Chemical Kinetics: Rate and Mechanistic Models (CHE 505) M.P. ...

Chemical Kinetics: Rate and Mechanistic Models (CHE 505) M.P. Dudukovic Chemical Reaction Engineering Laboratory (CREL), Washington University, St. Louis, MO

ChE 505 – Chapter 4N

Updated 01/28/05

CHAPTER 4. Chemical Kinetics: Rate and Mechanistic Models 4.1

Introduction

By listing the plausible reaction stoichiometries, we can calculate the composition of the system in its final state, i.e. at equilibrium, provided that we are given the thermodynamic properties such as heats (enthalpies) of formation, Gibbs free energies of formation, temperature and pressure of the system, as well as the composition of its initial state. As important as this may be, it does not tell us at all how long will it take for the system to transit from its initial to the final state. Chemicals kinetics deals with the rates of chemical reactions and with how these rates are affected by the composition of the system, temperature and pressure. Kinetic studies are important, since based on them empirical and semi-empirical rate expressions are obtained which then can be used for predictive purposes in analysis of system's dynamic behavior or in design of chemical reactors. Kinetic studies deal with reactions of wide variety of characteristic times. Reactant half-life can vary from a picosecond to a period longer than the age of the universe. Reaction speed will often depend on the conditions of the system. For example, O2 and H2 at room temperature and pressure left alone at stoichiometric ratio for water production would have a half-life of 1025 years or 3 x 1032 seconds. However, a spark causes the half-life to be reduced to 10-6 s! In summary, the scope of kinetic studies is to quantify the rate of reaction as a function of system intensive properties (composition, temperature, etc.) and to suggest a fundamental theory (if possible) for estimation of kinetic parameters. The knowledge of kinetics can answer many important questions related to the environment: For example, -

how fast a species will disappear in a given compartment discussed in Chapter 3. how much time it is needed to oxidize a pollutant in treatment facilities how fast processes take place in natural water, and thus will help understanding aquatic kinetics.

In the next section, we start with the definition of reaction rate and investigate how concentration, temperature and other system properties affect these rates. The rate constants can be calculated in different ways, such as empirically, mechanistically, and by performing molecular and quantum level calculations. In this book, we will focus on estimating these rate constants using empirical and mechanistic approaches. These calculations are provided in a later chapter. The present chapter mainly focuses on how reaction rate can be expressed as a function of system variables. 4.2

Reaction Rate

Consider an arbitrary mass of the reaction system in its initial state and place an invisible envelope around all the molecules that comprise that mass. This envelope creates the control volume within it. The control volume of the system should be large enough to contain a statistically significant number of molecules so that the concept of molar concentration is meaningful. The control volume of the system should also be small enough so that there are no spatial gradients of composition or temperature within this volume. Let us assume that we can observe this control volume at time t and count the molecules N j of each species j in it. We repeat the observation at time t + ∆ t when the number of molecules of the same species is N j + ∆ N j . During time ∆ t the number of molecules crossing the boundary of the control volume is negligible compared to the change in molecular form due to reaction within the system. The volume of the system can be evaluated at some time 1


Updated 01/28/05

1 t + ∆t ∫ V (t ) dt is the mean control volume of the system. (Note that ∆t t while the control volume may change the mass within it is constant). We can relate the number of Nj molecules to the number of moles via Avogradro's constant L, n j = . (L = 6.022.. x 1023 L molecules/mol). Then the rate of reaction (rate of formation) of species j is defined as the number of moles of j produced by reaction per unit time and unit volume of the system and is, therefore, given by

θ (t ≤ θ ≤ t + ∆t) so that V (θ ) =

Rj=

lim 1 ∆ n j 1 dn j = ∆ t → 0 V (θ ) ∆ t V dt

(1)

Please, note that while our above verbal statement is indeed the general definition of the reaction rate, eq (1) is not - it is subject to the assumptions and scenarios depicted above for an isolated, closed, batch control volume. The point is that the rate of reaction of j is always the change in the number of moles of j per unit time and unit volume of the system caused by reaction, but it is only represented mathematically by eq (1) for a closed batch system without spatial gradients. One should remember that in flow systems, or in macroscopic batch systems with spatial gradients, the number of molecules of a species within an element of volume changes not only due to reaction but also due to species being brought in and out of this element of volume by flow or diffusion. Hence, one must take considerable precautions in obtaining reaction rates from experimental data in order not to attribute rates of other processes to the rate of reaction. Details of including flow and diffusion is consider in later chapters. Our above definition defines reaction rate as an intensive property of the system. If only a single reaction takes place in the system then stoichiometry requires that the rates of various components participating in the reaction are tied as shown by equation (2). R

j

νj

=r

(2)

where r is the intrinsic kinetic rate of reaction. Unfortunately, even r is not uniquely defined for a reaction system since stoichiometric coefficients can by multiplied by any arbitrary constant, while the number of moles of j converted per unit time and unit volume, R j, only depends on the conditions of the system not on how stoichiometry is written. Thus when we define, r, it is usually with reference to a given stoichiometry for the reaction. Experience shows that reaction rate is a function of composition, i.e concentrations, temperature, catalyst activity, etc. We will talk about the dependence of rate on concentration in the next section. 4.2.1 Dependence of rate on concentration

Often, within a narrow range of composition and temperature, the rate can be correlated with a power law dependence on concentration and an exponential dependence on temperature. Hence, for an irreversible reaction a A + b B → p P with K → ∞ an n-th order rate form is frequently fitted to data

r = k o e − E /R T C A α C B β

(3)

