte law

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