1

CHAPTER 20 – Chemical Kinetics I. Introduction Why we study rates? Fundamental: Leads to an interpretation of the mechanism of reaction, the

sequence of elementary steps leading to reaction. Pragmatic: enables us to predict how concentrations or rates change over

time. II. Rate of Reaction - Definition:

A. Rates of consumption and formation. Consider rxn with following stoichiometry: N2O + (3/2)O2 → 2NO2 How do we define its rate? υ As rate of consumption of reactant N2O? υN2O = -d[N2O]/dt Rate of consump of O2 = υO2 υO2 = -d[O2]/dt Rate of formation of product NO2 = υN2O υNO2 = d[NO2]/dt But is υN2O = υO2 = υNO2 ? NO! Problem: How to uniquely define rate of reaction? B. Unique Definition of Rate – divide by the stoichiometric coefficient For Generalized Reaction: aA + bB + ... → yY + zZ

€

Rate ≡ υ ≡ −1a

d[A]dt

= −1b

d[B]dt

= +1y

d[Y]dt

= +1z

d[Z]dt

All differentials are equivalent after a steady state is reached

2

So:

€

υ = −d[N2O]

dt= −

13/2

d[O2]dt

= +12

d[NO2]dt

or

υ = υN2O=

23υO2

=12υNO2

III. The Rate υ is time-dependent

Since υ = -d[N2O]/dt = -(slope of [N2O] curve) υ is a quantity changing in time. Very often we measure initial slopes or initial rate. IV. Empirical Rate Law or Rate Equation:

A. Expresses dependence of rate on the concentration of species involved. Rate Law: an equation that expresses rate of reaction as a function of

concentration of all species that affect the rate.

υ = k[A]α[B]β... B. The Rate Law expression is determined by rate experiment, not

stoichiometry. 1. exponents α, β... are not necessarily equal to coefficients a, b... in

balanced chemical equation. 2. α, β... need not be integers. 3. α, β... may be different under different reaction conditions for the

same rxn.

3

4. Reaction Order n: overall order = α + β + ... = n order with respect to [A] = α etc. 5. Rate constant = k

a. units = conc(1-n)

time-1

b. highly T-dependent c. involves microscopic nature of the reactants and products

V. Analysis of Kinetic Results:

A. First main task - determine Rate Law (in principle, this requires

monitoring the rate with time and all conc’s of species involved).

B. Method of isolation

1. Measure rate under conditions in which all reactants except one are present in large excess.

2. Example: crystal violet (CV) reaction with hydroxide ion:

υ = k[CV]α[OH-]β

a. Find α by measuring rate with large excess of hydroxide. During course of rxn, [OH-] hardly changes, so:

υ = k’[CV]α

where k’ = k[OH-]β b. Analyze the rate vs [CV] to determine α Rxn is found to be 1st order in [CV], so k’ is said to be a pseudo-

first-order rate constant and υ = k’[CV] is a pseudo-first-order rate law in [CV] c. Next find β by doubling [OH-] with hydroxide still in large excess.

4

C. Method of Initial Rates

• rapidly mix rxn mixture with known set of initial concentrations, measure initial reaction rate.

• change the conc of one species at a time and measure initial reaction rate.

Example Problem: The following experimental data have been determined for the reaction: NH4

+(aq) + NO2-(aq) → N2(aq) + 2H2O(l) (in acid solution) [NH4

+] (M) [NO2-] (M) Rate = d[N2]/dt[mol/liter)/s]

0.0092 0.098 34.9 x 10-8 0.0092 0.049 16.6 x 10-8 0.0488 0.196 315 x 10-8 0.0249 0.196 156 x 10-8 Determine the rate law for the reaction, and evaluate the rate constant.

VI. Integrated Rate Laws (depend on Rxn order).

Integrate the differential equation to obtain the time-varying concentration of

species. A. 1st-order reaction dynamics.

1. Examples: radioactive decay, unimolecular decomposition, isomerization.

A → products

Rate law:

€

−d[A]dt

= k[A]

2. Find [A] as a function of time given that at t = 0, [A] = [A]o

5

€

Rearrange :

−d[A][A]

= kdt

Integrate :

−d[A][A][A]o

[A]t

∫ = kd $ t 0

t

∫

−ln[A][A]o

[A]t

= kt

− ln[A]t

[A]o

= kt

[A]t = [A]oe−kt

t

3. To determine whether rxn is 1st order, plot ln[A] vs. t (should be

straight line).