2


Updated 01/28/05

This rate form is a product of a function of temperature with a function of composition. At constant temperature we speak of a (specific) rate constant k = k o e − E /R T where k o is the frequency factor, E is the activation energy, R the ideal gas constant and T absolute temperature. This is an Arrhenius form for the temperature dependence which will be discussed later. The order of reaction with respect to reactant A is α and the order with respect to reactant B is β . The overall reaction order is n = α + β . If we measure concentrations in (mol/L) then the units of the rate constant for an n-th order reaction are {(mol/L)1-ns-1]. For a reversible single n-th order reaction (aA + bB = pP) the rate is given by

r = k fo e − E1

RT

α

β

C A C B − k bo e − E2

RT

Cp

γ

(4)

The rate constant for the forward reaction is k f = k fo e − E1

RT

, where Ef is the activation energy for the

− Eb RT

forward reaction, and k b = k bo e is the rate constant for the reverse reaction and Eb the activation energy for the reverse reaction. Reaction order for the forward reaction is n = α + β and for the reverse reaction is γ . At equilibrium the net rate of reaction is zero and the forward rate equals the reverse rate. Hence, the rate form must be compatible with thermodynamics and this requires that

⎛ C pγ =⎜ k b ⎜⎝ C A α C B β

k

f

⎞ ⎟ =K 1/ s c ⎟ ⎠ eq

so that p = sγ , a = sα , b = sβ

(5)

and

and E f − Eb =

∆ Hr s

(6)

where s is the stoichiometric number of the rate limiting step to be discussed later. One should note that reaction rate is a much stronger function of temperature than of concentration. Activation energies are typically in the range of 10 to 60 kcal/mol (about 40 to 250 kJ/mol). For example, for a irreversible second order reaction, with an activation energy of 40 kcal/mol, doubling the reactant concentration quadruples the reaction rate but raising the temperature by 10oC, from 20oC to 30oC, raises the rate almost tenfold (9.66 times). Many reaction rates cannot be fitted by the power law form and more complex forms are required to tie reaction rate, temperature and concentrations. Often when the functional dependence on temperature and composition cannot be separated in a convenient product of two functions, then a reaction order and activation energy cannot be separately defined. For example the simplest rate form of the HougenWatson or Langmuir-Hinshelwood type is given by

3


Updated 01/28/05 − E / RT

r=

k oe 1 C A K o e −∆ E /R T + C

(7) A

where E 1 , ∆ E , k o and K o are positive constants. Figure 1a illustrates that reaction order cannot be defined as it is readily seen that the apparent reaction order varies between order one (when at small − ∆ E / RT C A , C A > Ko e − ∆ E / R T ). We also see from Figure 1b that the apparent activation energy varies between E1 at low temperature, when − ∆ E/ R T − ∆ E/ RT C A >> Ko e , and (E 1 − ∆ E ) at sufficiently high temperature when C A k 2 and Rp=

k1k 2 [ A] = k k −1

1 ∞

[ A]

(34)

If pressure is sufficiently low k - 1 [A ] k

2

[M ]

− R R* = 2 K 1 k 2 ⎡⎣ R* ⎤⎦

2

(58)

[M ]

and third order termination is observed. Here K 1 = k 1/ k 2 is the equilibrium constant for the first step. Combination and termination of free radicals most likely follows the above mechanism. The combination of atoms represents an extreme situation. If energy transfer applies two atoms will come together and separate within the period of first vibration ~ 10-13 s unless an effective chaperon molecule arrives and collides with the complex within such short a time. Only at gas pressures of 104 to 105 atm are collision frequencies high enough for them to be likely. The combination rates are low and 3rd order. However, an alternative mechanism often plays a role, that is the atom-molecule complexation mechanism. R* + M

k

1 ⎯⎯→ ←⎯⎯ ⎯ k

RM *

(59a)

−1

20


RM * + M

Updated 01/28/05

2 ⎯⎯→ ←⎯⎯ ⎯ k k

RM + M

(59b)

−2

k2 RM + R* ⎯⎯ → R2 + M

(59c)

The predicted rate is − RR* = 2k3 K1 K 2 ⎡⎣ R* ⎤⎦

2

[M ]

(60)

where K 1 = k1 k −1 and K 2 = k 2 k − 2 . The kinetics remains third order at all pressures. This mechanism is favored if k1 is large, R M * is a strong complex. How to derive rate forms from postulated mechanism in a systematic way will be discussed in the next Chapter. Elementary reactions in solution Elementary reactions also occur in the liquid phase. They can be reactions between two species dissolved in a solvent or reactions of solute with the solvent. These reactions are quite different from gas phase reactions, since the medium (solvent) is considerably denser and more viscous. The solvent may and may not participate in the stoichiometry of the reaction. In either event, since the concentration of the solvent is much greater than that of the solute reactants, the solvent concentration does not change due to reaction. Some new concepts, yet to be introduced, will be needed, to handle even elementary reactions in solution. This will be postponed for a later chapter, after our treatment of elementary gas phase reactions via collision and transition complex theory. 4.4

Non-elementary Reactions Composite or complex reactions consist of a sequence of elementary steps and speaking of the molecularity for the overall reaction has no meaning. Reaction order and stoichiometry are now unrelated. Many reactions proceed via chain mechanisms that induce a closed loop catalytic sequence. Catalysts can be generated by initiation steps of the chain reaction or are unchanging species that speed up reaction rates. Inhibition of the rate, i.e slow down, is catalyzed by inhibitors - negative catalysts. We will consider next how to evaluate rate forms for gas phase reactions based on proposed mechanisms consisting of sequences of elementary steps.

21

February 9, 2007 Addendum to Chapter 4 For the spontaneous decomposition of ozone triggered by activation caused by UV radiation, we have seen that the following mechanism is in agreement with the observed rate form. O3

k1 ⎯⎯ →

← ⎯k2⎯

O + O2

(1)

k3 O + O3 ⎯⎯ → 2O2

(2)

The overall stoichiometry is

2O3 = 3O2

(3)

The first step, shown is reversible, is essentially the UV triggered initiation step, from left to right, and extinction (termination) step for the free radical, from right to left.

