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Chapter 25: Rates of Chemical Reactions

Chapter 25: Rates of Chemical Reactions. Homework: Excercises 25.1(a) to 25.18 (a) {do any 9}; Problems 25.2; 25.4;25.6; 25.15 {do two}. Empirical Chemical Kinetics. Chemical kinetics is the study of reaction rates Experimental methods

Chapter 25: Rates of Chemical Reactions

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1. Chapter 25: Rates of Chemical Reactions Homework: Excercises 25.1(a) to 25.18 (a) {do any 9}; Problems 25.2; 25.4;25.6; 25.15 {do two}

2. Empirical Chemical Kinetics • Chemical kinetics is the study of reaction rates • Experimental methods • Monitoring the progress of a reaction - use a method reflective of change in one of the reactants • Pressure change • Change in absorption at a given wavelength • Weight loss of solid • Gas chromatography, nmr, esr, mass spectrometry, etc.

3. Methods in Kinetics • Real-time analysis - bulk analysis or sampling • Quenching - stop the reaction (e.g. temperature change) • Quench whole reaction or sample • Suitable only if rate is slow compared to quenching rate • Flow methods - usually uses optical analysis methods • Reactants mixed as they flow together • Observations taken at various points down the tube • Distance equivalent to time • Large volumes required • Stopped flow - stopping syringe allows smaller volumes , spectrometer fixed • Flash photolysis - short duration light flash followed by analysis vs time • Usually use with lasers ns to ps pulse

4. Rate of Reaction • Consider a reaction: aA + bB -> cC +dD • Instantaneous rate is the slope of the concentration of reactant or product curve vs. time at a point. • UNITS: mol L-1s -1 • Rate of consumption: -d[Reactants]/dt • Fact that it is consumption is expressed by (-), rate itself is a positive # • Rate of formation : +d[Products]/dt • From stoichiometry : (1/c)d[C]/dt = (1/d)d[D]/dt = -(1/a)d[A]/dt (1/c)d[C]/dt • Unique rate of reaction, v combines these concepts • v = (1/vj) d[J] dt where is the stoichiometric factor of the j th species • vj is positive for products and negative for reactant

5. Example: Self Test 25.2 • For the reaction, 2CH3(g) -> CH3CH3 (g) the rate of change in CH3 d[CH3]/dt as given as -1.2 mol/L/s what is • a) the rate of reaction • v = (-1/2) (-1.2 mol/L/s) = 0.6 mol/L/s • b) the rate of formation of CH3CH3 • v = (-1/2) (-1.2 mol/L/s) = 0.6 mol/L/s = (1/1) d[CH3CH3]dt or • d[CH3CH3]dt = 0.6 mol/L/s

6. Rate Laws • Rates are often proportional to concentration of reactants raised to a power, i.e., v =f([A]n, [B]m,…) • Coefficient preceding the concentration is called the rate constant • Rate laws are determined experimentally • Powers may or may not reflect stoichiometry • May be conicidence or may reflect something about mechanism (how reaction occurs • Example: Formation of HBr from hydrogen and bromine • v = {k[H2][Br2]1.5}/{[Br2] +k’[HBr]} • Reaction: H2(g)+ Br2(g)->2HBr(g) • Mechanism must be consistent with the rate law • Rate law can be used to predict compositions as a function of time if rate constant known

7. Reaction Order • The order of a reaction refers to the power to which a concentration is raised • Over all order is the sum of the individual orders • Example v =k[A][B]2[C]3 • Reaction is first order in A,second order in B, third order in C • Overall order is sixth • Reaction may not have an overall order if it cannot be expressed in a simple form, e.g. v = {k[H2][Br2]1.5}/{[Br2] +k’[HBr]} • 1st order in hydrogen, -1order in HBr no overall order • If a reaction is not dependent on concentration it is said to be zero order (either overall or w.r.t. a given component) • v= k

8. Three Problems in Kinetics • Identify the rate law and obtain the rate constant (Chapter 25) • Construct reaction mechanisms consistent with the rate law (Chapter 25 & mainly 26) • Account for values of rate constants and explain their temperature dependence (Chapter 25 & 27)