4. k units are time-1. 5. Method for finding k other than finding slope (if you already know it’s

first order): measure [A] at 2 different times: t1 and t2.

− ln

[A]t2

[A]t1

= k(t2− t

1)

solve for k

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6. Half-life t1/2 of 1st order rx: t1/2 = time interval in which half of the reactant is used up.

€

− ln

12[A]0

[A]0= kt1 / 2

− ln12

= kt1 / 2

ln2 = kt1 / 2

t1 / 2 =ln2k

Note: t1/2 independent of [A]o. Means that in each interval of duration t1/2 the concentration

decreases to half of what it was at the beginning of the interval, throughout the entire course of the reaction.

Graphical picture of half life:

7. The time constant τ Defined simply as τ =1/k

t/t1/2

1 2 3

.125

.25

.5

1

[A]t [A]o

7

B. 2nd-Order Reactions - simplest type (involving only one species conc).

€

−d[A]dt

= k[A]2

Solve for [A]t

€

Rearrange :

d[A]1

[A]2[A]o

[A]t

∫ =0

t

∫ kdt

1[A] [A]o

[A]t

= kt

finally :1

[A]t

−1

[A]o

= kt

also :

[A]t =[A]o

1 + kt[A]o

Careful: simply a plot of [A] vs time looks very much like first order

reaction decay curve. However, if the following plot is linear, it reveals second order:

Homework: derive relationship between t1/2 and k for second order rxn

Now depends on [A]o

€

t12

=1

k A[ ]o

Half-life is less meaningful for second-order reactions, since it changes

with concentration.

8

Study t1/2 at 2 different initial concentration conditions (1) and (2):

€

t12

1( )

t12

2( )=

A[ ]o

2( )

A[ ]o

1( )

C. General 2nd-Order (involving 2 species). Stoichiometry = νAA + νBB → prod

Rate =

€

−1νA

$

% &

'

( )

d A[ ]dt

$

%

& &

'

(

) )

= k A[ ] B[ ] (2nd order)

1. Case (1): initial [A]o = [B]o and A and B are used up at same rate. i.e. 1A + 1B → Z Then [A] = [B] at all times Rate = k[A][B] = k[A]2 = k[B]2 Same treatment as before. 2. Case (2): initial [A]o ≠ [B]o

Suppose we have: Stoichiometry: A + B → Z and Rate Law: υ = k[A][B]

Can be shown, by method of integration, that:

€

ln

A[ ]t

A[ ]o

"

#

$ $

%

&

' '

B[ ]t

B[ ]o

"

#

$ $

%

&

' '

= A[ ]o− B[ ]

o( )kt

So you see, it can get very complicated, very quickly. See ref. Capellos and Bielski, Kinetic Systems.

9

D. Higher Orders. 1. 3rd-Order, 1 species: 3A → Z Homework: show that if -d[A]/dt = k[A]3

€

1

A[ ]t

2−

1

A[ ]o

2= 2kt

2. In general, higher orders very complicated but can be worked out.

E. Zero Order Reactions. Rate unaffected by concentrations.

Example: rate may be determined by amount of catalyst only, except at

very low concentration of reactant. A → Z -d[A]/dt = k k would depend on surface area of catalyst in heterogeneous catalysis. Table 20B.3 Integrated rate laws

F. No Simple Order.

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Most rxns have multi-step mechanisms (composite rxns). These can have complicated rate laws which only simplify to first or

second order under certain limiting conditions. Example: Enzyme catalyzed rxns. E + S →ES ES → Z + E net E + S → E + Z or just S → Z E = Enzyme; S = Substrate; ES = complex; Z = Product Found that:

Rate Law

€

d Z[ ]dt

=V S[ ]

Km + S[ ]( )

where V and Km are constants. At low [S]

Rate ≈

€

VKm

"

# $

%

& ' S[ ] appears 1st order

At high [S] Rate ≈ V appears zero order G. Pseudo-Orders. Suppose have rx: A + B → Z with rate law: -d[A]/dt = k[A][B] 2nd order Suppose rx were studied under conditions in which B is present in great

excess: [B]o >> [A]o such that [B] hardly changes while [A] is completely used up. Then: -d[A]/dt ≈ k’[A] where k’ = k[B]o So ln[A] vs. t gives straight line, giving an apparent 1st order behavior

with rate constant k’.

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Studied again at different [B]o gives another straight line, but different k’. Rx behaves dynamically (i.e., integrates like) 1st order → Pseudo-1st

order under these conditions. H. Reversible Reactions.

