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1、Chapter 3b. Chemical Kinetics: Reaction RateOutline1.The Rate of a Chemical Reaction2.Theoretical Models for Chemical Kinetics3.Effect of Concentration on Reaction Rates: The Rate Law4.Concentration (reactants and products) and Time5.Effect of Temperature on Reaction Rate6.CatalystTwo questions may
2、be asked in the consideration of a proposed chemical process:1.“Is the reaction spontaneous under the given conditions?” - it is answered by chemical thermodynamics2.“Under what conditions will the reaction proceed rapidly enough for it to be practicable?” - it is answered by chemical kineticsThermo
3、dynamics says NOTHING about the rate of a reaction.Thermodynamics : Will a reaction occur ?Kinetics : If so, how fast ?A reaction may have a large, negative Grxn, but the rate may be so slow that there is no evidence of it occurring.Kinetic Vs. ThermodynamicConversion of graphite to diamonds is a th
4、ermodynamic favor process ( G -ve ).C (graphite) C (diamond) Kinetics makes this reaction nearly impossible (Requires a very high pressure and temperature over long time)Ammonia ProductionAutomobile ExhaustReaction between H2 and O21.The Rate of a Chemical Reactionu The reaction rate is defined as t
5、he change in the concentration of a reactant (or a products) in a unit of time.Units: mol dm3 s1, mol dm3 min1 or mol dm3 h1Average RateaA + bB yY + zZFor a general reactionGeneral Rateu If the time interval is very short, the equation gives the instantaneous rate - that is, the rate at a particular
6、 of timeInstantaneous Rate can only be obtained by experimental determination.2N2O5 4NO2+O2 (45 )400 800 1200 1600 2000 时间时间 (s)v=5.410-4 mol dm-3 s-1v=2.710-4 mol dm-3 s-10.201.000.800.600.40c(N2O5)/mol dm-3400 800 1200 1600 2000 时间时间 (s)v=5.410-4 mol dm-3 s-1v=2.710-4 mol dm-3 s-10.201.000.800.600
7、.40c(N2O5)/mol dm-3A AB Bc(N2O5)/moldm-3 v/mol dm-3 s-1 0.90 5.410-4 0.45 2.710-4Example 3b-1： Reaction 2W+X Y+ZWhich expression is correct for the rate of reaction?2. Theoretical Models for Chemical Kinetics1.Collision Theory - 1918, LewisFundamental points The molecules must collide in order for a
8、 reaction to take place between molecules. Most collisions do not result in reaction. Only a fraction of collisions among molecules lead to chemical reaction- effective collision (?). u Molecules with sufficient energy whose molecular collisions are most likely lead to chemical reaction - “activated
9、” moleculesu Molecules must collide in a certain orientationKinetic EnergyFraction of molecules having a certain KEFor a reaction to occur on the collision of two molecules, the molecules must collide: with sufficient energy with a certain orientation Ee: The average kinetic energy Ec: The minimum e
10、nergy of “activated” moleculesEa: Activation energyMolecular collisions and chemical reactions(a) Favorable collision(b) Unfavorable collisionNO2(g) + CO(g) NO(g) + CO2(g)Molecular collisions and chemical reactions2. Transition State TheoryA + B-C A B C* A-B + Creactants(initial state)activated comp
11、lex (transition state) products(finial state)NO2(g) + CO(g) NO(g) + CO2(g)EtONOCOE*NO2 + COEreactantEaEaNO + CO2EproductProgress of reactionPotential EnergyABCH The energy of reactants must pass the potential energy barrier (activated energy). （1） + （2）：）： NO2 + CO NO + CO2 NO2 + CO OONOC rH（1）= Ea
12、NO + CO2OONOC rH（2）= Ea rH = rH（1）+ rH（2） = Ea Ea t E rH = rH（1）+ rH（2） = Ea Ea If Ea Ea, rH 0endothermicEt If Ea Ea, rH 0exothermicTransition state The energy of reactants must pass the potential energy barrier (activated energy). Even for an exothermic reaction (rH0), the minimum energy (Ea) is st
13、ill needed.Example 3b-2: In the following figure, which point represent the reaction intermediates? Which point represent the activated complex?1234P Po ot te en nt ti ia al l e en ne er rg gy yProgress of Reaction3. Effect of Concentration on Reaction Rates: The Rate Law3.1 Rate law - the rate law
14、for a chemical reaction is an equation which links the reaction rate with concentrations or pressures of reactants.aA + bB cCvi = ki Aa Bb1.Rate law for an elementary reactionu k (rate constant) - depends on the specific reaction and the reaction temperatureu a and b stand for coefficients in the ba