We have seen that application of PSSA to the above mechanism leads to the following rate form: 2

−RO3 =

2k1k3CO 3 k 2CO2 + k 2CO3

(4)

Since the first term in the denominator dominates at most conditions, the rate can be represented as 2

C −RO3 = 2k3K1 O 3 CO2

(5)

where K1 = k1/k2 is the equilibrium constant for the first step. Observing our previous result for the concentration of the active intermediate, CO, we note that CO = K1

CO 3 CO 2

(6)

so that the rate can be represented by −RO3 = 2 k3 ( CO )(CO 3 )

(7)

which is the rate for step 2, which is the rate-limiting step (RLS). 1

The value of the rate constant is reported as: k3 = 1.9x10−11 e−2300 / T (cm 3 / molecule s)

(8)

[Convert this to usual molar units such as (cm3/mol s) or (L/mol s)]

We are interested in estimating by how much can this rate be accelerated in the presence of the catalyst such as chlorine atoms or radicals that occur in the stratosphere due to the presence of fluorochlorocarbons like freons.

This catalyzed decomposition of ozone can be represented by a simplified RowlandRobinson mechanism CF2Cl 2

hυ → < 220 nm

Cl + CF2Cl

(9)

The above is the initiation step that gives rise to the following catalytic sequence (cycle) c1 Cl + O3 ⎯⎯ → O2 + ClO

(10)

c2 → Cl + O2 ClO + O ⎯⎯

(11)

k

k

The sum of (10) and (11) leads to Cl O + O3 ⎯⎯ → 2O2

(12)

and we want to compare the rate of the catalyzed ozone decomposition (reactions (10) and (11)) to step (2) of the uncatalyzed mechanism. It is known that kc1 = 5 x10−11 e−140 / T ( cm 3 / molecule s )

(13)

kc 2 = 1.1x10−10 e−2.20 / T ( cm 3 / molecule s )

(14)

Notice the much lower activation energy for the rate constants of the catalyzed steps.

Using PSSA we set the net rate of formation of ClO to zero. RClO = k c1CCl CO 3 − kc 2CClO CO = 0

(15)

Since the sequence is catalytic we recognize that the total concentration of Chlorine atoms available (Cl and ClO) is constant CCl + CClO = CCol

(16)

2

Since from eq (15) it follows that

CCl =

kc 2CO CClO k c1CO 3

(17)

Upon substitution for CClO from eq(16) into eq (17) we get

CCl =

o kC 2CO CCl k c1CO 3 + kc 2CO

(17a)

and

CClO =

O kC1C03CCl k c1CO 3 + kc 2CO

(17b)

The catalyzed rate of ozone decomposition becomes

−RO3C = kc1CCl CO 3 =

o CO CO 3 k1kc 2CCl kc1CO 3 + kc 2CO

(18)

Typically, kc 2CO > k2 CO3 the rate simplifies to: − RO3 ≈

2 k1 f k2 CO23 k1b

(12)

CO2

6


8.

Updated 01/31/05

The experimentally found rate under these conditions of low O3 is RO3 = k CO23 CO−21 . Agreement seems to exist between the proposed mechanism and data.

Consider another example. The overall reaction stoichiometry is given as: A+B=R

(13)

and the experimentally determined rate is − RA = k C A2

(14)

It is desired to determine whether the mechanism outlined below is consistent with the observed rate. 1.

k1 A + A ⎯⎯→ I

(15a)

k2 I + B ⎯⎯→ R+A

(15b)

2.

The mechanism is obviously consistent with the overall stoichiometry since a straight forward addition of steps leads to it. The number of intermediates is one, I.

3.

The net rate of formation of the intermediate is: R I = k1 C A2 − k 2 C I C B = 0

4.

(16)

And according to PSSA is equal to zero. The concentration of I is then k1 C A2 (17) k2 CB Since A appears in both steps of the mechanism and R only in one, set up the rate of formation of R for convenience CI =

5.

RR = k 2 C I C B

(18)

Eliminating C I we get R R = k1 C A2 6.

(19)

From stoichiometry − R A RR = = k1 C A2 1 1

7.

(20)

The rate agrees with the experimentally determined one. (Compare eq. (20) and (14)). The mechanism may be right.

7


Updated 01/31/05

NOTE: In writing the net rate of formation of I , RI , in step 3 we used the equivalent rate and thus based the rate constant k1 on I. If we were to write the rate of disappearance of A, − R A = 2k1 C A2 − k 2 C I C B

(21)

notice that the first term has to be multiplied by the stoichiometric coefficient of A in the 1st step, which is two, since the rate of disappearance of A in the 1st step is twice the rate of appearance of I in that step. If we followed a different convention we would have based k1 on the left hand side from where the arrow originates i.e. we would have based it on a component A. In that case a factor1/2 would have appeared with k1 in step 3 and there would be no 2 but 1 in front of k1 in the expression for − R A . This seemingly trivial point often causes a lot of errors and grief. Just notice that if we forgot the proper stoichiometric relationship, and the expression for − R A in eq. (21) did not have a 2 k1 but k1 , while R I was as given, they would be identically equal to each other and thus asserting R I = 0 would be asserting that − R A = 0 , which is absurd since − R A = RR which is a respectable expression. This demonstrates the importance of not forgetting the stoichiometric coefficients in setting up rates. At the end it should be mentioned that the PSSA fortunately does not only rest on the verbal arguments presented at the beginning of this section (can you find any fault with this?) but has solid mathematical foundations originating in the theory of singular perturbations. This will be illustrated in an Appendix since the same theory is applicable to may other situations. The application of PSSA to multiple reactions is a straightforward extension of the above procedure.

