Presentation on theme: "Ch #16: Kinetics Kinetics: Rates and Mechanisms of Chemical Reactions."— Presentation transcript:
1 Ch #16: KineticsKinetics: Rates and Mechanismsof Chemical Reactions
2 Chemical kinetics:Study of reaction rates, changes in concentrations of reactants or products as a function of time.
3 Factors that influence reaction rate: 1.Concentration - molecules must collide to react;Rate α collision frequency α conc2.Physical state - molecules must mix to collide;Smaller the particle size, greater the surface area and more the collisions.3.Temperature - molecules must collide with enough energy to react;Rate α collision frequency α conc.
4 The effect of surface area on reaction rate. Figure 16.3The effect of surface area on reaction rate.
5 Collision energy and reaction rate. Figure 16.4Collision energy and reaction rate.
6 Factors that influence reaction rate: 4.The use of a catalyst.
7 Expressing the Reaction Rate reaction rate - changes in the concentrations of reactants or products per unit timereactant concentrations decrease while product concentrations increaseFor a reaction A → BRate of reaction = -∆[A]/∆t[A]= conc of A in mol/L∆=change
8 Expressing the Reaction Rate Units= moles per liter per second.Mol L-1s-1 or mol/ L.sChange I product conc is positive so the rate is Rate of reaction = ∆[B]/∆tRate decreases during course of reaction.Instantaneous rate: Rate at a particular instant: considering closer values.
9 Expressing the Reaction Rate Use initial rate as soon as the reactants are present( no products at this time)For a reaction aA + bB →cC+ dD,Rate =-1/a∆[A]/∆t=-1/b∆[B]/∆t==1/c∆[C]/∆t==1/d∆[A]/∆t
10 Table 16.1 Concentration of O3 at Various Time in its Reaction with C2H4 at 303KTime (s)Concentration of O3 (mol/L)0.03.20x10-5 (conc A)-t10.02.42x10-520.01.95x10-530.01.63x10-540.01.40x10-550.01.23x10-560.01.10x10-5
11 The concentrations of O3 vs. time during its reaction with C2H4 Figure 16.5The concentrations of O3 vs. time during its reaction with C2H4rate =- [C2H4]t=- [O3]t
12 Plots of [C2H4] and [O2] vs. time. Figure 16.6Plots of [C2H4] and [O2] vs. time.Tools of the Laboratory
13 Problem 1P.1.2H2 (g) +O2 (g)→ 2H2O (g) Write the rate equation in terms of reactants and products. If [O2] is decreasing at 0.23 mol/L .s at what rate is [H2O] increasing?
14 The Rate Law and its components: Rate = k[A]m [B]nK= rate constant(does not change as reaction proceeds)M and n are reaction orders.Coefficients are not related to reaction orders.Rate constant and orders can only be found by experimental data.
15 Determining the Initial Rate: By performing expt and collecting data.
16 Reaction order: Rate=k[A]- 1st order Rate = k[A]2- 2nd order Rate = k[A]0- Zero order OR Rate=k(1)=k (not dependent on conc of A)
17 Problem 2P.2.For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law.2NO(g) + O2(g) → 2NO2(g); rate = k[NO]2[O2]CH3CHO(g) → CH4(g) + CO(g); rate = k[CH3CHO]3/2
19 Determining Reaction Orders: P.3.NO2(g) + CO (g) → NO (g) + CO2(g) rate=k[NO2]m [CO]n
20 Sample Problem 16.3Determining Reaction Order from Initial Rate DataPROBLEM:Many gaseous reactions occur in a car engine and exhaust system. One of these israte = k[NO2]m[CO]nUse the following data to determine the individual and overall reaction orders.ExperimentInitial Rate(mol/L*s)Initial [NO2] (mol/L)Initial [CO] (mol/L)1230.00500.0800.100.400.20PLAN:Solve for each reactant using the general rate law using the method described previously.SOLUTION:rate = k [NO2]m[CO]nFirst, choose two experiments in which [CO] remains constant and the [NO2] varies.