9. Determining the Rate Law • Isolation method - all reactants in large excess except one • This means concentration of all reactants except one are constant • The other values would be lumped into the rate constant determined • Order thus determined is called psuedo- nth order • Example say rate is v = k [A][B]2 • If B is in large excess, v = k’[A] pseudo-first order • k’ = k[B0]2 • If A is in large excess, v= k’[B]2 psuedo 2nd order • Method of initial rates - initail rate of rxn (esp. using isolation method) given by initial concentration, vinitial = k[A0]a • Taking log, log vinitial = log k + a log A0 • Plot of Log of initial rate vs. log of range of intial concentrations gives a straight line of slope “a” and intercept log k

10. Example Self test 25.3 • Data • Concentration/10-3 mol/L:5.0 8.2 17 30 • Rate/(10-7 mol/L/s):3.6 9.6 41 130 1) Plot log rate vs. log concentration 2) Slope = order = 2.0 3) Intercept = log k = -1.84 k= 0.0014 (mol/L)/s

11. Comments on Method of Initial Rate • Might not reveal full rate law • Products might participate in the reaction (e.g. HBr in earlier reaction) • Use results to fit data throughout reaction • Predict concentration at any time and compare with actual data • Check to see whether addition of products effects rate law • Look for surface effects

12. Integrated Rate Laws • Rate laws are differential equations, if you want to use them to find concentrationas as function of time you must integrate • Exactly - simple, but useful cases • Numerically • The integrated form of a rate law is called the integrated rate law

13. Integrated Rate Laws - 0th Order Rxns • For a zeroth order reaction, v = d[A]/dt = -k or • d[A] = -kdt • Integrating from [A0 ] to [A] and 0 to t, [A] - [A0 ] = -kt or A = [A0 ] - kt • For a zeroth order reaction, a plot of [A] vs. time gives a straight line of slope -k and intercept of [A0 ]

14. First Order Reactions • For a 1st order rxn, v = k[A] or d[A]/dt= -k[A] for consumption of A • Rearranging we have: d[A]/[A] =-kdt • Integrating between A0 and A and 0 and t, ln(A/ A0) = -kt or A = A0 exp(-k/t) • For a first order reaction, a plot of ln(A/ A0) vs. t is a straight line of slope -k • For a first order reaction, concentration decreases exponentially with time • If A = 0.5 A0) then ln (1/2) =-kt or t = -ln(1/2)/k = ln(20/k = 0.693/k • That time, t 1/2, at which the concentration drops by a factor if two is called the half-life • Half life is independent of initial concenrtration • 2 half-lives gives 1/4 the concentration, 3, 1/8 n, 1/2n

15. Second Order Reactions: v =-k[A]2 • For 2nd order rxn, d[A]/dt =-k[A]2 or d[A]/ [A]2 =-k dt • Integrating, ∫ d[A]/ [A]2 = -1/[A] + 1/[A0] = -kt • -1/[A] + 1/[A0] = -kt • 1/[A] = kt + 1/[A0] = (kt [A0] +1)/ [A0] or • [A] = [A0]/(kt [A0] +1) • Plot of 1/[A] vs. t gives straight line with slope of k and intercept of 1 /[A0]

16. Half-Lives for Second Order Reactions: v =-k[A]2 • [A] = [A0]/(kt [A0] +1) or [A]/ [A0] = 1/(kt [A0] +1) • If [A]/ [A0] = 0.5 then 2 = (kt [A0] +1) or 1 = kt [A0] • Thus, t1/2 = 1/{k [A0]} • This means that for a second order reaction, unlike first order reaction, the half-life does depend upon the initial concentration

17. Half-Lives for Second Order Reactions: v = d[A]/dt =-k[A][B] • To integrate reactions of this type, you need to know the relationship between [A] and [B] via stoichiometry • For a simple reaction: A + B -> Products • If intial concentration falls by x then by stoichiometry [A] = [A0] - x and A] = [A0] - x • d[A]/dt =- dx/dt = -k[A][B] = -k([A0] - x )([B0] - x ) • Thus {1/ ([A0] - x )([B0] - x )} dx = kdt and • ∫ {1/ ([A0] - x )([B0] - x )} dx = kt • {1/ ([B0]-[A0])}{ln{[([A0] / ([A0] - x )} -ln{([B0] /([B0] - x )}}=kt • But ([A0] - x ) = [A] and ([B0] - x ) = [B] so • ln [([B] /[B0])/ ([A] / ([A0])] = ([B0] -[A0]) kt • If [B0] = [A0] then you have the case solved earlier • Other Integrated Rate Laws can be found in a similar manner and are given in Table 25.3

