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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}

- 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.

- Monitoring the progress of a reaction - use a method reflective of change in one of the reactants

- 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

- 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

- Instantaneous rate is the slope of the concentration of reactant or product curve vs. time at a point.
- 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

- v = (1/vj) d[J] dt where is the stoichiometric factor of the j th species

- 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

- v = (-1/2) (-1.2 mol/L/s) = 0.6 mol/L/s = (1/1) d[CH3CH3]dt

- 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

- Powers may or may not reflect stoichiometry

- 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

- 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)

- 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

- If B is in large excess, v = k’[A] pseudo-first order

- This means concentration of all reactants except one are constant
- 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

- Taking log, log vinitial = log k + a log A0

- 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

- 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

- 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

- 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 ]

- 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

- That time, t 1/2, at which the concentration drops by a factor if two is called the half-life

- If A = 0.5 A0) then ln (1/2) =-kt or t = -ln(1/2)/k = ln(20/k = 0.693/k

- 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]

- [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

- 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

- 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.

- As t goes to infinity [A]eq = {k’ /(k+k’} [A0] and since [B] = [A0] - [A],

- A-> B v =k[A] and B->A v=k’[B] so d[A]/dt = -k[A] + k’[B]

- 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

- t is the relaxation time

- 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
- Pressure change (P-jump or pressure jump) also possible

- Temperature change (T-jump) rapid changes in temperature 5-10K/µs

- 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.

- 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

- Activation Energy (Ea)
- Minimum kinetic energy reactants must have to form products
- Fraction is defined by Boltzman distribution exp(- Ea /RT)

- Minimum kinetic energy reactants must have to form products
- Pre-exponential factor (A)
- Measure of rate of collisons irrespective of their energy
- A x exp(- Ea /RT) gives the rate of successful collisions

- Measure of rate of collisons irrespective of their energy

- 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

- Unimolecular reaction - one species dissociates or rearranges

- 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

- 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]

- Each step can have a different rate constant

- 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

- 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

- The height and time of maxima depend on the relative rate constants involved

- This says that [A] decays exponentially, [I] rises to maximum then falls to zero and [P} rises from zero to [A0]

- 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

- 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

- 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)

- ka [O3] = ka’[O2][O] + kb [O3][O]

- dO/dt = 0 or ka [O3] - ka’[O2][O] - kb [O3][O] = 0
- 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)

- 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}

- d[P]/dt = k[I] = kK[A][B]

- Occurs when rate of formation and reverse reaction for intermediate is faster than rate of reaction of intermediate to products

- 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)

- 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]

- Steady State Approximation: d[ES]/dt = 0 or ka[E][S] - ka’[ES] - kb[ES]. Therefore [ES] = ka[E][S] /(ka’ + kb)

- 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

- Define KM = (ka’ + kb)/ ka )
- 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

- 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

- 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

- 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)

- 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

- 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)