15 Chemical KineticsChemical kinetics: the study of reaction rate, a quantity conditions affecting it, the molecular events during a chemical reaction (mechanism), and presence of other components (catalysis).Factors affecting reaction rate: Concentrations of reactants Catalyst Temperature Surface area of solid reactants or catalystWhat quantities do we study regarding chemical reactions?

Reaction Rate DefinedReaction rate: changes in a concentration of a product or a reactant per unit time.[ ] concentration Reaction rate = t[ ]t[ ]tchangeDefine reaction rate and explain Average reaction rate Instantaneous reaction rate (2 tangents shown) Initial reaction rate

Expressing reaction ratesFor a chemical reaction, there are many ways to express the reaction rate. The relationships among expressions depend on the equation.Note the expression and reasons for their relations for the reaction 2 NO + O2 (g) = 2 NO2 (g) [O2] 1 [NO] 1 [NO2] Reaction rate = = = t 2 t 2 tMake sure you can write expressions for any reaction and figure out the relationships. For example, give the reaction rate expressions for2 N2O5 = 4 NO2 + O2How can the rate expression be unique and universal?

Calculating reaction rateThe concentrations of N2O5 are 1.24e-2 and 0.93e-2 M at 600 and 1200 s after the reactants are mixed at the appropriate temperature. Evaluate the reaction rates for2 N2O5 = 4 NO2 + O2Solution: (0.93 1.24)e-2 0.31e-2 M Decomposition rate of N2O5 = = 1200 600600 s= 5.2e-6 M s-1.Note however, rate of formation of NO2 = 1.02e-5 M s-1. rate of formation of O2 = 2.6e-6 M s-1.The reaction rates are expressed in 3 formsBe able to do this type problems

Determine Reaction RatesTo measure reaction rate, we measure the concentration of either a reactant or product at several time intervals.The concentrations are measured using spectroscopic method or pressure (for a gas). For example, the total pressure increases for the reaction:2 N2O5 (g) 4 NO2 (g) + O2(g)Because 5 moles of gas products are produced from 2 moles of gas reactants. For the reactionCaCO3 (s) CaO(s) + CO2 (g)The increase in gas pressure is entirely due to CO2 formed.barometer

Differential Rate LawsDependence of reaction rate on the concentrations of reactants is called the rate law, which is unique for each reaction.For a general reaction, a A + b B + c C productsthe rate law has the general formorder wrt A, B, and C, determined experimentally reaction rate = k [A]X [B]Y [C]Z the rate constantFor example, the rate law israte = k [Br-] [BrO3-] [H+] for 5 Br- + BrO3- + 6 H+ 3Br2 + 3 H2OThe reaction is 1st order wrt all three reactants, total order 3.Use differentials to express rates

Variation of Reaction rates and OrderFirst order, rate = k [A]k = rate, 0th order[A]rate2nd order, rate = k [A]2The variation of reaction rates as functions of concentration for various order is interesting.Mathematical analysis is an important scientific tool, worth noticing.[A] = ___?

Differential Rate Law determinationEstimate the orders and rate constant k from the results observed for the reaction? What is the rate when [H2O2] = [I-] = [H+] = 1.0 M?H2O2 + 3 I- + 2 H+ I3- + 2 H2OExprmt[H2O2][I-][H+]Initial rate M s-1 10.0100.0100.00501.15e-6 20.0200.0100.00502.30e-6 30.0100.0200.00502.30e-6 40.0100.0100.01001.15e-6Solution nextLearn the strategy to determine the rate law from this example.Figure out the answer without writing down anything.