1. Simplest example. (1st order in both directions)

€

A kf← → $ $ kr

B

One-way forward rate = kf[A] - call this the “chemical flux” in forward direction One-way reverse rate = kr[B] - Chemical flux in rev direction Rate Law: -d[A]/dt = +kf[A] – kr[B] = net rate = 0 at equilib So: kf[A]eq = kr[B]eq

€

B[ ]eq

A[ ]eq

=kf

kr

= Keq

2. Homework: Show that if no B present initially:

€

A[ ]eq

=kr

kf + kr

"

# $

%

& ' A[ ]

o

and

€

B[ ]eq

=kf

kf + kr

"

# $

%

& ' A[ ]

o

3. How do we determine kf, kr?

a. One way: determine d[A]/dt at t=0 when [B]o = 0.

€

−d[A]dt

#

$ %

&

' ( t= 0

= kf[A] − kr[B] ≈ kf[A]

Then: kr = kf/Keq b. Another way, find an integrated form: Find way to plot [A] vs

time data to extract k’s. (find time-dependence [A])

12

Let’s solve:

€

−d[A]dt

#

$ %

&

' ( = kf[A] − kr[B]

Now if no B present initially :

−d[A]dt

#

$ %

&

' ( = kf[A] − kr([A]o − [A])

€

−d[A]dt

#

$ %

&

' ( = (kf + kr)[A] − kr[A]o

−d[A]dt

#

$ %

&

' ( = (kf + kr){[A] −

kr

kf + kr

[A]o}

d[A]dt

#

$ %

&

' ( = −(kf + kr){[A] − [A]eq}

Now subtract d[A]eq

dtfrom left side, since it is zero.

And now define the difference from equilibrium:

δ[A]= [A]-[A]eq

Then:

dδ[A]dt

#

$ %

&

' ( = −(kf + kr)δ[A]

Now it is really easy to integrate to get :

δ[A] = δ[A]oe−(kf +kr )t

Logarithmic form:

ln(δ[A]δ[A]o

) = −(kf + kr)t

13

So we see that this is similar to irreversible 1st order decay, but decay is in displacement from equilibrium δ[A] instead

€

Now it is easy to find [A]t :

[A]=kr + kfe

−(kf +kr )t

kf + kr

[A]o

VII. Experimental Methods for Studying Fast Reactions.

A. Fundamental Requirements for Kinetic Study:

1. Preparation of a system which is not in chem equilib. 2. Must create homogeneous (well-mixed) non-equilibrium system in a

time short compared to reaction time. Example: Conventional bench chem mixing takes seconds (τmix = 1 sec) O.K. if rx takes minutes (τrx > 60 sec) Must have τmix << τrx In best designed circumstances, τmix ≥ 10-3 sec. 3. Need means of monitoring reaction as it occurs (recording device

must have response time rapid compared to reaction time) B. Continuous Flow Device.

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Above, mixing is fast by gravitational (hydrostatic) pressure. Can also force rapid mixing by driving syringes: C. Stopped-flow device: Reaction mixture is rapidly prepared by driving and

stopping syringe, and then spectrometer monitors changing concentration of absorbing species.

D. Flash photolysis: reaction is initiated by a brief flash of light, then

contents of the reaction mixture are monitored spectroscopically. E. Quenching methods: All of the above are examples of real-time methods. Quenching methods

abruptly stop the progress of reaction after it has been allowed to run for variable times.

Quenching can be done chemically, or by freezing.

15

F. Relaxation Methods: For studying very fast rxns (τrx < 10-3 sec), start with system in

equilibrium, then instantly perturb equilibrium with: 1. Rapid T change (T-jump exp).

Keq is fct of T (e.g. pulse of microwave energy into rx vessel τheat ≈ 10-6 sec)

Must not produce T-gradients, need even heating. 2. Rapid Pressure change (gas phase rx). N2O4(g) ! 2NO2(g) if P↑ rx will shift (←) if P↓ rx will shift (→)

Measure rate of return to equilibrium (e.g. reduce pressure by

allowing high-P gas to flow thru a ruptured disk) G. Analysis of Relaxation Data:

Suppose have reversible system, second order in forward direction, first order in reverse:

€

A + B⇔k−1

k1

C

-with rate law: d[C]/dt = k1[A][B] – k-1[C] -initially at equilibrium at T1: k1[A]o[B]o = k-1[C]o -now jump temperature. Then Keq changes,

system not in equilibrium anymore. [C]o/[A]o[B]o ≠ Keq(@T2) [A] = [A]f - δ[A] [B] = [B]f - δ[B] [C] = [C]f - δ[C] ↑ ↑ ↑ actual conc at displacements conc final from final equilib after equilib perturbation [C]f/[A]f[B]f = Keq(T2)

