《化学原理Chemistry课件-post+7+kinetics+》由会员分享，可在线阅读，更多相关《化学原理Chemistry课件-post+7+kinetics+（31页珍藏版）》请在金锄头文库上搜索。
1、Chemical Kinetics & Equilibrium Chapter 13 & 14p How fast or slow is the reaction? p Which one is the rate-determining step? Nosaka et al. 2006. J. Phys. Chem. B.Kubo et al. 2006. J. Am. Chem. Soc.1. Reaction rate:rate = -DADtrate = DCDtaA + bB cC + dDrate = -DADt1a= -DBDt1b=DCDt1c=DDDt1dCH4 (g) + 2
2、O2 (g) CO2 (g) + 2H2O (g)rate = -DCH4Dt= -DO2Dt12=DH2ODt12=DCO2DtBr2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)393 nmlightDetector393 nmBr2 (aq)I0ItUV-visible absorption spectroscopic spectrumA: absorbance e e : molar coefficient (cm-1.M-1)l : light path length (cm) C: concentration (M)It = I0
3、 e - e e l CLambert-Beer Lawaverage rate = -DBr2Dt= -Br2final Br2initialtfinal - tinitialslope oftangentslope oftangentslope oftangentinstantaneous rate = a rate for specific instance in timeInitial rate is the rate measured in the initial period of time, so as to avoid possible interference from re
4、action intermediates or products. rate = k Br2k = 3.50 x 10-3 s-1 rate constantSlope = 3.50 x 10-3 s-1This reaction is called as first-order to Br2.This is called the Rate law.aA + bB cC + dDRate = k Ax ByIt means that the reaction is the (x + y)th order, the xth in A, and the yth in B.Note that:p T
5、he rate laws are always determined by experiment.p X or y can be any value, but it does not necessarily equal to a or b (x a, y b) In general,F2 (g) + 2ClO2 (g) 2FClO2 (g)As F2 is constant, quadruples ClO2, the rate is quadrupled.y = 1Assume: rate = k F2x ClO2yAs ClO2 is constant, doubles F2, the ra
6、te is also doubled.y = 1Therefore: rate = k F2x ClO2yDetermine the rate law and calculate the rate constant for the following reaction from the following data:S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq)ExperimentS2O82-I-Initial Rate (M/s)10.080.0342.2 x 10-420.080.0171.1 x 10-430.160.0172.2 x 10-4
7、Assume: rate = k S2O82-x I-yWe see that x= 1; y = 1k = rateS2O82-I-=2.2 x 10-4 M/s(0.08 M)(0.034 M)= 0.08/Msrate = k S2O82-I-The reaction is the 2rd order.First-Order ReactionsA productrate = -DADt= k ADADt= k A-A0 and A are the concentration of A at t = 0 and t = t, respectively.A = A0exp(-kt)The h
8、alf-lifelnA0A0/2k=tln2k=lnA = lnA0 - ktThe reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?lnA = lnA0 - ktt =lnA0 lnAk= 66 sA0 = 0.88 MA = 0.14 MlnA0Ak=ln0.88 M0.14 M2.8 x 10-2 s-1=What is the half-life
9、of the reactant A in this reaction ?tln2k=Second-Order ReactionsA productrate = -DADtDADt= k A2-A0 and A is the concentration of A at time t =0, and t = t, respectively.1A=1A0+ ktt =1kA0IfZero-Order ReactionsA productrate = k A0 = kA = A0 - ktDADt= k-t =A02kSummary of Common Reaction KineticsOrderRa
10、te LawConcentration-Time EquationHalf-Life012rate = krate = k Arate = k A2lnA = lnA0 - kt1A=1A0+ ktA = A0 - kttln2k=t =A02kt =1kA02. Effect of Temperature lnk = -1TB + CThe rate of most chemical reactions increases as the temperature rises. The experimental data of such effect can be well fitted to
11、the following equation. p The value of Ea varies from reaction to reaction (40 400 kJ/mol).p Larger the value of Ea, stronger the temperature effect on the reaction rate. Ea is the minimum amount of energy required to react. It is now called as activation energy .A is the frequency factor.R is the g
12、as constant (8.314 J/Kmol).T is the absolute temperature (K).In 1888, Arrhenius proposed an empirical equation to describe the temperature effect.How to explain the concentration and temperature effect?(1) The collision modelThe reacting molecules must collide with one another. The colliding molecul
13、e should have a certain minimum energy (Ea).Eap As temperature increases, the fraction of active molecules increases. So the reaction rate increases with T.p As concentration or pressure increases, the total number of active molecules increases, and so the reaction rate increases.The colliding molec
14、ules must form an activated complex in a preferable direction. In this transition state, the effective energy transfer and chemical bond rearrangement occur at one time. (2) Transition state modelCO2(g)+NO(g)CO(g)+NO2(g)OCONOProgress of reactionPotential energy kJEa = 134 kJEa = 360 kJD DH = - 226 k
15、JExothermic ReactionEndothermic ReactionA + B C + D3. Reaction MechanismsAt molecular level, the observed chemical reaction (called also as complicated reaction ) is actually composed of a series of simple elementary steps or reactions.Elementary step or reaction is the step or reaction that really
16、occurs during the chemical reaction.2NO (g) + O2 (g) 2NO2 (g)N2O2 is detected during the reaction, called as the intermediateElementary step:NO + NO N2O2Elementary step:N2O2 + O2 2NO2Overall reaction:2NO + O2 2NO2+Complicated reaction:Based on this information, one proposed the following reaction me
17、chanism:Unimolecular reactionA productsrate = k ABimolecular reactionA + B productsrate = k ABBimolecular reactionA + A productsrate = k A2For elementary steps, the rate raw can directly be written as. Writing plausible reaction mechanisms:The sum of the elementary steps must give the overall balanc
18、ed equation for the reaction.The rate-determining step should predict the same rate law that is determined experimentally.The rate-determining step is the slowest step in the sequence of steps leading to product formation.The experimental rate law for the reaction between NO2 and CO to produce NO an
19、d CO2 is rate = kNO22. The reaction is believed to occur via two steps:Step 1:NO2 + NO2 NO + NO3Step 2:NO3 + CO NO2 + CO2What is the equation for the overall reaction?NO2+ CO NO + CO2What is the intermediate?NO3What can you say about the relative rates of steps 1 and 2?rate = kNO22 is the rate law f
20、or step 1 so step 1 must be slower than step 24. Chemical equilibriumequilibriumN2O4 (g) 2NO2 (g)Rate(+)Rate(-) At the equilibrium, p Rate(+) = Rate (-) p Both N2O4 and NO2 remain constant.p it is a dynamic equilibriumKc = NO22N2O4Equilibrium constantKp = NO2P2N2O4PKc KpHere P (pa), Po = 101 325 paH
21、ere C (mol/L), Co = 1 mol/LaA + bB cC + dDIt is quite often written as If pressure is in a unit of atm,Kp = Kc(RT)DnFor a gas reaction, here Dn= (c + d) (a + b)The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are CO = 0.012 M, C
22、l2 = 0.054 M, and COCl2 = 0.14 M. Calculate the equilibrium constants Kc and Kp.CO (g) + Cl2 (g) COCl2 (g)Kc = COCl2COCl2=0.140.012 x 0.054= 220Kp = Kc(RT)DnDn = 1 2 = -1R = 0.0821T = 273 + 74 = 347 KKp = 220 x (0.0821 x 347)-1 = 7.714.2The concentration of solids and pure liquids are taken as a uni
23、t, and not included in the K expression.14.2CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)Kc =CH3COO-H3O+CH3COOHH2OSince H2O = constantCaCO3 (s) CaO (s) + CO2 (g)Kc =CaOCO2CaCO3Since CaCO3 = constant CaO = constantKc = CO2Kp = PCO2Kc = CH3COO-H3O+CH3COOH= Kc H2OConsider the following equilibrium at
24、 295 K:The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction?NH4HS (s) NH3 (g) + H2S (g)Kp = PNH3H2SP= 0.265 x 0.265 = 0.0702Kp = Kc(RT)DnKc = Kp(RT)-DnDn = 2 0 = 2T = 295 KKc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-414.2The factors influencing the chemical equilibrium:
25、 (1) Increasing A or B will shift the reaction to the right, until a new equilibrium is established to keep K.aA + bB cC + dD(2) Increasing system pressure will shift the reaction to the right if Dn 0, or to the left if Dn 0, or to the left if DH 0.If an external stress is applied to a system at equ
26、ilibrium, the system will make adjustment to offset such stress, consequently resulting into a new equilibrium position. Le Chteliers Principleincreasing or decreasing the reaction rate, without loss of catalyst, only due to a change in the reaction pathway. 5. catalysisk = A exp( -Ea/RT )Eak Cataly
27、st does not change the equilibrium constant, and also does not shift the equilibrium. Catalyst lowers both Ea of forward and reverse reactions.Homogeneous catalysisheterogeneous catalysisN2 (g) + 3H2 (g) 2NH3 (g)Fe/Al2O3/K2OcatalystCO + Unburned Hydrocarbons + O2CO2 + H2Ocatalyticconverter2NO + 2NO22N2 + 3O2catalyticconverterExercises:13.3713.4213.7313.114 14.6214.8817.1 kJ2.1 kJ0.4 kJO3 + OO3 + O + ClD DH = - 392 kJProgress of reaction2O22O2 + ClO2CFCl3UVUVDepletion of Ozone in the stratosphere Chapter 17