5.2 RATE LIMITING STEP ASSUMPTION (RLSA) This approach is less general than the previous one, in the sense that we must know more about the mechanism than just its form in order to apply it. The rate expression obtained from RLSA represents thus a limiting case of the one that could be obtained using PSSA. The basic hypothesis of the RLSA is as follows: 1.

One step in the mechanism is much slower than the others and thus that rate limiting step determines the overall rate.

2.

With respect to the rate limiting step all other steps may be presumed in equilibrium.

This is best explained based on an example. Consider a reaction 2A + B = R. Suppose that we want to find its rate form based on the mechanism shown below. In addition, it is known that the 1st step is the slowest. If we then depict the magnitude of the forward and reverse rates for each step with , and the magnitude of the net forward rate by a solid arrow, and if we keep in mind that the net rate must be of the same magnitude in all the steps (which is required by PSSA) because otherwise the active intermediates would accumulate someplace, we get the picture presented to the right of the hypothesized mechanism below.

1.

A

k1 f

←⎯ ⎯⎯ ⎯⎯ ⎯→ k

A*

1b

8


2.

A* + B

Updated 01/31/05

k2 f

←⎯⎯⎯ ⎯⎯ ⎯→ k

AB *

2b

3.

AB * + A*

k3 f

⎯⎯⎯→ ←⎯ ⎯⎯ k

R

3b

Clearly the magnitude (length)of the arrows indicating forward and reverse rates isthe smallest in step 1, this step is the slowest and limits the rate. At the same time the magnitude of the net rate forward is almost 1/2 of the total rate forward in step 1 and that step clearly is not in equilibrium while the ) in comparison to the total rate forward ( ) and total reverse magnitude of the net rate forward ( rate (←) in steps 2 and 3 is negligible. Thus, in these two steps rates forward and backward are approximately equal, and in comparison to step 1 these two steps have achieved equilibrium.

Thus, the procedure for applying the RLSA to a single reaction can be outlined as follows: 1.

Write down the hypothesized mechanism and make sure that it is consistent with stoichiometry.

2.

Determine which is the rate determining step. This should be done based on experimental information. Often various steps are tried as rate limiting due to the lack of information in order to see whether the mechanism may yield at all a rate form compatible with the one found experimentally.

3.

Set up the net rates of all the steps, other than the rate limiting one, to be zero, i.e. set up equilibrium expressions for all other steps.

4.

Set up the rate form based on the law of mass action for the rate determining step, and eliminate all concentrations of intermediates using the expressions evaluated in step 3.

5.

Using the stoichiometric number of the rate determining step relate its rate to the desired rate of a particular component.

6.

Check the obtained rate form against the experimentally determined rate.

7.

Remember that the agreement or disagreement between the derived and experimental rate form does not prove or disprove, respectively, the validity of the hypothesized mechanism and of the postulated rate limiting step. If the two rate forms disagree try another rate limiting step and go to step 3. If the two rate forms agree use the rate form with caution.

The greatest limitation of the rate forms based on RLSA is that they are much less general than those developed from PSSA and they do not test the mechanism in its entirety. The rate forms based on RLSA may be valid only in narrow regions of system variables (e.g., T, concentrations) since with the change in variables (i.e. concentrations, T) the rate limiting step may switch from one step in the mechanism to another. Clearly, the magnitude of our arrows representing the rates depends on concentration levels, temperature, pressure etc. and may change rapidly as conditions change. An example in the shift of the rate limiting step is the previously covered Lindemann's-Christensen mechanism for unimolecular reactions. Suppose that we want to evaluate the rate of formation of R for the reaction given above and under the assumptions made.

9


1.

Updated 01/31/05

We first test the compatibility of the proposed mechanism with stoichiometry. By inspection we find: 2 A = 2 A* A* + B = AB * AB * + A = R

v1 = 2 v2 = 1 v3 = 1

(22a) (22b) (22c)

2A + B = R 2.

Step 1 is rate limiting (given)

3.

Set up equilibrium expressions for step 2 and 3

4.

K C2 =

k2 f

K C3 =

k3 f

k 2b k 3b

= =

C AB*

(23a)

C A* C B CR CR = C AB* C A* K C2 C A2 C B

(23b)

Set up the rate for the rate limiting step

rl = k f C A − k1b C A*

(24)

Using eq. (23b) and eliminating C A* we get

C A* =

K C2

CR K C3 C B

rl = k1 f C A −

5.

K C 2 K C3

C R1 2 C B1 2

(24a)

The stoichiometric number of the rate limiting step is v l = s = 2 as per (eq. 22a). Thus, since v RR = Rl rl v

RR =

6.

k1b

k1 f k1b C R1 2 1 rl = CA − 12 2 2 2 K C 2 K C3 C B

(25)

Information not available. Now suppose that in the same mechanism step 2 was rate limiting (i.e. arrows for the rate of step 2 now being much shorter than those in step 1 and 3). We can quickly write the equilibrium relationships for step 1 and 3:

10


Updated 01/31/05 *

k1 f C = A = C A k1b

K C1

k3 f

KC3 =

=

k 3b

(26a)

CR *

C AB C A

(26b)

*

and from these obtain the concentrations of the intermediates

C A* = K C1 C A C AB =

(27a)

CR KC3 C A

*

K C1

(27b)

The rate for the rate limiting step (step 2) is:

rl =k 2 f C A* C B −k 2b C AB*

(28)

2 k 2b C R k 2 f K C1 K C 3 C A C B − k 2 b C R = K C1 K C 3 C A K C1 K C 3 C A 2

re = k 2 f K C1 C A C B −

RR = re since v l = 1

(28a) (29)

Clearly this is an entirely different rate form than obtained previously based on step 1 being rate limiting.

NOTA BENE 1: It would be quite tedious to find the rate form for the above mechanism based on PSSA since when the net rates of formation of intermediates are set to zero we get nonlinear equations * * due to the product C AB C A resulting from the rate forward in step 3. Try it anyway for an exercise.