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By stoichiometry: -δ[A] = -δ[B] = +δ[C] d[C]/dt = k1[A][B] – k-1[C] By substitution: (d/dt)[[C]f - δ[C]] = k1{[A]f + δ[C]}{[B]f + δ[C]} – k-1{[C]f - δ[C]} But d[C]f/dt = 0 and expand { } - dδ[C]/dt = k1[A]f[B]f – k-1[C]f + k1[A]fδ[C] +k1[B]fδ[C] + k1(δ[C])2 + k-1δ[C] First two terms: k1[A]f[B]f – k-1[C]f = 0 dδ[C]/dt = -k1δ[C][A]f - k1δ[C][B]f + k1δ[C]2 – k-1δ[C] dδ[C]/dt = -{k1[A]f + k1[B]f + k-1}δ[C] + k1δ[C]2 ↑ δ[C]2≈ 0 (square of small perturbation) If δ[C] is small perturbation, (δ[C])2 is really small. Then: dδ[C]/dt ≈ -{k1[A]f + k1[B]f + k-1}δ[C] ↑ ↑ approx 1st order {constant of relaxation} decay back to = 1/τrelax equilib (when perturbation is slight) τrelax = “relaxation time”

= time required for perturbation from equilib to decay to 1/e of its original value.

Integrate to obtain:

δ[C] = δ[C]o exp(-t/τrelax) looks just like 1st order decay of the perturbation from equilibrium.

17

Plot lnδ[C] vs. t

Measure τrelax = (k1{[A]f + [B]f} – k-1)-1 = (k1{[A]f + [B]f - Keq

-1})-1 Have similar (but simpler) analysis for rxn:

€

A↔k−1

k1

B:

Gives τrelax = (k1 + k-1)-1 relation of τrelax to t1/2 t1/2 = 0.693/(k1 + k-1) ↑ ln 2 t1/2 = 0.693 τrelax (for A ! B type rx) H. Problem:

Rxn

€

HAc↔k−1

k1

H+ + Ac−

Has τrelax = 8.5 ns and Keq = 1.73 x 10-5 M at 25°C for a 0.1 M HAc

solution. Find k1 and k-1. Solution: By analogy with rxn, A + B ! C which had τrelax = (k1{[A]eq +

[B]eq} + k-1)-1 This rx has: τrelax = (k1 +k-1{[H+]eq + [Ac-]eq})-1

[H+]eq = [Ac-]eq = (0.1 Keq)1/2

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1/τrelax = k1(1 + Keq-1 2*(0.1 Keq)1/2 )

1/τrelax = k1[1 + 152] 1/(8.5 x 10-9 sec) = k1 x 153 k1 = 7.68 x 105 s-1 k-1 = k1 Keq

-1 = 4.44 x 1010 M-1s-1 I. Temperature dependence of reaction rates: Arrhenius theory.

1. It has been empirically observed for thousands

of chemical reactions that plotting ln k vs 1/T gives nearly a straight line over a modest temperature range. This implies:

€

k = Ae−Ea / kT

Ea = activation energy ~ energy required to

react.

€

e−Ea / kT ≈ probability of having enough energy to react.

This factor appears in all microscopic theories of the rate constant. A = frequency factor (units s-1 in 1st order rxns) = frequency of attempts to cross activation barrier unpredictable without more complicated theory Pre-exponential factor contains much interesting microscopic

information.

Table 20D.1 Arrhenius parameters

19

2. The Arrhenius parameters may be regarded as purely empirical quantities that enable us to discuss the variation of rate constant with T.

3. Even when the Arrhenius plot is not strictly linear, we can still define

the activation energy of any reaction as:

Ea= −R dlnk

d(1 / T)

"

#$$

%

&'' = RT

2 dlnkdT

"

#$$

%

&''

4. Interpretation of the Arrhenius parameters, the “reaction profile” concept.

Reaction profile shows how the

potential energy of reacting molecules changes with the configuration of the molecules.

Reaction coordinate is a

combination of interatomic distances or angles which specify a path from reactant to product configuration.

Activation barrier is the hill in

energy that must be crossed along the path to products.

Reaction Coordinate Activated complex is the

configuration of atoms or molecules at the top of the barrier. Also called the transition state.