18. Kinetics of Reactions Near Equilibrium • Most reactions are far from equilibrium so reverse reactions aren’t important • As you near equilibrium you should consider reverse rxn so for 1st order • A-> B v =k[A] and B->A v=k’[B] so d[A]/dt = -k[A] + k’[B] • If initial conditions are [A] = [A0] and [B]=0 then at all times [A] + [B] = [A0] • So d[A]/dt = -k[A] + k’[B]= -k[A] + k’([A0]-[A])= -(k-k’)[A] + k’[A0] • Solving this 1st order differential equation: [A] = ({k’ + k x exp(-k +k’)t}/(k+k’}) [A0] • As t goes to infinity [A]eq = {k’ /(k+k’} [A0] and since [B] = [A0] - [A], [B]eq = {k /(k+k’} [A0] • Thus the equilibrium constant K = [B]eq / [A]eq= k/k’ or the equilibrium constant is the ratio of the rate constant for the forward and reverse reactions • If you know one rate constant and the equilibrium constant you can calculate the other • For multi step reactions K is the ratio of forward and reverse rates for each step.

19. Relaxation methods • Relaxation means a return of a system to equilibrium • In kinetics if an external influence shifts the equilibrium suddenly, relaxation occurs when the concentrations adjust to an equilibrium characteristic of the new conditions • Temperature change (T-jump) rapid changes in temperature 5-10K/µs • For a simple A<-> equilibrium which is 1st order in each direction, if x is the departure from equilibrium at the new temperature t and x0 is the immediate departure from equilibrium after the T-jump then x = x0 exp(-t/t) where 1/ t = ka + kb • t is the relaxation time • Expression depends on forward and reverse kinetics • Pressure change (P-jump or pressure jump) also possible

20. Derivation of 1st order Relaxation Constant • If you have a 1st order forward and reverse reaction, d[A]/dt = -k[A] +k’[B] and at equilibrium k[A] = k’[B] • After a T-jump x is the deviation from equilibrium • [A] = Ae +x and [B] = Be-x, where Ae and Be are equilibrium values • Thus, d[A]/dt = -k(Ae + x) +k’(Be-x) =-kAe +k’Be -kx - k’x • -kAe +k’Be = 0 so d[A]/dt = -kx +k’x = -(k +k’)x • But [A]/dt = d x/dt = -(k + k’)x • This differential equation has the solution x = x0e-at where a = 1/t = k + k’ • For more complicated reactions you need only define the rate equation in terms of the kinetics of forward and backward reactions and solve differential equations.

21. Temperature Dependence of Reaction Rates • For many reactions, it is observed that a plot of ln(k) vs. 1/T(K) is a straight line : ln(k) = ln(A) -(Ea /RT) or k = A exp(-(Ea /RT) • Arrhenius equation • Slope (-Ea/R) where Ea is the activation energy • The greater Ea, the more strongly the rate constant depends on temperature • If Ea =0, there is no temperature dependence • If Ea <0, rate decreases with temperature - indicates complex mechanism • Assumes is Ea independent of temperature. • Ea =RT2(dlnk/dT) • Commonly a sign tunneling is involved in rxn • Intercept, (A) is called the pre-exponential or frequency factor • Together they are called the Arrhenius parameters

22. Interpretation of Arrhenius Parameters {k = Aexp(- Ea /RT)} for Gas Phase Rxns • Activation Energy (Ea) • Minimum kinetic energy reactants must have to form products • Fraction is defined by Boltzman distribution exp(- Ea /RT) • Pre-exponential factor (A) • Measure of rate of collisons irrespective of their energy • A x exp(- Ea /RT) gives the rate of successful collisions

23. Accounting for Rate Laws • Elementary reactions are those involving a small number of ions during each step • May be many steps • Molecularity is the number of species which come together in a elementary reaction • Unimolecular reaction - one species dissociates or rearranges • Isomerization • Radiocative decay • Bimolecular reactions - a pair collide & exchange energy or groups of atoms undergo change

24. Molecularity vs. Reaction Order • Reaction order is an emperircal quantity • Molecularity refers to the mechanism of a step • Rate Laws • Unimoleular Rxn - first order because the number that decay in any one time is proportional to the number there are availble to decay • Bimolecular Rxn - second order because the rate is proportional to the rate at which the two species meet which is proportional to their concentrations • Just because a reaction is second order doesn’t mean its mechanism is bimoluclar