NOTA BENE 2:

(

)

If the first step v 1 = 2 was rate limiting our rate form is given by RR =

k1 f

2

CA −

k1b

2 K C 2 K C3

= k f C A − kb

C R1 2 C B1 2

(25)

C R1 2 C B1 2

At equilibrium RR = 0 so that

kf kb

=

Recall that

k1 f k1b k1 f k1b

KC 2 KC3 =

C R1 2 C A C B1 2

(30)

= K C1

(31a) 11


Updated 01/31/05

and that for the reaction 2A + B = R the equilibrium (concentration units) constant K C is given by CR C A2 C B

KC =

(31b)

Substituting eqs. (31a) and (31b) into eq. (30) we get

kf kb

= K C1

K C 2 K C 3 = K C1 2

kf

Recall that

(32)

1

= K C = K C s where s is the stoichiometric number of the rate determining step. P

kb

Here s = v = 2 . Hence, 1

K C = K C1 K C 2 K C 3 2

(

(33)

)

If the 2nd step v 2 = 1 is rate limiting the rate form is given by eq. (28a) R R = k 2 f K C1 C A C B − = k f C A C B − kb

k 2b CR K C1 K C 3 C A

(28a)

CR CA

so that at equilibrium RR = 0

kf kb and K C =

=

CR 2

C A CB

k2 f k 2b

= KC

'

(33)

K C1 K C 3 = K C 1 K C 2 K C 3 2

2

(33)

Let us go back to the reaction of decomposition of ozone 20 3 = 30 2 . The mechanism was given before and let us assume that the 2nd step is rate limiting. Then

K C1 =

k1 f k1b

C O* = K C1

=

C O 2 C O*

(34a)

CO3

CO3 CO 2

(34b)

12


Updated 01/31/05 2

C rl =k 2 CO* CO 3 =k 2 K C1 O 3 CO 2

(35)

Now v l = 1 but vO3 = 2

Thus, − RO 3 = 2k 2 K C1

2 2 k1 f K 2 C O 3 2 CO3 = CO 2 k1b CO 2

(36)

Notice that this is the rate form obtained when using PSSA after certain additional assumptions were made. The rate form generated by the use of PSSA without additional assumptions is much more general. The above expression is its limiting case. For multiple reactions, rate forms can be obtained by RLSA by the straightforward extension of the above rules.

5.3 HALF LIFE AND CHARACTERISTIC REACTION TIME Based on either PSSA or RLSA we are usually able to derive a rate form for a particular reaction. Often n-th order rate form is obtained , say for reactant j:

(− R ) = kC j

n j

(37)

Then the characteristic reaction time is defined as τ R =

1

kC jo

n −1

and it is the time that it would take

in a close system (batch) for the concentration C j to decay to e −1 of its original value C jo . The balance for reactant j in a closed system is

dC j dt

= − R j . So when the reaction is not n-th order the

characteristic reaction time can be defined as τ R =

C jo R jo

where the rate R jo is evaluated at initial

conditions.

Half-life is the time needed for the reactant (species) concentration to be reduced to half of its original value in a closed system. Integration of the species balance yields t1 2

t1 2 =

C jo

∫ dt = ∫ o

C jo 2

dC j Rj

=

C jo R jo

1

dc ∫1 R = τ R 2

1

dc

∫R

(38)

1 2

where c = C j C jo and R = R j R jo . The integration is performed by substituting into the above expression all concentrations in terms of C j using the stoichiometric relations. For an n-th order reaction we get 13


t1 2 =

Updated 01/31/05

2 n −1 −1

k (n − 1) C jo

n −1

=

2 n −1 − 1 τ (n − 1) R

(39)

For a first order process

t1 2 =

ln 2 = τ R ln 2 k

(39a)

5.4 COMPARTMENTAL MODELING OF SINGLE PHASE SYSTEMS

We should note that in developing the rate forms based on mechanisms we have so far assumed that we deal with a system of constant mass consisting of fixed amounts of various atomic species. These atomic species are constituents of chemical species (components) that participate in elementary chemical reactions taking place in the system. These chemical species are either stable, and present in measurable concentrations (e.g. reactants and products), or are active intermediates present in much smaller concentrations. The volume of the system (i.e. of the invisible envelope that engulfs our constant mass) is small enough that molecules are free to interact by molecular motion so that concentration or temperature gradients can never develop in the system. Hence, the system is well mixed. Moreover, the system is closed since no exchange of mass occurs across system’s boundaries. We extend this approach now to modeling of isothermal, single phase systems or arbitrary volume. We assume that the volume of the system is perfectly mixed at all times so that there never are any spatial composition gradients in the system. We can apply then the basic conservation law to quantities of the system that are conserved (e.g. mass, species mass, energy etc.) that states:

(Rate of accumulation ) = (Rate of input ) − (Rate of output ) + (Rate of generation )

(40)

Such a well mixed system, to which we apply equation (40) often is called a compartment - hence, the name compartmental modeling. We can have closed systems (no exchange of mass across system’s boundary) and open systems (that exchange mass with the surroundings). Here we focus only on single phase systems and we start by developing the governing equation for a closed system. A sketch of a closed system is shown below: SYSTEM OF - VOLUME V - MASS M

At time t = 0 c j = c ji At time t = t

c j (t )

- Total mass of the system M (kg ) = const - Total volume of the system V m 3 may vary in time ⎛ mol j ⎞ = n j V molar concentration of species j in the system at time t - cj⎜ 3 ⎟ ⎝ m ⎠

( )

14


Updated 01/31/05

⎛ mol ⎞ - R j ⎜ 3 ⎟ = reaction rate of j ⎝m s⎠

For a well mixed closed system the species mass balance for species j can be written as: d (VC j ) dt

