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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

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## Chapter 25: Rates of Chemical Reactions

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**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 • 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.**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**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**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**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**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**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)**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**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**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**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**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 ]**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**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]**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**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**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.**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**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.**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**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**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**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**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]**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**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**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**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)**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}**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)**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]**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**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**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**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)**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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