25. Consecutive Elementary Reactions • Reactions which proceed through the formation of one or more intermediates are consecutive elementary reactions • Each step can have a different rate constant • Radioactive decay chain • Pyrolysis of acetone: • (CH3)2CO -> CH2=CO (ketene) + CH4 • CH2=CO -> 1/2 C2H4 + CO • Rate A -> I -> P( ka, kb) • d[A]/dt = - ka[A] • If A is not replenished, d{I]/dt = ka[A] - kb[I] and d[P]/dt = kb[I]

26. Integrated Rate Law for Consecutive Rxn • At all times [A0] = [A0] + [I] + [P] if [I0] and [P0] are 0 • For step 1, [A] = [A0] exp(-kat) • Substituting this for [A] in d[I]/dt and integrating [I] = (ka/ka +kb )(exp(-kat) - exp(-kbt)) [A0] • Substituting for [I] into d[P]/dt and intergrating [P] = { 1 + [(kaexp(-kbt) - (kbexp(-kat)]/[kb - ka ] [A0] • This says that [A] decays exponentially, [I] rises to maximum then falls to zero and [P} rises from zero to [A0] • The height and time of maxima depend on the relative rate constants involved • For example by differentiating d[I]/dt and setting the result = 0 (max in fn) you get max when kaexp(-kat) = kbexp(-kbt) or by taking the ln of this eqn when t =tmax= (1/(ka-kb)ln (ka/kb) • At that point [I] = (ka/kb)c [A0] where c = kb/kb -ka

27. Rate Determining Step • Recall, [P] = { 1 + [(kaexp(-kbt) - (kbexp(-kat)]/[kb - ka ] [A0] • If one rate constant much greater than the other this becomes [P] = { 1 + - exp(-ksmallert)}[A0] • That’s because denominator reduces to the larger term which cancels with the pre-exponential of the smaller term and the exponential of the large k goes to zero • Thus the [P] depends on the slowest rate, the rate determining step • This holds for more complicated reaction mechanisms

28. Steady State Approximation • Made to simplify calculations • After an induction period, assume d[intermeditaes]/dt =0 • Concentration of intermediates don’t change • Thus d[I]/dt = 0 = ka[A] - kb[I] or [I] = ka /kb[A] • Then d[P]/dt = kb[I] = ka [A] • Rate determining step is the decay of A • Integrating : [P] = ka [A0]∫exp (- ka t) = (1- exp (- ka t) ) A0

29. Self Tests 25.8 • dO3/dt = -ka[O3]- kb [O3][O] + ka’[O2][O] • dO/dt = +ka [O3] - ka’[O2][O] - kb [O3][O] • dO/dt = 0 or ka [O3] - ka’[O2][O] - kb [O3][O] = 0 • ka [O3] = ka’[O2][O] + kb [O3][O] ka [O3] = [O]( ka’[O2] + kb [O3]) [O] = ka [O3] /( ka’[O2] + kb [O3]) = ka [O3] /( A) • dO3/dt = -ka[O3]- kb [O3][O] + ka’[O2][O] • dO3/dt = {-ka[O3]{( ka’[O2] + kb [O3]) /( ka’[O2] + kb [O3])}}-kb [O3][O] + ka’[O2][O] = {(-ka[O3]ka’[O2] - ka[O3] kb [O3]) /(A)}}-kb [O3] ka [O3] /( A) + ka’[O2] ka [O3] /( A) = 2 kb ka [O3] [O3] /( A)

30. Pre-Equilibria • S/S Approximation assumes that there is no change in intermediate concentration suppose intermediate reaches equilibrium with reactants • Occurs when rate of formation and reverse reaction for intermediate is faster than rate of reaction of intermediate to products • It is rate determining step • Reaction A+ B <->I -> P where I is in equilibrium(K) and rate of formation of P is k • d[P]/dt = k[I] = kK[A][B] • Rate law is 2nd order and rate constant kK is composite = k {kf/kr}

31. Self Test 25.9 • For Rxn: 2A <-> I (K) I +B-> P (k) show that rate is 3rd order • d[P]/dt = k[I][B] • But [I] = K[A]2 ( K = [I]/ [A]2 ) so • d[P]/dt = k K[A]2 [B] or a third order reaction (2nd order in A)

32. Pre-Equilibria:Michaelis-Menton Mechanism • For enzyme catalyzed reaction, the rate of product(P) depends on the amount of enzyme(E) as well as the amount of substrate (S) • Enzyme undergoes no net change • Proposed mechanism: E + S « ES -> P +E ka,ka’, kb • The rate of formation of bound enzyme d[ES]/dt = ka[E][S] - ka’[ES] - kb[ES] • Steady State Approximation: d[ES]/dt = 0 or ka[E][S] - ka’[ES] - kb[ES]. Therefore [ES] = ka[E][S] /(ka’ + kb) • If total enzyme = [E0] then [E0] = [E] + [ES] • Thus [ES] = ka([E0] - [ES])][S] /(ka’ + kb) if only a little substrate is added so [S] doesn’t change • Solving for [ES], [ES] = ka[E0] [S] /(ka’ + kb) + ka[S]

33. Michaelis-Menton Mechanism (cont.) • d[P]/dt = kb[ES]= kb ka[E0] [S] /(ka’ + kb) + ka[S] = [E0] {kb ka [S] /(ka’ + kb) + ka[S]} = [E0] {kb [S] /((ka’ + kb)/ ka )+ [S]} • Define KM = (ka’ + kb)/ ka ) • KM is the Michaelis Constant • Therefore d[P]/dt = [E0] {kb [S] /(KM)+ [S]} or d[P]/dt = k [E0], where k = {kb [S] /(KM)+ [S]} • Means the rate varies linearly with enzyme concentration • If [S] >> /KM then k » kb so d[P]/dt = kb [E0] • This means rate of product formation is zero-order in S when large amounts are present • Rate is at maximum so kb [E0] is the maximum velocity of enzymolysis and kb is the maximum turnover number

34. Michaelis-Menton Mechanism (cont.) • If [S] >> /KM then k = {kb [S] /(KM)+ [S]} = kb [S] /(KM) and d[P]/dt = kb [E0][S] /(KM) • Rate is now proportional to both the concentration of enzyme and substrate • Lineweaver-Burke Plot • k = {kb [S] /(KM)+ [S]} so 1/k = (KM)+ [S]} /kb [S] = (KM /kb [S] )+ 1 /kb • Plot of 1/k versus [S] gives a straight line with slope of KM /kb and intercept of 1 /kb • KM can be determined directly

35. Unimolecular Reactions - Lindemann-Hinselwood - 1 • Issue with uni-molecular reactions is that most of them are first order but in order for molecule to gain eneough energy to react it must collide with another molecule which is a bi-molecular event • Called uni-molecular because they involve a uni-molecular step • Fredrick Lindemann (1921) proposed a two step reaction, the first step (collision) is bimolecular the second step is uni-molecular • If second step is slow enough to be rate determining, then overall reaction is uni-molecular

36. Unimolecular Reactions - Lindemann-Hinselwood - 2 • Step 1: A + A -> A* + A d[A*]/dt = ka[A]2 (second order) • Reverse reaction possible d[A*]/dt = -ka’[A][A*] • Step 2: A* ->P d[A*]/dt =- kb[A*] • Thus d[A*]/dt = ka[A]2 - ka’[A][A*] - kb[A] • Using the steady state approximation (d[A*]/dt=0), [A*] = ka[A]2 /( ka’[A] + kb) • So d[P]/dt = kb[A*] = kb ka[A]2 /( ka’[A] + kb) • If ka’[A] >> kb then d[P]/dt = kb ka[A] / ka’ or the rate is first order in A • Predicts that kinetics should switch to second order when kb >> ka’[A] (low partial pressures of A • Then d[P]/dt = ka[A]2 • Rate determining step is the bimolecular reaction to produce A* • This mechanism does not agree in quantitatively at low pressures • This is because it assumes the lifetimes of all A* states are the same which they aren’t (lifetime depends on excess energy)

37. Activation Energy of Composite Reactions • If each of the rate constants have Arrhenius temperature dependence, then you can write the composite • For d[P]/dt = kb ka[A] / ka’ or k = kb ka/ ka’, you can write k = (A(a) exp-Ea(a)/RT) x (A(b) exp-Ea(b)/RT) / (A’(a) exp-E’a(a)/RT) • Thus k= {A(a)A(b)/A’(a)}{exp-{Ea(a) +Ea(b) - E’a(a)}/RT or • k =A”exp-E”/RT where E” = Ea(a) +Ea(b) - E’a(a)} • E” can be + or - depending on value of E’(a) reverse reaction • If reverse activation energy is large { (-) temperature dependence}, then deactivation is favored at higher temperatures

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