Activation energy is the minimum kinetic energy that the reactants

must have to surmount the barrier and form products. The pre-exponential factor A may be interpreted as the frequency

at which the reactants make an attempt to cross the activation barrier.

20

Simple illustration of A for a unimolecular reaction - rate of desorption of a particle (H-atom) adsorbed on a surface.

Particle exchanges energy with surface through "collision" with

surface atoms. Maintains equilibrium distribution of vibrational energy.

Fraction of adsorbed atoms with Ea (energy to react) ≈

€

e−Ea / kT . Now:

€

k = νe−Ea / kT Now estimate ν as vibrational frequency of HCl: νHCl = 8 x 1013 sec-1

k = 8 x 1013 e-Ea/kT ↑ turns out to be correct order of magnitude for many chem rxns!!

21

VIII. Problem Involving Arrhenius Law Temperature Dependence.

A first order reaction has a rate constant of 1.3 x 10-3 sec-1 at T=25°C and 2.5 x 10-3 sec-1 at T=35°C. What is the Ea of the rxn, and what is the pre-exp factor?

k(T1) = 1.3 x 10-3 sec-1 =

€

Ae−Ea / k∗298

k(T2) = 2.5 x 10-3 sec-1 =

€

Ae−Ea / k∗308

(1.3 x 10-3)/(2.5 x 10-3) =

€

e−Ea / k( ) 1/ 298( )− 1/ 308( )[ ]

€

k(T1) = Ae−Ea / RT1

k(T2) = Ae−Ea / RT2

k(T2)k(T1)

= exp −Ea

R1T2

−1T1

#

$ %

&

' (

)

* + +

,

- . .

lnk(T2)k(T1)

#

$ %

&

' ( = −

Ea

R1T2

−1T1

#

$ %

&

' (

Ea is activation energy per mole here

Ea = 11.9 kcal/mole (typical)

A = 1.3 x 10-3 sec-1/e-11.9/1.987 x 10-3 x 298

A = 6.95 x 105 sec-1

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IX. Accounting for the Rate Laws. Second stage of analysis of kinetic data, their explanation in terms of a postulated reaction mechanism. A. Composite Reaction Mechanisms.

1. Elementary vs. composite reactions:

a. Elementary reactions are 1-step (for example, one step in a complex series of steps); For it, the Rate Law can be written directly from stoichiometry:

Molecularity of an elementary step refers to the number of

molecules coming together to react. Here is a unimolecular reaction: e.g. N2O5 ! NO2 + NO3 -d[N2O5]/dt = k[N2O5] Here is a bimolecular reaction: e.g. H + Br2 → HBr + Br -d[H]/dt = k[H][Br2] b. Composite reactions are multiple step; composed of several

successive elementary steps in a scheme called a mechanism. No obvious correlation between stoichiometry and Rate Law.

c. This chapter will deal with analyzing kinetics of composite

reactions. 2. Consecutive elementary reactions Consider the dynamics of the following irreversible reaction:

€

A ka" → " " I kb" → " " P How do the concentrations vary with time?

Rate equation for [A] is: d[A]/dt = - ka[A] for [I] d[I]/dt = ka[A]- kb[I]

23

for product [P] d[P]/dt = kb[I] Find time dependence of every species: [A]=[A]oexp(-kat) d[I]/dt = ka[A]oexp(-kat) - kb[I] If [I]o = 0

€

[I] =ka

kb − ka

exp(−kat) − exp(−kbt)( )[A]o

Then:

€

[P] = 1 +ka exp(−kbt) − kb exp(−kat)

kb − ka

# $ %

& ' ( [A]o

Note: because the reaction is irreversible in all steps, the intermediate never reaches a steady state.

3. Relationship between overall rate laws and reaction mechanism.

e.g. Suppose several different mechanisms all predict the same rate

Law. Thus, an experiment showing this to be the correct rate Law cannot distinguish which mechanism is correct.

We will see examples of this.