15、lance equation. u a+b, the overall order of reactionLaw of Mass Action2.Rate Laws for non-elementary reactionsp Most reaction do not proceed in a single step; instead, they go through a series of steps. For example, H2(g) + I2(g) =2HI(g), this reaction occurs in two elementary steps: I2(g) = I(g) +
16、I(g)H2(g) + I (g) + I(g) = HI(g) + HI(g)pFind the rate determining step via experiments A2+B A2B A2 2A Slow reaction 2A+BA2B Fast reactionNon-Elementary Rate Laws can only be established using experimental data! v = k c(A2)Example 3b-3. The conversion of ozone (O3) to oxygen (O2) is represented as:
17、2O3(g) 3O2(g). This reaction occurs in two elementary steps: O3 O2+O （fast） O+O3 2O2（slow）Which one is the rate expression of this reaction?a.kc2 (O3)(O3)c(O)b.c. kc(O3)2c(O2) d. kc2 (O3) c-1(O2)2 Na(l) + 2 H2O(l) 2 NaOH(aq) + H2(g)C(s) + O2(g) CO2(g) k (rate constant)elementary reaction aA + bB cC
18、vi= ki Aa Bbu k is a function of Tu The larger the value of k, the faster a reaction goes. u The unit of k depend on the order of reaction.Keep in mind about the Rate Law The concentration of pure solid and liquid is a constant which dont express in rate law.Rate of reaction = kRate of reaction = k
19、c(O2) For the reaction involving gaseous substance, we can sometimes work directly with the partial pressure to replace the mole concentration in the rate Law .Rate of reaction = k p(O2)（k is different with k）C(s) + O2(g) CO2(g) Whether a reaction is elementary process or not can only be determined
20、by experiments. For a non-elementary process, the rate laws can only be established using experimental data!aA + bB dD +eEExperimentally determined rate law： V k (A)m(B)n3.2 Order of reactionm is the order of with respect to An is the order of with respect to B m + n is the overall orderma, n b, m +
21、 n abFor elemental processDefinition of reaction orderUnit of k depend on the order of the reaction u Zero-Order Reaction: = k(cA)0 = k Unit: moldm-3s-1u First-Order Reaction : = kcA Unit: s-1u Second-Order Reaction : = k(cA)2 Unit: mol-1dm3s-1u 2/3-Order Reaction : = k(cA)3/2 Unit: ？Can you determi
22、ne the order of reaction via the unit of k?Example 3b-4: For elemental reaction CO (g) + NO2 (g) CO2(g) + NO (g), the rate law is represented as: V k (CO)(NO2)Question： What is the reaction order with respect to CO? What is the reaction order with respect to NO2? What is the overall reaction order?H
23、ow do we establish the rate law? - experimental data Method of initial ratesExample 3b-5: Use the following data establish the rate law and the order of reaction with respect to H2 and NO and also the overall order of the reaction.2H2 +2 NO = 2H2O + N2experimentInitial c/moldm-3initial rate of N2v/m
24、oldm-3s-1c(NO)c(H2)1234566.0010-3 1.0010-36.0010-3 2.0010-36.0010-3 3.0010-31.0010-3 6.0010-32.0010-3 6.0010-33.0010-3 6.0010-3 3.1910-36.3610-39.5610-30.4810-31.9210-34.3010-3Solutionp In comparing experiment 1, 2 and 3: N2 H2p In comparing experiment 4, 5 and 6: N2 NO2N2 H2 NO2N2 = k H2 NO2= 8.86
25、104 dm6mol-2s-14. Concentration (reactants and products) and Time1. Zero-Order reactionA B The A t plot is a straight line Slope = -k, intercept = A0 d A = k dtd A = k dtA = A0 ktIntegrated rate lawdifferential rate law A = = kd AdtHalf-life (t1/2) of a reaction is the time required for one-half of
26、a reactant to be consumed. whenHalf-lifeA = A0 ktIntegrated rate lawt = A0 2k2. First-Order reactionA BA = k A = k Ad Adt= k dtd A A= k dtd A AlnA lnA0 = kt lgA lgA0 = t k2.303 The lgA t plot is a straight line lg = tAA0k2.303whenlgA lgA0 = t k2.303A B3. Second-Order reaction The plot is a straight
27、line k c0 1t = 4. Third-Order reactionA B The plot is a straight line t =2 k c02 3Table A Summary of Reaction Kinetic for reaction A B c0interceptslope- slope- slopek lnA tA tstraight linelnA lnA0 = kt A = A0 ktintegrated rate equation k A2k Akrate LawSecondFirstZeroOrderExample 3b-6: The data liste
28、d below were obtained for the decomposition reaction (A products). (a) Establish the order of the reaction. (b) Whats the rate constant, k? (c) What is the half-life if CA0 = 1.00 M?t/min05101525c/(moldm-3)lnc/(moldm-3)(1/c)/(mol-1dm3)1.000.001.000.63-0.461.60.46-0.782.20.36-1.022.80.25-1.394.0Solut
29、ion(a) Plot the following three graphs c t (if a straight line, reaction is zero order.) lnc t (if a straight line, reaction is first order.) 1/c t (if a straight line, reaction is second order.)c tlnc t1/c tOnly plot (c) is a straight line. The reaction is second order. rate = k c2(A) (b) The slope
30、 of graph (c)(c) 5. Effect of Temperature on Reaction Rate5.1 Arrhenius (阿仑尼乌斯阿仑尼乌斯) equation In 1889, Svante Arrhenius demonstrated that the rate constants of many chemical reactions vary with temperature in accordance with the expression:l A is called frequency factor l Ea is the activation energy
31、 l R is the universal gas constant 8.314 Jmol-1K-1l T is the Kelvin temperature Graphical Method5.2 Application of Arrhenius equation Determination Ea of a reaction at any temperatureY = aX + blgk 1/T(lgk1, 1/T1)(lgk2, 1/T2) Determination based on calculationThe rate constants are k1 and k2 for a re
32、action at two different absolute temperatures, T1 and T2, respectively.Eliminating the constant lgAEaAExample 3b-7. Use data for reaction 2N2O5(g) 4NO2(g) + O2(g) to determine the activation energy Ea. T() T(K) 1/T(K-1) k(s-1) ln(k) 20 293 3.4110-3 2.0 10-5 -10.82 30 303 3.30 10-3 7.3 10-5 -9.53 40
33、313 3.19 10-3 2.7 10-4 -8.22 50 323 3.10 10-3 9.1 10-4 -7.00 60 333 3.00 10-3 2.9 10-3 -5.84 Solutionlnk 1/T Ea = - R (slope) = - (8.314JK-1mol-1)(-1.2 104 K) =1.0 105Jmol-16.1 Catalysis is the increase in rate of a reaction as the result of addition of a catalyst, which is not consumed itself.6. Ca
34、talyst The compose and properties of a catalyst will not change before and after reaction. A catalyst can not effect the spontaneity of a reaction. It can only speeds up a reaction with G0. A catalyst increase (decrease) the rate of forward and reverse reaction with the same multiple. However, it do
35、es not affect the enthalpy changeH. The selectivity of a catalyst The catalyst take part in the reaction mechanism and provide a new pathway for the reaction with a lower activation energy (for both forward and reverse reaction).6.2 Homogeneous Catalysis and Heterogeneous Catalysis Heterogeneous Cat
36、alysis : The catalyst is present in a different phase of matter than are the reactant and products. u Homogeneous catalysis is a sequence of reactions that involve a catalyst in the same phase as the reactants. Most commonly, a homogeneous catalyst is co-dissolved in a solvent with the reactants. N2
37、(g) + 3 H2(g) 2 NH3(g) N2(g) N N 2N / / Molecules of H2 and N2 are adsorbed on the Fe surface resulting in the breaking of N-N and H-H followed by the formation of Fe-N and Fe-HNH3 desorbs into the gaseous state.Then the adsorbed atoms combine to form NH3.H2(g) HH 2H / / N + H NH / / /NH + H NH2 / /
38、 /NH2 + H NH3/ / /NH3 NH3/ The hydrogenation involving ethylene 废气从汽车引擎是经过废气从汽车引擎是经过 接触反应的转换到最小化环境损害接触反应的转换到最小化环境损害Principle of converterobjectCatalyst converter for treatment of mobile Emission. Amazingly efficient very specific body temperature Enzymes as catalystsEnzyme catalysis is the catalysis
39、 of chemical reactions by specialized proteins known as enzymes. HomeworkH3b-1: The initial rate of a hypothetical reaction, A+BC, was measured using different initial concentrations of A and B. Results are summarized in the table below. (a) What is the rate law for the reaction? Determine the value
40、 of the rate constant, k for the reaction. (b) What will be the initial rate of the reaction if the initial concentration of A = 0.050 moldm-3 and the initial concentration of B = 0.020moldm-3experimentInitial c/moldm-3initial rate of Cv/moldm-3s-1AB1230.030 0.0100.060 0.0100.030 0.020 1.710-86.810-
41、84.910-8H3b-2: At 338 K, for the first-order decomposition of 2 N2O5 (g) = 2NO2 + O2, the rate constant, k is 0.292 min-1. (a) What is the half-life for this reaction at 338 K? (b) How long will it take for the concentration of N2O5 to decrease to 14% of its original value? (c) What will the remaining percentage of N2O5 be after 1020s ?H3b-3: The first-order reaction A products has a half-life, t1/2, of 46.2 min at 25 and 2.6 min at 102 . (a) Calculate the activation energy of this reaction. (b) At what temperature would the half-life be 10.0 min?
