Professor's Notes on Chemical Kinetics and Rates

RATES OF REACTIONCHEMICAL KINETICS:-STUDY REACTION RATES-HOW THESE RATES CHANGE DEPEND ONCONDITIONS-DESCRIBES MOLECULAR EVENTS THAT OCCUR DURING THE REACTION.VARIABLES EFFECTING REACTION RATES:-REACTANT-CATALYST-TEMPERATURE-SURFACE AREAFACTORS THAT INFLUENCE REACTION RATESI.CONCENTRATION:MOLECULES MUST COLLIDE IN ORDER FOR A REACTION TO OCCUR.II.PHYSICAL STATE:MOLECULES MUST BE ABLE TO MIX IN ORDER FOR COLLISIONS TO HAPPEN.III.TEMPERATURE:MOLECULES MUST COLLIDE WITH ENOUGH ENERGY TO REACT.VARIABLE WHICH AFFECT REACTION RATES-REACTANTS:rate as [ ] general[ ] no effect on rate-CATALYST:a substance that increases the rate of Rx without being consumed in overall RxMnO42H2O22H2O + O2-TEMPERATURE:rate as T, thus less time to boilan egg @sea level than in mountains-SURFACE AREA OF SOLID REACTANT/CATALYST:rate as surface area, pieces of wood will burn faster than whole trunks, area = rate of RxRATE OF REACTION-DESCRIBES THE INCREASE IN MOLAR P(PRODUCTS)OF A REACTION PER UNIT TIME-DESCRIBES THE DECREASE IN R(REACTANTS) PERUNIT TIMER = [P]R = [R]t tQ. 2H2O2 2H2O + O2R = ?-RATE OF REACTION CAN BE REFERRED TO AS THEINSTANTANEOUS OR AVERAGE RATESCHEMICAL KINETICSThe study of reaction rates that is, the study of reactant(and/or product) concentrations as a function of time.For example: Given 2 O3(g)302(g)the rate of disappearance of ozone is related how to therate of formation of oxygen? Give the rate law.The rate law is dependent on stoichiometry.For example: If the rate of appearance of O2is[O2]= 6.0 x 10-5M/s at a particular instant,t what is the value of the rate of disappearance of O3at the same time?CHEMICAL KINETICSREACTION RATES & STOICHIOMETRY1.How is the rate of disappearance of ozone relatedto the rate of appearance of oxygen in the followingequation:203(g)302(g)R = -1[O3] = 1[O2]2 t 3 t2.If the rate of appearance of O2; [O2] = 6 x 10-5M/stat a particular instant, what is the value of the rateof disappearance of O3; -[O3] at the same time?t-[O3] = 2[O2] = 2(6.0 x 10-5M/s)t 3 t 3 = 4 x 10-5M/sQUESTION:The decomposition of N2O5, proceeds 2N2O5(g)4NO2(g)+ O2(g)If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7M/s, what is the rate of appearance of NO2?What is the rate of appearance of O2?RATE LAW (RATE EQUATION)R = k [A]m[B]n….For aA + bB + …. = cC + dD +….k = rate constant (at constant temperature; the rate constant does not change as thereaction proceeds.)m, n = reaction orders (describes how the rate is affected by reactant concentration)note: a & b are not related to m & nnote: R, k, & m/n are all found experimentallyREACTION ORDER1. What are the overall reaction orders for:A. 2N2O5(g)4NO2(g)+ O2 (g)R = k[N2O5]B. CHCl3(g)+ Cl2(g)CCl4(g)+ HCl(g)R=k[CHCl3] [Cl2] 1/2The overall reaction order is the sum of the powers to whichall the [reactants] are used in the rate law.A. Is 1st order & 1storder overallB. 1storder in [CHCl3], 1/2order in [Cl2]; overall = 3/22. What are the usual units of the rate constant for the rate law for a? Units of rate = (units of k) (units of [ ])units of k = units of rate= M/s= s-1unit [ ] MQ: what is the reaction order of H2?H2(g)+ I2(g)2HI(g)TR=k [H2][I2]Q: what is the units of the rate constant?THE EXPERIMENTAL RATE1.Calculated by measuring the [Products] as the reaction proceeds.2.Calculated by measuring the change in pressure if one of the products is a gas.3.Colorimetry uses Beer’s law: A= -Log 1/TEXPERIMENTAL DETERMINATION OF RATE1.Calculate [product] as Rx proceeds (slow Rx)2.If a Gas, use P (manometer)3.ColorimetryDEPENDENCE OF RATE ON CONCENTRATIONAn equation that relates the Reaction to the [reactants] or to a [catalyst] raised to a powerRate = k [H2O2]nINITIAL RATE METHOD1. The initial rate of a reaction AB was measured for severaldifferent starting concentrations of A & Btrail[A][B]R(m/s)10.1000.1004 x 10-520.1000.2004 x 10-530.2000.10016 x 10-5a. Determine the rate law for the reactionb. Determine the magnitude of the rate constant.C. Determine the rate of the reaction when [A] = 0.030M & [B] = 0.100MA particular reaction was found to depend on the concentration of the hydrogen ion, [H+]. The initialrates varied as a function of [H+] as follows:[H+]R0.05006.4 x 10-70.10003.2 x 10-70.20001.6 x 10-7a. What is the order of the reaction in [H+]b. Predict the initial reaction rate when [H+] = 0.400MHOW DOES CONCENTRATION CHANGE WITH TIME?A B + CR = k [A] is the rate lawso the rate of decomposition of A can be written as:-d [A]= k [A]dtINTEGRATED RATE LAWSFirst-order reaction: A B R = k[A]1N [A]t= -kt[A]oSecond-order reaction: R = k[A]21 -1 = +kt[A]t[A]oZero-order reaction: R = k[A]t-[A]o= -ktCONCENTRATION WITH TIME1.The first-order rate constant for the decomposition of certain insecticide in water at 12°C is 1.45 year-1. A quantity of this insecticide is washed into a lake in June, leading to a concentration of 5.0 x 10-7g/cm3 of water. Assume that the effective temperature of the lake is 12°C.A. What is the concentration of the insecticide in June of the following year?B. How long will it take for the [Insecticides] to drop to 3.0 x 10-7g/cm3?Cyclopropane is used as an anesthetic. The isomerization of cycloproprane () to propene is first order with a rate constant of 9.2 s-1@1000°C.A. If an initial sample of has a concentration if 6.00 M, what will the concentration be after 1 second?B. What will the concentration be after 1 second if the reaction was second order.HALF-LIFE-The time it takes for the reactant concentration to decrease to half it’s initial value.1st order2nd ordert1/2= 0.693t1/2= 1kk[A].Q1. The thermal decomposition of N2O5to form NO2& O2is 1st order with a rate constant of 5.1 x 10-4s-1at 313k. What is the half-life of this process?Q2. At 70°C the rate constant is 6.82 x 10-3s-1suppose we start with 0.300mol of N2O5, how many moles of N2O5will remain after 1.5 min.?Q3. What is the t1/2of N2O5at 70 °C?HALF LIFE answers2 N2O54 NO2+ O21. k313= 5.1 x 10-4s-1t 1/2 = ?t1/2 = .693/k -.693/5.1 x 10-4s-1t1/2 = 1358.8t 1/2 = 1.4 x 103s2. k70 = 6.82 x 10-3s-1[N2O5]I = 0.300mol[N2O5]t = ?t = 1.5min (60 s)min In[A] = -kt[A]. [A] = [A.] e-kt(.300)e -6.82 x 10-3(90s)= .693/6.82 x 10-3s-1 = 0.162mol = [A]t3. = .693/6.82 x 10-3s-1= 102 sec = 1.69 minRATE AND TEMPERATUREArrhenius Equationk= Ae-Ea/RTR = 8.31 J/K molEa = activation energyT = absolute temperatureA = frequency factorIf two temperatures are compared:In k1= Ea(1-1)k2R T2T1H3C-N =C: H3C -C=N:methyl isonitrile acelonitrileFor the conversion of methyl isonitrile to acetonitrile, thetable below shows the relationship between temperature and the rate constant.Tk1.98.9°C5.25 x 10-5230.3°C6.30 x 10-4251.2°C3.16 x 10-31. Calculate Ea.2. What is k at 430.3 K?COLLISION THEORYA theory that assumes that Reactant particles must collide with an energy greater than some minimum value and with proper orientation.Ea -Activation EnergyMinimum energy of collision required for 2 particles toreactk = zfpz = collision frequencyf = fraction of collisions w/e > Eap = fraction of collisions w/proper orientationNO(g)+ Cl2(g)NOCl(g)+ Cl-(g)Experimentally observed rate constantsk25°C= 4.9 x 10-6L/molsk35°C= 1.5 x 10-5L/mols* Generally a 10°C will double or triple the rate. Thereexists a strong dependence on temperature.1.The collision frequency (z) is proportional to 3RT/MM (rms) temperature dependent.2. The fraction of collisions greater than Ea (f) xe-Ea/RTtemperature dependentTRANSITION STATE THEORYExplains the reaction resulting from the collision of 2particles in terms of an activated complex.Activated Complex-an unstable group of atoms which break upto form the products of a chemical reaction.O = N + Cl -Cl [O = N….Cl….Cl] O = N -Cl + ClThe energy transferred from the collision (KE) is localizedin the bonds (….) of the activated complex as vibrational motion. At some point the energy in the (….) bond becomes so great resulting in the (….) bond breaking.ELEMENTARY REACTIONS-Describes a single molecular event such as a collisionof molecules resulting in a reaction.REACTION MECHANISM-A set of elementary reactions whose overall effectis given by the Net Chemical equation.REACTION INTERMEDIATE-A species produced during a reaction that does notappear in the Net equation. The species reacts in a subsequent step in the mechanism.MOLECULARITYThe number of molecules on the reaction side of an elementary reaction.Unimolecular:1 reactant moleculeA PBimolecular:2 reactant moleculesA + B PTermolecular:3 reactant molecules2A + B P1. Br + Br + Ar Br2+ Ar*2. O3* O2+ O3. NO2+ NO2NO3+ NOC Cl2F2decomposes in the stratosphere from irradiationwith short UV light present at that altitude. The decompositionyields chlorine atoms. This atom catalyzes the decomposition of O3in the presence of O-atoms.Classify the following:I.l. C Cl2F2CF2Cl • + Cl•2. Cl(g)+ O3(g)ClO•(g)+ O2(g)ClO(g)+ O(g)Cl•(g)+ O2(g)O3(g)+ O(g)2O2(g)II.H2O2(l)+ FeCl3(ag)H2O(l)+ FeO+FeO++ H2O2 H2O + O2+ Fe3+2H2O2 2H2O + O2REACTION MECHANISM1. The elementary steps must add up to the overall equation.2. The elementary steps must be physically possible.Termolecular is rare3. The mechanism must correlate with the rate law.Rate-determining step:This is the elementary step that is slowest and therefore limits the rate for the overall reaction.The rate law for the rate determining step isthe rate law for the overall reaction.THE RELATIONSHIP BETWEEN THE RATE LAW AND MECHANISMThe actual mechanism can not be observed directly. It must be devised from experimental evidence and scientific method.Q1.2O3(g)3O2(g)overall Rxproposed mechanism:O3k1O2+ Ofastk-1k2O3+ O 2O2slowwhat is the rate law?Q2.H2O2+ I-H2O + IO-IO-+ H2O2H2O + O2+ I-What is the rate law?Q3.Q2 is the mechanism at 25°C but at 1000°Cthe first equation is faster than the second.Now what is the rate law?Q1. overall reaction:Mo(CO)6+ P(CH3)3Mo(CO)5P(CH3)3+ COProposed mechanism:MO(CO)6 Mo(CO)5+ COMO(CO)5+ P(CH3)3MO(CO)5P(CH3)31. Is the proposed mechanism consistent with the equation for the overall reaction?2. Identify the intermediates?3. Determine the rate law.Q2. A) Write the rate law for the following reaction assumingit involves a single secondary step.2NO(g)+ BR2(g)2 NOBr(g)B) Is a single step mechanism likely for this reaction?CATALYSISA Catalyst speeds up the reaction without being consumed.-biological catalyst EnzymesHow does a catalyst work?-A catalyst is an active participant to a reaction. It either affects the frequency of collisions (A) orit may decrease the activation energy (Ea)Homogeneous catalyst:-The catalyst is in the same phase as the reactant.Heterogeneous catalyst:-The catalyst is in a different phase from the reactants. Physical Absorption:-Weak intermolecular forcesChemisorption:-Binding of species to surface by Intramolecular forces