24

4. Example: 2-step, first step in rapid pre-equilibrium.

a. Stoichiometry of the net or composite reaction: 2NO + H2 → N2O + H2O

b. Proposed mechanism (Mech I) consists of 2 elementary steps:

(step i)

€

2NO↔k2

k1

N2O2 (suppose k1 and k2 are very large, such

that a rapid equilibrium is established on overall rxn time scale)

Keq = k1/k2 (step ii)

€

N2O2 + H2k3" → " N2O + H2O (slow step)

c. Now, derive the Rate Law predicted by this mechanism. Rate = d[N2O]/dt = k3[N2O2][H2] (for Step 2) ↑ eliminate [N2O2] (an intermediate species) in terms of either reactant or products [ ]) Since step (i) is in equilibrium: k1[NO]2 ≈ k2[N2O2] or [N2O2] = (k1/k2)[NO]2 Therefore: Rate = d[N2O]/dt = (k3k1/k2)[NO]2[H2] = k[NO]2[H2] d. This Rate Law (2nd order in NO, 1st order in H2) happens to be

the experimentally observed Rate Law. So, is this the true reaction mechanism? Not necessarily, all we can say is that it is a plausible one. Why? Because I can propose a different mechanism that predicts the same correct Rate law.

25

5. Example (part 2): a. Here is a different proposed mechanism consists of these 2

elementary steps:

(i)

€

NO + H2↔k2

k1

NOH2 (rapid equilibrium)

(ii)

€

NO + NOH2k3" → " " N2O + H2O (slow step)

b. Deduce rate Law predicted by this mechanism. Rate = d[N2O]/dt = k3[NOH2][NO] ↑ intermediate k1[NO][H2] ≈ k2[NOH2] so: [NOH2] = (k1/k2)[NO][H2] Rate Law: Rate = d[N2O]/dt = (k3k1/k2)[NO]2[H2] ↑ k c. Same Rate Law. So which is true mechanism? → Can't say,

without additional experiments. 6. Steady State analysis of mechanism:

a. Previous examples of mechanisms are special cases involving

assumption that 1 step is slow and another is fast and maintains a quasi-equilibrium. Here we introduce the more general situation.

b. Consider this more general mechanism:

(i)

€

2NO↔k2

k1

N2O2 (no rapid equilib assumption made)

(ii)

€

N2O2 + H2k3" → " N2O + H2O (not necessarily slow step)

c. Derive rate law using Steady State Approximation.

26

S.S. Approximation: after a short "induction" time, the conc. of all intermediates becomes approximately constant for most of the duration of the rxn.

d. Derivation: (i) First write down rate equation for each intermediate: d[N2O2]/dt = k1[NO]2 - k2[N2O2] – k3[N2O2][H2] (ii) Make steady state approx to solve for [intermediates]: d[N2O2]/dt ≈ 0 So: 0 = k1[NO]2 - k2[N2O2] – k3[N2O2][H2]

And: N2O2!"

#$ =

k1NO!"

#$2

k2+k

3H2

!"

#$

(iii) Plug into rate equation for formation of product: d[N2O]/dt = k3[N2O2][H2] General rate law (steady state rate Law)

Rate = d[N2O]/dt =

€

k3k1 NO[ ]2

H2[ ]"

# $

%

& '

k2 + k3 H2[ ]( )

(iv) Could now examine special limiting cases of interest -

suppose 2nd step is slow compared to 1st (rapid equilibrium case examined earlier) or: k2 >> k3[H2]

Rate would become =

€

k3k1 k2( ) NO[ ]2

H2[ ]"

# $

%

& '

27

Note that by studying rxn at various [H2] over wide range of [H2], could see a change in the Rate Law.

7. The rate-determining step concept: 8. Another Steady State application - unimolecular reactions.

a. Example: gas phase dissociation of a diatomic: H2(g) ! 2H b. Observed forward chemical flux: -d[H2]/dt = k[H2] (unimolecular) c. Problem: How can dissociation occur without an activating

collision with another gas molecule (say M), and so why don't we have:

Rate = k[H2][M] d. Explained by Lindemann Mechanism:

€

H2 + M ka" → " H2 *+ $ M (H2* is excited H2)

€

H2 *+M kd" → " H2 + $ M

€

H2 *kuni" → " " 2H(decomp of excited species)

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e. Derive S.S. Rate Law: d[H2*]/dt = ka[H2][M] – kd[H2*][M] – kuni[H2*] ≈ 0 [H2*] = (ka[H2][M])/(kd[M] + kuni) Rate = +

€

12d[H]/dt = kuni[H2*]

Rate =

€

kunika H2[ ] M[ ]( )kd M[ ] + kuni( )

f. Usually M is an inert buffer gas (air molecules) in high

concentration, so: kd[M] >> kuni

So: Rate ≈

€

kunika

kd

"

# $

%

& ' H2[ ]

↑ k∞ ( ∞ means high pressure limit) g. Interpretation: at high pressure, activating and deactivating

collisions maintain an equilibrium population of H2* species (equilibrium constant ka/kd) available for dissociation.

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Notes: