Chapter 25 rates of chemical reactions l.jpg
Sponsored Links
This presentation is the property of its rightful owner.
1 / 37

Chapter 25: Rates of Chemical Reactions PowerPoint PPT Presentation


  • 148 Views
  • Uploaded on
  • Presentation posted in: General

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

Download Presentation

Chapter 25: Rates of Chemical Reactions

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


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


  • Login