= VRj

;

t = 0, Cj = C jo

(41)

where ⎛ mol j ⎞ Cj⎝ = molar concentration of j at time t L ⎠ V(L) = volume of the system t (s) = time ⎛ mol j ⎞ Rj ⎜ = reaction rate of production of species j (if j is a reactant Rj < 0) ⎝ Ls ⎠ ⎛ mol j ⎞ C jo ⎝ = initial concentration of species j L ⎠ Note: units of time instead of seconds could be minutes, hours or days; volume could be measured in m3, etc. If the volume of the system is constant V = const, eq (41) is reduced to: dC j

= Rj

dt

;

t = 0, Cj = C jo

(41a)

If we have R reactions occurring in the system, then their stoichiometry is given by: s

∑υ A j =1

ij

j

= 0 ; i = 1,2,...R

(42)

here υij = stoichiometric coefficient of species j in reaction i (recall υij > 0 products, υij < 0 reactants) and S = total number of stable species in the system. Often, in chemical as well as in environmental systems such as encountered in air or water pollution, the set of R reactions is extended to include all active intermediate species (their total number is S ' ) that are linked among themselves and with stable species via a large set of R' elementary reactions based on the proposed mechanism. (Note R’ > R). This augmented set of R' elementary reactions arising from the mechanism can be written as S +S '

∑υ j =1

m ij

Aj = 0

; i = 1, 2,... R '

(43)

where υijm is the stoichiometric coefficient chemical species of j in the mechanistic step i.

15


Updated 01/31/05

There are two ways now to proceed. a) An exact solution can be obtained by solving equation (41a) for all the S + S' species, stable as well as active intermediates, i.e for j = 1,2,3...S,S+1, S+2...S+S'. This can be stated as: R' dC j m = Rj = ∑ υij ri dt i= 1

; t = 0, C j = Cjo

(44)

for j = 1,2,...S, S + 1... (S' + S) where ri is the reaction rate of the elementary reaction i in the proposed mechanism for which the rate constant is evaluated from transition state theory (TST) or by other appropriate means. For each of the stable compounds (j = 1,2...S) the initial concentration, Cjo, is known, for all active intermediates (j = S + 1, S + 2...S + S') the initial concentration is zero, Cjo = 0. This (eq (44)) results in a stiff system of differential equations which must be solved by an appropriate computer package. Numerical solution is almost always required. The stiffness (vastly different eigenvalues) is caused by a wide range of time constants. The results can be displayed for active intermediates as well as stable compounds as functions of time. b)

An approximate solution can be obtained by using PSSA. For all active intermediates we set R'

m Rj = ∑ υij rj = 0 for j = S + 1, S + 2,...S + S'

(45)

i =1

From the resulting S' equations (often nonlinear) we evaluate the S' concentrations of active intermediates which then can be substituted into the rate forms for stable species. S differential equations need to be solved now (instead S+S’): R dC j o = Rj = ∑ υijri ; t = 0, C j = Cjo dt i= 1

(46)

only for the S stable species (j = 1,2,3...S). The reaction rate for reaction i among stable species, ri o , is now obtained from the mechanism by applying PSSA and eliminating the concentrations of active intermediates. These expressions for ri o are not rates of elementary reactions necessarily, and can be of rather complex form. Upon solution of equation (46) for the concentrations of the stable species these can be displayed as function of time. The concentrations of the active intermediates can then be calculated from the solution of equation (45) and are constant in time. Nota bene: More complex dynamic behavior with limit cycles, etc., can also be encountered. Concentrations of active intermediates are always small. 2.

Open System

A sketch of the open system is shown below: Qin

C jo

SYSTEM OR

Qout

MASS M VOLUME V

Cj

16


Updated 01/31/05

( ) (m s ) - volumetric flow rate out

-

Qin m 3 s = volumetric flow rate in

-

Qout 3 C j out ≡ C j for all j. Outlet of perfectly mixed system is representative of system’s content.

In the open system all the species are brought in and out across the system’s boundary. We will assume here that this happens by convection only, and that the volumetric flow rate in and out is the same, Q (L s ) = Qin = Qout . The species balance which is based on the basic conservation equation (40) can be written as: V

dCj dt

(

)

= Rj V + Q C jin − Cj ; t = 0, Cj = C jo

(47)

⎛ mol j ⎞ where C jin ⎝ is the concentration of species j in the incoming (feed) flow. Due to the assumption L ⎠ of perfect mixing the concentration of j in the exit flow is the same as the concentration Cj of j at that time in the system. The rate Rj is given as before. Again one can go for the exact solution by applying eq (47) to all species, j = 1, 2, 3...S + S'. Or one can attempt a PSSA solution by setting Rj = 0 for j = S + 1, S + 2,...S + S' and by solving eq (47) for j = 1, 2...S for the stable species only. dC j = 0 and then An open system can reach steady state when dt

(

(− R j ) V = Q Cjin − Cj

)

(47a)

In complex reaction networks true steady state may never be reached as cycles set in. We can apply now the compartmental single phase model to gaseous systems contained in a vessel, in a room, urban atmosphere or in the whole atmosphere. The underlying assumption is then that in the volume we have chosen mixing is perfect so that no composition spatial gradients occur at any time. Otherwise we must divide the system into a number of compartments and describe the flows between them. Let us consider some gas phase atmospheric reactions next. 5.5

GAS-PHASE ATMOSPHERIC REACTIONS

Atmosphere is an oxidizing environment with abundance of solar energy (during day time hours) so that the pollutants are continuously brought to a higher and higher oxidation state. Hydrocarbons are reacted to aldehydes, then acids, then CO2. Sulfur containing compounds are oxidized to SO2 and ultimately to H2SO4. Ammonia and other nitrogen containing species are oxidized to NO, then NO2 and HNO3. 5.5.1