Differential Rate Law determination - continueEstimate the orders from the results observed for the reaction H2O2 + 3 I- + 2 H+ I3- + 2 H2OExprmt[H2O2][I-][H+]Initial rate M s-1 10.0100.0100.00501.15e-6 20.0200.0100.00502.30e-6 1 for H2O2 30.0100.0200.00502.30e-6 1 for I- 40.0100.0100.01001.15e-6 0 for H+1.15e-6 = k [H2O2]x [I-]y [H+]z 1.15e-6 k (0.010)x(0.010)y(0.0050)z exprmt 111 ----------- = ------------------------------------- ----= --- 2.30e-6 k (0.020)x(0.010)y(0.0050)z exprmt 22 2 x = 1( )x

Differential Rate Law determination - continueOther orders are determined in a similar way as shown before. Now, lets find k and the rateThurs, rate = 1.15e-6 = k (0.010)(0.010) from exprmt 1 k = 1.15e-6 M s-1 / (0.010)(0.010) M3 = 0.0115 M-1 s-1And the rate law is therefore, d [H2O2] k rate == 0.0115 [H2O2] [I-]a differential rate law d ttotal order 2The rate when [H2O2] = [I-] = [H+] = 1.0 M:The rate is the same as the rate constant k, when concentrations of reactants are all unity (exactly 1), doesnt matter what the orders are.

Differential Rate Law determination continueThe reaction rate d[H2O2]/dt = 0.0115 [H2O2] [I], for H2O2 + 3 I- + 2 H+ I3- + 2 H2O What is d[I]/dt when [H2O2] = [I] = 0.5?Solution: Please note the stoichiometry of equation and how the rate changes. d[I]/dt = 3 d[H2O2]/dt = 3* 0.0115 [H2O2] [I] = 0.0345 * 0.5 * 0.5 = 0.0086 M s-1In order to get a unique rate constant k, we evaluate k for the reaction a A + b B product this way rate = -1/a d[A]/dt = -1/b d[B]/dt = k [A]x [B]yNote the reaction rate expression and the stoichiometry of equation.

Differential Rate Law determination - continueFrom the following reaction rates observed in 4 experiments, derive the rate law for the reaction A + B + C products where reaction rates are measured as soon as the reactants are mixed.Expt 1 2 3 4 [A]o 0.100 0.200 0.200 0.100 [B]o 0.100 0.100 0.300 0.100 [C]o 0.100 0.100 0.100 0.400 rate 0.100 0.800 7.200 0.400Solution nextThis example illustrates the strategy to determine, and a reliable method to solve rate-law experimentally.

Differential Rate Law determination - continueFrom the following reaction rates, derive the rate law for the reaction A + B + C products where reaction rates are measured as soon as the reactants are mixed.Expt 1 2 3 4 order [A]o 0.100 0.200 0.200 0.100 3 from expt 1 & 2 [B]o 0.100 0.100 0.300 0.100 2 expt 1, 2 & 3 [C]o 0.100 0.100 0.100 0.400 1 expt 1 & 4 rate 0.100 0.800 7.200 0.4000.800 k 0.2x 0.1y 0.1z ----- = ---------------------- 0.100 k 0.1x 0.1y 0.1zTherefore 8 = 2x log 8 = x log 2 x = log 8 / log 2 = 3Assume rate = k [A]x[B]y[C]z

Integrated Rate Laws concentrations as functions of timeOne reactant A decomposes in 1st or 2nd order rate law.Differential rate lawIntegrated rate law d[A] / dt = k[A] = [A]o k t d[A] = k [A][A] = [A]o e k t or ln [A] = ln [A]o k t d t d[A] 1 1 [A] conc at t = k [A]2 = k t d t [A] [A]o [A]o conc at t = 0Describe, derive and apply the integrated rate lawsLearn the strategy to determine rate-law

Concentration and time of 1st order reactionDescribe the features of plot of [A] vs. t and ln[A] vs. t for 1st order reactions. Apply the technique to evaluate k or [A] at various times.[A]tln[A]t[A] = [A]o e k tln [A] = ln [A]o k tt

Half life & k of First Order DecompositionThe time required for half of A to decompose is called half life t1/2.Since[A] = [A]o e k t or ln [A] = ln [A]o k tWhen t = t1/2, [A] = [A]oThus ln [A]o = ln [A]o k t1/2 ln 2 = k t1/2k t1/2 = ln 2 = 0.693 relationship between k and t1/2Radioactive decay usually follow 1st order kinetics, and half life of an isotope is used to indicate its stability.Evaluate t from k or k from t

1st order reaction calculationN2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t and percent decomposed in 500 s.Solution nextIf the rate-law is known, what are the key parameters?

1st order reaction calculationN2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t and percent decomposed in 500 s.Solution: Assume [A]o = [N2O5]o = 1.0, then [A] = 0.9 at t = 30 s or 0.9 = 1.0 e k t apply [A]o = [A] e k t ln 0.9 = ln 1.0 k 30 s 0.1054 = 0 k * 30 k = 0.00351 s 1 t = 0.693 / k = 197 s apply k t = ln 2 [A] = 1.0 e 0.00351*500 = 0.173 Percent decomposed: 1.0 0.173 = 0.827 or 82.7 %After 2 t (2*197=394 s), [A] = ()2 =, 75% decomposed. After 3 t (3*197=591 s), [A] = ()3 =1/8, 87.5% decomposed.Apply integrated rate law to solve problems

Typical Problem wrt 1st Order ReactionThe decomposition of A is first order, and [A] is monitored. The following data are recorded:t / min0248 [A]/[M]0.1000.09050.08190.0670Calculate k (What is the rate constant? k = 0.0499) Calculate the half life (What is the half life? Half life = 13.89) Calculate [A] when t = 5 min. (What is the concentration when t = 5 min?) Calculate t when [A] = 0.0100. (Estimate the time required for 90% of A to decompose.)Work out all the answers

A 2nd Order ExampleDimerization of butadiene is second order:2 C4H6(g) = C8H12(g).The rate constant k at some temperature is 0.100 /min. The initial concentration of butadiene [B] is 2.0 M. Calculate the time required for [B] = 1.0 and 0.5 MCalculate concentration of butadiene when t = 1, 5, 10, and 30.Solution nextIf the rate-law is known, what are the key parameters?Apply the right model and work out all the parameters.

A 2nd Order ExampleDimerization of butadiene is second order:2 C4H6(g) = C8H12(g).The rate constant k at some temperature is 0.100 /min. The initial concentration of butadiene [B] is 2.0 M. Calculate the time t required for [B] = 1.0 and 0.5 MCalculate concentration of butadiene when t = 1, 5, 10, and 30.1 1 = k t [B][B]o1 1 [B][B]o t = k[B]o [B] = [B]o k t + 1 t = 1510153035[B] = 1.671.00.670.500.290.25Work out the formulas and then evaluate values

Half life of 2nd Order Chemical Kineticst = 1510153035[B] = 1.671.00.670.500.290.251/[B]How does half life vary in 2nd order reactions?

Plot of [B] vs. t & 1/[B] vs. t for 2nd Order Reactionst = 1510153035[B] = 1.671.00.670.500.290.25[B]1 [B]tt1 1 = k t [B][B]o[B]o [B] = [B]o k t + 1 What kind of plot is linear for 1st and 2nd reactions?

Chemical Reaction and Molecular CollisionMolecular collisions lead to chemical reactions. Thus, the reaction constant, k is determined by several factors.k = Z f pZ: collision frequency p, the fraction with proper orientation f, fraction of collision having sufficient energy for reaction f is related to the potential energy barrier called activation energy, Ea.f e Ea / RT or exp ( Ea / R T)Thus, k = A e Ea / RTconstantEaPotential energyreactionconstantHow does temperature affect reaction rates?Explain energy aspect in a chemical reaction

Energy in chemical reactionsPotential energyProgress of reactionR + A P + DH exothermicEa Ea for reverse reactionEndothermic rxnRA-PD activated complexExplain the various terms and energy changes in a chemical reaction

The Arrhenius Equation1903 Nobel Prize citation in recognition of the extraordinary services he has rendered to the advancement of chemistry by his electrolytic theory of dissociationThe temperature dependence of the rate constant k is best described by the Arrhenius equation:k = A e Ea / R T or ln k = ln A Ea / R TIf k1 and k2 are the rate constants at T1 and T2 respectively, thenk1 Ea 1 1 ln = k2 R T1 T2 How does temperature affect reaction rates? Derive and apply these relationship to solve problems, and recall the Clausius-Clapeyron equation.

Application of Arrhenius EquationFrom k = A e Ea / R T, calculate A, Ea, k at a specific temperature and T.The reaction:2 NO2(g) -----> 2NO(g) + O2(g)The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11?Method: derive various versions of the same formulak = A e Ea / R TA = k e Ea / R T A / k = e Ea / R T ln (A / k) = Ea / R TMake sure you know how to transform the formula into these forms.

Application of Arrhenius Equation (continue)The reaction:2 NO2(g) -----> 2NO(g) + O2(g)The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11?Use the formula derived earlier:A = k eEa / R T = 1e-10 s-1 exp (111000 J mol-1 / (8.314 J mol-1 K 1*300 K)) = 2.13e9 s-1k = 2.13e9 s-1 exp ( 111000 J mol-1) / (8.314 J mol-1 K 1*273 K) = 1.23e-12 s-1T = Ea / [R* ln (A/k)] = 111000 J mol-1 / (8.314*46.8) J mol-1 K-1 = 285 K

The Effect of Temperature on Reaction Rates Reaction rate = k [A}x[B]y[C]z (Concentration effect at constant T)k = A exp ( Ea / RT)(Temperature effect)Use graphic method to discuss the variation of k vs. T variation of k vs 1 / T variation of ln(k) vs T variation of ln(k) vs 1 / T

See a potential multiple choice question in an exam?

Elementary Reactions and MechanismElementary reactions are steps of molecular events showing how reactions proceed. This type of description is a mechanism. The mechanism for the reaction between CO and NO2 is proposed to beStep 1NO2 + NO2 NO3 + NO(an elementary reaction)Step 2NO3 + CO NO2 + CO2(an elementary reaction)Add these two equations led to the overall reactionNO2 + CO = NO + CO2(overall reaction)A mechanism is a proposal to explain the rate law, and it has to satisfy the rate law. A satisfactory explanation is not a proof.Explain terms in red

Molecularity of Elementary ReactionsThe total order of rate law in an elementary reaction is molecularity.The rate law of elementary reaction is derived from the equation. The order is the number of reacting molecules because they must collide to react.A molecule decomposes by itself is a unimolecular reaction (step); two molecules collide and react is a bimolecular reaction (step); & three molecules collide and react is a termolecular reaction (step).O3 O2 + Orate = k [O3]NO2 + NO2 NO3 + NOrate = k [NO2]2Br + Br + Ar Br2 + Ar*rate = k [Br]2[Ar]Caution: Derive rate laws this way only for elementary reactions.

Molecularity of elementary reactions - ExampleSome elementary reactions for the reaction between CH4 and Cl2 are Cl2 2 Cl 2 Cl Cl2 2Cl + CH4 Cl2 + CH4* Cl + CH4 HCl + CH3 CH3 + Cl CH3ClCH3 + CH3 CH3-CH3CH3Cl + Cl HCl + CH2ClCH2Cl + Cl CH2Cl2* * * (more)Write down the rate laws and describe them as uni- bi- or ter-molecular steps yourself, please.

Elementary Reactions are Molecular EventsN2O5 NO2 + NO3 Explain differences between elementary and over reaction equations

Rate Laws and MechanismsA mechanism is a collection of elementary steps devise to explain the the reaction in view of the observed rate law. You need the skill to derive a rate law from a mechanism, but proposing a mechanism is task after you have learned more chemistry For the reaction, 2 NO2 (g) + F2 (g) 2 NO2F (g), the rate law is,rate = k [NO2] [F2] .Can the elementary reaction be the same as the overall reaction?If they were the same the rate law would have been rate = k [NO2]2 [F2],Therefore, they the overall reaction is not an elementary reaction. Its mechanism is proposed next.

Rate-determining Step in a MechanismThe rate determining step is the slowest elementary step in a mechanism, and the rate law for this step is the rate law for the overall reaction.The (determined) rate law is,rate = k [NO2] [F2], for the reaction, 2 NO2 (g) + F2 (g) 2 NO2F (g), and a two-step mechanism is proposed: i NO2 (g) + F2 (g) NO2F (g) + F (g) ii NO2 (g) + F (g) NO2F (g) Which is the rate determining step?Answer: The rate for step i is rate = k [NO2] [F2], which is the rate law, this suggests that step i is the rate-determining or the s-l-o-w step.Explain rate determining step in a mechanism and use it to derive the rate law.

Deriving a Rate Law From a Mechanism - 0The decomposition of H2O2 in the presence of I follow this mechanism,iH2O2 + I k1 H2O + IO slow ii H2O2 + IO k2 H2O + O2 + I fastWhat is the rate law? EnergySolve the problemEaiEaiireaction

Deriving a rate law from a mechanism - 1The decomposition of H2O2 in the presence of I follow this mechanism,iH2O2 + I k1 H2O + IO slow ii H2O2 + IO k2 H2O + O2 + I fastWhat is the rate law?SolutionThe slow step determines the rate, and the rate law is:rate = k1 [H2O2] [I ]Since both [H2O2] and [I ] are measurable in the system, this is the rate law.

Deriving a rate law from a mechanism - 2Derive the rate law for the reaction, H2 + Br2 = 2 HBr, from the proposed mechanism: i Br2 2 Brfast equilibrium (k1, k-1) iiH2 + Br k2 HBr + H slow iii H + Br k3 HBr fastSolution: The fast equilibrium condition simply says that k1 [Br2] = k-1 [Br]2 and[Br] = (k1/k-1 [Br2]) The slow step determines the rate law, rate = k2 [H2] [Br] Br is an intermediate = k2 [H2] (k1/k-1 [Br2]) = k [H2] [Br2] ; k = k2 (k1/k-1) M- s -1 total order 1.5explain

Deriving a rate law from a mechanism - 3The decomposition of N2O5 follows the mechanism: 1N2O5 NO2 + NO3fast equilibrium 2NO2 + NO3 k2 NO + O2 + NO2slow 3NO3 + NO k3 NO2 + NO2fast Derive the rate law.Solution:The slow step determines the rate, rate = k2 [NO2] [NO3] NO2 & NO3 are intermediate From 1, we have [NO2] [NO3] = KK, equilibrium constant [N2O5] K differ from kThus, rate = K k2 [N2O5]

Deriving rate laws from mechanisms steady-state approximationThe steady-state approximation is a general method for deriving rate laws when the relative speed cannot be identified. It is based on the assumption that the concentration of the intermediate is constant.Rate of producing the intermediate, Rprod, is the same as its rate of consumption, Rcons.Rprod = Rcons[Intermediate]timeRprod < RconsRprod > RconsBe able to apply the steady-state approximation to derive rate laws

Steady-state approximation - 2Lets assume the mechanism for the reaction. H2 + I2 2 HI as follows. Step (1)I2 k1 2 I Step (1)2 I k-1 I2 Step (2) H2 + 2 I k2 2 HI Derive the rate law. Derivation: rate = k2 [H2] [I] 2 (cause this step gives products) but I is an intermediate, this is not a rate law yet. Since k1 [I2](= rate of producing I) = k-1 [I]2 + k2 [H2] [I]2(= rate of consuming I)Thus, k1 [I2] [I]2 = k-1 + k2 [H2] rate = k1 k2 [H2] [I2] / {k-1 + k2 [H2] }Steady state

Steady-state approximation - 3From the previous result:k1 k2 [H2] [I2] rate = {k-1 + k2 [H2] }Discussion:(i) If k-1

Steady-state approximation - 4In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S2O82- + SO32- + 2 OH- 3 SO42- + H2O. The following mechanism has been proposed: i S2O82- + SO32- k1 S2O72- + SO42- ii S2O72- + H2O k2 2 SO42- + 2 H+ iii H+ + OH- k3 H2O (fast equilibrium to be discussed)Steady-state approximation follows these steps:What is or are the intermediates I? Use which step to give the rate law that may involve [I]? Express the rates of producing and consuming intermediate(s) Express [I] of intermediate(s) in terms of [Reactants] Derive the rate law in terms of [Reactants]DiscussSee page 607 PHH Text

CatalysisA catalyst is a substance that changes the rate of a reaction by lowing the activation energy, Ea. It participates a reaction in forming an intermediate, but is regenerated. Enzymes are marvelously selective catalysts.A catalyzed reaction, NO (catalyst) 2 SO2 (g) + O2 2 SO3 (g) via the mechanism i2 NO + O2 2 NO2(3rd order) iiNO2 + SO2 SO3 + NOUncatalyzed rxnCatalyzed rxnrxnEnergy

Catalyzed decomposition of ozoneR.J. Plunkett in DuPont discovered carbon fluorine chlorine compounds.The CFC decomposes in the atmosphere: CFCl3 CFCl2 + Cl CF2Cl3 CF2Cl + Cl.The Cl catalyzes the reaction via the mechanism: iO3 + h v O + O2, ii ClO + O Cl + O2 iii O + O3 O2 + O2. The net result or reaction is 2 O3 3 O2 Scientists sound the alarm, and the CFC is banned now.

Homogenous vs. heterogeneous catalystsA catalyst in the same phase (gases and solutions) as the reactants is a homogeneous catalyst. It effective, but recovery is difficult.When the catalyst is in a different phase than reactants (and products), the process involve heterogeneous catalysis. Chemisorption, absorption, and adsorption cause reactions to take place via different pathways.Platinum is often used to catalyze hydrogenationCatalytic converters reduce CO and NO emission.

Heterogeneous catalystsCeryx's vision is to design, produce, and commercialize advanced systems that balance Cost, Performance, Emissions Reduction, and Fuel Penalty to make the economics of pollution control viable. We explore new ways to look at the air quality challenges faced by industry and search for potential solutions by combining proven technologies with state-of-the-art science.Catalyzed reactions:CO + O2 CO2 2 NO N2 + O2

Enzymes selective catalystsEnzymes are a long protein molecules that fold into balls. They often have a metal coordinated to the O and N sites. Molecules catalyzed by enzymes are called substrates. They are held by various sites (together called the active site) of the enzyme molecules and just before and during the reaction. After having reacted, the products P1 & P2 are released.Enzyme + Substrate ES (activated complex)ES P1 + P2 + E Enzymes are biological catalysts for biological systems.

X-ray 3-D structure of fumarate reductase. It reduces fumerate, an important role in the metabolism of anaerobic bacteria, from Max Planck Inst.

Chemical Kinetics - SummaryExplain how the various factors affect reaction rates.Define reaction rates, average rates, initial rates and rate constants. Evaluate rate law from experimentsProperly apply 1st and 2nd differential rate laws and integrated rate laws.Interpret elementary reactions and mechanisms. Derive rate laws from a given mechanism. Apply the steady-state method to derive the rate law of a given mechanism, and discuss the results.Explain the action of catalysts in terms of chemistry and in terms of energy of activation.