21 Sample Problem 16.3Determining Reaction Order from Initial Rate Datacontinued[NO2] 2[NO2] 1m=rate 2rate 1k [NO2]m2[CO]n2k [NO2]m1 [CO]n1=The reaction is 2nd order in NO2.0.400.10=m0.0800.0050;16 = 4m and m = 2[CO] 3[CO] 1n=rate 3rate 1=k [NO2]m3[CO]n3k [NO2]m1 [CO]n1The reaction is zero order in CO.0.200.10n0.0050=;1 = 2n and n = 0rate = k [NO2]2[CO]0 = k [NO2]2
22 Determining the Rate constant: Calculated from collected data.Units of Rate constant depend on order.
23 Table 16.4 An Overview of Zero-Order, First-Order, and Simple Second-Order ReactionsZero OrderFirst OrderSecond OrderRate lawrate = krate = k [A]rate = k [A]2Units for kmol/L*s1/sL/mol*s[A]t =k t + [A]0Integrated rate law in straight-line formln[A]t =-k t + ln[A]01/[A]t =k t + 1/[A]0Plot for straight line[A]t vs. tln[A]t vs. t1/[A]t = tSlope, y-interceptk, [A]0k, 1/[A]0-k, ln[A]0Half-life[A]0/2kln 2/k1/k [A]0
24 Integrated rate laws and reaction order Figure 16.7Integrated rate laws and reaction order1/[A]t = kt + 1/[A]0[A]t = -kt + [A]0ln[A]t = -kt + ln[A]0
25 Problem 5P.5At 1000 o C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87s-1, to two molecules of ethylene (C2H4).a) If the initial C4H8 concentration is 2.00M, what is the concentration after s?(b) What fraction of C4H8 has decomposed in this time?
26 Rules for orders through graphs If a straight line is produced with the foll:Ln[reactant] vs time-1st1/[reactant] vs time-2nd[reactant]vs time -0 order.
27 Half LifeHalf Life: Time required for the reactant concentration to reach half of its initial value.
28 A plot of [N2O5] vs. time for three half-lives. Figure 16.9A plot of [N2O5] vs. time for three half-lives.t1/2 =for a first-order processln 2k0.693=
29 Problem 6P.6 Cyclopropane is the smallest cyclic hydrocarbon. Because its 600 bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 10000C via a first-order reaction:The rate constant is 9.2s-1; (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value?
30 Effect of Temperature: Temperature affects rate by affecting rate constant.Arrhenius equ: k=Ae-Ea/RTK=rate const, e=base of natural logs, T=absolute temp, R= Universal gas const, Ea=activation energy (minimum energy that molecules must have to react)
31 Effect of Temperature: As T increases, k increases, rate increases.Ln k =ln A- Ea/R(1/T)A plot of ln k vs 1/T gives a straight lineSlope=-Ea/RY-int= ln ALn K2/k1=-Ea/R(1/T2-1/T1)
32 The Effect of Temperature on Reaction Rate The Arrhenius Equationwhere k is the kinetic rate constant at TEa is the activation energyR is the energy gas constantT is the Kelvin temperatureln k = ln A - Ea/RTA is the collision frequency factorlnk2k1=EaRT-1T2T1
33 Figure 16.11Graphical determination of the activation energyln k = -Ea/R (1/T) + ln A
34 Problem 7P.7.The decomposition of hydrogen iodide 2HI(g) → H2(g) + I2(g)has rate constants of 9.51x10-9L/mol*s at 500. K and 1.10x10-5 L/mol*s at 600. K. Find Ea.
35 Figure 16.12Information sequence to determine the kinetic parameters of a reaction.Series of plots of concentra-tion vs. timeInitial ratesDetermine slope of tangent at t0 for each plotCompare initial rates when [A] changes and [B] is held constant and vice versaReaction ordersRate constant (k) and actual rate lawSubstitute initial rates, orders, and concentrations into general rate law: rate = k [A]m[B]nFind k at varied TActivation energy, EaRearrange to linear form and graphRate constant and reaction orderFind k at varied TUse direct, ln or inverse plot to find orderIntegrated rate law (half-life, t1/2)Plots of concentration vs. time
36 Effects of concentration and temperature: Collision Theory: Reactant particles must collide with each other.Concs are multiplied in rate law.Ea: Energy required to activate the molecules into a state from which reactant bonds can change into product bonds.Only those collisions with enough energy to exceed Ea can lead to reaction.
37 The dependence of possible collisions on the product Figure 16.13The dependence of possible collisions on the productof reactant concentrations.AB4 collisionsABABAdd another molecule of A6 collisionsABAABAdd another molecule of BABAB
38 Effects of concentration and temperature: Rise in temp enlarges the fraction of collisions with enough energy to exceed Ea.F= e-Ea/RT e-base of natural log, T=temp, R=gas const.In exothermic reaction: Eaf > EarIn endothermic reaction: Eaf < Ear: k=Ae-Ea/RT A is the frequency factor, A=pZ, Z=collision frequency and p=orientation probability factor.
39 The effect of temperature on the distribution of collision energies Figure 16.14The effect of temperature on the distribution of collision energies
40 Energy-level diagram for a reaction Figure 16.15Energy-level diagram for a reactionCollision EnergyCollision EnergyACTIVATED STATEEa (forward)Ea (reverse)REACTANTSPRODUCTSThe forward reaction is exothermic because the reactants have more energy than the products.
41 Figure 16.17The importance of molecular orientation to an effective collision.NO + NO NO2A is the frequency factorZ is the collision frequencyp is the orientation probability factorA = pZ where
42 Transition state theory: Transition state/activated complex: neither reactant nor product present, a transitional species with partial bonds is present.Ea is the energy required to stretch and deform bonds in order to reach transition state.Reaction energy diagram : TB. Pg.699
43 Nature of the transition state in the reaction between CH3Br and OH-. Figure 16.18Nature of the transition state in the reaction between CH3Br and OH-.CH3Br + OH CH3OH + Br -transition state or activated complex
44 Reaction energy diagram for the reaction of CH3Br and OH-. Figure 16.19Reaction energy diagram for the reaction of CH3Br and OH-.
45 Reaction energy diagrams and possible transition states. Figure 16.20Reaction energy diagrams and possible transition states.
46 Molecularity of a reaction: Elementary steps make up the reaction mechanism.An elementary step is not made up of simpler steps.Molecularity means the number of reactant particles involved in the step.
47 Molecularity of a reaction: In an elementary step equation coeff= order.Reaction order= molecularityA→ product Unimolecular Rate = k[A]2A→ product Bimolecular Rate = k[A]2A + B→ product Bimolecular Rate = k[A][B]2A + B→ product Termolecular Rate =k [A]2[B]
48 Table 16.6 Rate Laws for General Elementary Steps REACTION MECHANISMSTable Rate Laws for General Elementary StepsElementary StepMolecularityRate LawA productUnimolecularRate = k [A]2A productBimolecularRate = k [A]2A + B productBimolecularRate = k [A][B]2A + B productTermolecularRate = k [A]2[B]
49 Problem 8P.8. The following two reactions are proposed as elementary steps in the mechanism of an overall reaction: NO2Cl(g) → NO2(g) + Cl(g) NO2Cl(g) + Cl(g)→ NO2(g) + Cl(g) (a) Write the overall balanced equation.NO2Cl(g) → NO2(g) + Cl(g)NO2Cl(g) + Cl(g)→ NO2(g) + Cl(g)
50 Problem 8 (a) Write the overall balanced equation. Determine the molecularity of each step.Write the rate law for each step
51 Rate-Determining step: Also called as rate limiting step.It represents rate law for overall reaction.Slow reaction is a rate determining reaction.The elementary steps must add up to the overall equation.The elementary steps must be physically reasonable.The mechanism must correlated with the rate law.
52 Catalysis: Each catalyst has its own specific way of functioning. In general a catalyst lowers the energy of activation.Lowering the Ea increases the rate constant, k, and thereby increases the rate of the reactionA catalyst increases the rate of the forward AND the reverse reactions.
53 Catalysis:A catalyzed reaction yields the products more quickly, but does not yield more product than the uncatalyzed reaction.A catalyst lowers Ea by providing a different mechanism, for the reaction through a new, lower energy pathway
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