Atmospheric Photochemical Reactions

These are initiated by absorption of a photon by an atom, molecule, radical or ion. k A + h ν ⎯⎯ → A*

17


Updated 01/31/05

The excited activated molecule (radical), A * , can participate in four different reaction types k1 dissociation A* ⎯⎯ → B1 + B2 k2 direct reaction A* + B ⎯⎯ → C1 k3 * fluorescence A ⎯⎯→ A + hν k4 collisional deactivation A* + M ⎯⎯ → A+M

(1) (2) (3) (4)

Reactions (1) and (2) lead to chemical change while (3) and (4) return the molecule to its ground state. The quantum yield, φi , of a process (i = 1 to 4) is defined as the ratio of the number of radicals A * reacting by that process to the number of photons absorbed. The active intermediate A * is short lived so that PSSA applies 0 = RA* = k [ A] − k1 ⎡⎣ A* ⎤⎦ − k2 ⎡⎣ A* ⎤⎦ [ B ] − k3 ⎡⎣ A* ⎤⎦ − k4 ⎡⎣ A* ⎤⎦ [ M ] k [ A] ⎡⎣ A* ⎤⎦ = k1 + k2 [ B ] + k3 + k4 [ M ]

The rate of formation of [A *] by absorption of light is k [ A ] where k is the specific absorption rate (k first order rate constant independent of [A ] ). Using the quantum yield definition for the above four processes we can write RB1 = k1 ⎡⎣ A* ⎤⎦ = φ1k [ A]

RC1 = k2 ⎡⎣ A* ⎤⎦ [ B ] = φ2 k [ A] RA = k3 ⎡⎣ A* ⎤⎦ = φ3 k [ A]

RA = k4 ⎡⎣ A* ⎤⎦ [ M ] = φ4 k [ A] 4

So that

∑φ i =1

i

=1.

The solar flux, or actinic irradiance, F , (photons/cm2s) is the radiation intensity striking a unit area of surface per unit time. dF = I ( λ ) d λ

where I ( λ ) d λ = actinic irradiance in the wave length range ( λ to λ + d λ ) so that the actinic irradiance density I ( λ ) is in (photons/cm3s). We are most interested in dissociation reactions for which the first order rate constant is given by

18


Updated 01/31/05

λ2

k1 = ∫ σ

A

(λ,T ) φA (λ,T ) I (λ ) d λ

λ1

where

σ A ( λ , T ) = absorption cross-section of A at wave length λ and temperature T φA (λ,T

)

= the quantum yield or probability that molecule A decomposes on absorbing radiation of wave length λ at temperature T.

The integral is usually approximated by summation. In the troposphere the wave lengths of interest are 280 nm < λ < 730 nm. Stratosphere O2 and O3 block out most small wave lengths λ < 280 nm. No chemistry occurs at λ > 730 nm. Absorption cross-sections and quantum yields are measured in laboratory. (Examples are shown in Figures 1 and 2). The major variable is the intensity change that occurs as a function of time of day, latitude, time of year and state of the atmosphere (e.g. humidity, cloudiness, particulates, etc.). Solar flux as a function of λ is shown in Figure 3. In the US maximum noontime, intensity in summer months is about 2 x 1016 (photons/cm2s) in the 300-400 nm range and is close to this value for 4-6 hours. In winter, 0.7 x 1016 to 1.5 x 1016 (photons/cm2s) are typical values depending on the latitude. Major photo dissociating species in the lower atmosphere are summarized in Table 1. FIGURE 1:

Absorption cross section for O3 and NO2 in the uv and visible regions of the spectrum (Luther and Gelinas, 1976)

FIGURE 2:

Experimental quantum yield data for the production of atomic oxygen (O) from the photolysis of NO2 as a function of wavelength. The dashed line represents the quantum yields suggested by Demerjian et al. 91980).

19


Updated 01/31/05

20


FIGURE 3:

Updated 01/31/05

Direct solar flux at 0 and 15 km as a function of wavelength (National Center for Atmospheric Research, 1982).

21


TABLE 1:

Updated 01/31/05

Major photodissociating species in the lower atmosphere.

22


Updated 01/31/05

TABLE 1CONTINUED:

From J. Seinfeld. Air Pollution

23


Updated 01/31/05

5.5.2 Ozone Formation - NOx Route When nitrogen oxides (NO or NO2 ) are present in sunlight, ozone formation occurs as the result of NO2 photolysis. The mechanism is as follows: k1 NO2 + hν ⎯⎯ → NO + O k2 O + O2 + M ⎯⎯ → O3 + M

(1) (2)

O3 + NO

(3)

k3 ⎯⎯ → NO2 + O2

This is an entirely closed cycle in which the energy is ultimately absorbed by the third body, M , and no overall change occurs. However, the dynamics of the system are of interest in determining if one starts with a certain concentration of [ NO ]o and [ NO2 ]o what would the peak concentration of ozone [O3 ] max be. Clearly PSSA implies RO = k1 [ NO2 ] − k2 [O2 ][ M ][O ] = 0

[O ] =

k1 [ NO2 ]

k2 [O2 ][ M ]

Since [O 2 ], [M ] are large and approximately constant we see that [O ] adjusts itself rapidly by following [N O 2 ]. A species balance for ozone in a closed system (i.e system of constant volume with no exchange with surroundings) yields d [O3 ] = RO3 = k2 [O ][O2 ][ M ] − k3 [O3 ][ NO ] dt Upon substitution of [O ] in the above we get d [O3 ] = RO3 = k1 [ NO2 ] − k3 [O3 ][ NO ] dt Similarly making a balance on N O and N O 2 we get d [ NO ] = RNO = k1 [ NO2 ] − k3 [O3 ][ NO ] dt d [ NO2 ] = RNO2 = − k1 [ NO2 ] + k3 [O3 ][ NO ] dt From the above it is evident that d {[O3 ] − [ NO ]} = 0 dt

24


Updated 01/31/05

so that

[O3 ] − [O3 ]o = [ NO ] − [ NO ]o It is also evident that d {[ NO] + [ NO2 ]} = 0 dt so that

[ NO ] − [ NO ]o = [ NO2 ]o − [ NO2 ] Using these relations we eliminate both concentrations of N O and N O 2 in terms of the concentration of ozone and of the initial concentrations indicated by subscript o

[ NO ] = [ NO ]o − [O3 ]o + [O3 ] [ NO2 ] = [ NO2 ]o + [O3 ]o − [O3 ] which yields the following rate of change of ozone concentration for any starting mixture d [O3 ] = k1 dt

{[ NO ] + [O ] − [O ]} − k [O ] ([ NO] − [O ] + [O ]) 2 o

3 o

3

3

3

3 o

o

3

[O3 ] = [O3 ]o

At t = 0

We are at the moment only interested in the maximum ozone concentration which occurs at the d [O3 ] pseudostationary state i.e when = 0. We can also look at it as the PSSA solution to the problem dt if we assume O3 also to be an active intermediate. This is justified since its concentration is always much smaller than that of oxides of nitrogen. Now we get:

[O3 ] =

k1 [ NO2 ] k3 [ NO ]

=

k1 ( [ NO2 ]o + [O3 ]o − [O3 ] )

(

k3 [ NO ]o − [O3 ]o + [O3 ]

)

Solving the resulting quadratic equation for [O 3 ] results in: 2

⎛ k ⎞ 4k 1 ⎜ [ NO ]o − [O3 ]o + 1 ⎟ + 1 [O3 ] = ⎝ k3 ⎠ k3 2 −

([ NO ] + [O ] ) 2 o

3 o

1⎛ k1 ⎞ ⎜ [ NO ]o − [O3 ]o + ⎟ 2⎝ k3 ⎠

If [ N O ]0 = [O 3 ] 0 = 0 then the above simplifies to the following: 25


Updated 01/31/05

2 ⎛ ⎞ 1 ⎜ ⎛ k1 ⎞ 4k1 k1 ⎟ = + − O [ NO ] [ 3] ⎜ ⎜ ⎟ 2 o 2 ⎝ k3 ⎠ k3 k3 ⎟ ⎝ ⎠

(*)

Since k 1 / k 3 = 0.01 ppm we get from equation (x)

[ NO2 ]o [O3 ]max

ppm ppm

0.1

1.0

10

0.027 0.095 0.311

If we take [ NO2 ]o = [O3 ]o = 0 then

[O3 ] = 0

since ozone formation cannot get started.

The maximum ozone concentration is achieved when the initial charge contains only NO2 in air. Nevertheless, the mechanism consisting solely of eqs (1-3) does not explain the level of ozone concentrations encountered (i.e it underpredicts them) and needs to be modified. 5.5.3 Photolysis Rate of NO2

We want to find the photolysis rate constant k1 from experimental measurement in a closed system. To understand the dynamics of N O 2 photolysis let us add the following 6 mechanistic steps to the previously considered three step mechanism: k4 O + NO2 ⎯⎯ → NO + O2 k5 O + NO2 + M ⎯⎯→ NO + M k6 NO + NO3 ⎯⎯→ 2 NO2

(4) (5) (6)

k7 O + NO2 + M ⎯⎯→ NO2 + M k8 NO2 + NO3 + M ⎯⎯→ N 2 O5 + M

(7) (8)

k9 N 2O5 ⎯⎯ → NO2 + NO3

(9)

Inclusion of a third body M indicates a third order reaction. Our experience and measurements in this system indicates that O , O3 , NO3 and N2O5 can be treated as active intermediates since their concentrations are much smaller than those of the other species. Applying PSSA to these 4 species we get RO = r1 − r2 − r4 − r5 − r7 =

k1 [ NO2 ] − k2 [O ][O2 ][ M ] − k4 [O ][ NO2 ] − k5 [O ][ NO2 ][ M ] − k7 [O ][ NO ][ M ] = 0

RO3 = r2 − r3 = k2 [O ][O2 ][ M ] − k3 [O3 ][ NO ] = 0

RNO3 = r5 − r6 − r8 + r9 = k5 [O ][ NO2 ][ M ] − k6 [ NO ][ NO3 ] − k8 [ NO2 ][ NO3 ][ M ] + k9 [ N 2O5 ] = 0

RN2O5 = r8 − r9 = k8 [ NO2 ][ NO3 ][ M ] − k9 [ N 2 O5 ] = 0

26


Updated 01/31/05

From RN2O5 = 0 we get k k9

[ N 2O5 ] = 8 [ NO2 ][ NO3 ][ M ] From RNO3 = 0 we get

[ NO3 ] =

k5 [O ][ NO2 ][ M ]

2

k6 [ NO ]

which substituting in the above yields

[ N 2O5 ] =

k5 k8 [O ][ NO2 ] 2 [ M ] k6 k9 [ NO ]

From before we knew that

[O3 ] =

k2 [O ][O2 ][ M ] k3 [ NO ]

and we get the active oxygen concentration from R O = 0 as:

[O ] =

k1 [ NO2 ]

k2 [O2 ][ M ] + k4 [ NO2 ] + k5 [ NO2 ][ M ] + k7 [ NO2 ][ M ]

Substitution of the expression for [O] into the equation for [O3 ] completes the problem of determining the concentrations of active intermediates in terms of concentrations of stable species. Let us now consider our goal, which is to monitor the rate of change of [NO2 ] in a closed system. The governing equation is: d [ NO2 ] = − r1 + r3 − r4 − r5 + 2 r6 + r7 − r8 + r9 dt Using the above PSSA relations for the rates we get d [ NO2 ] = − r2 − 2 r4 − 2 r5 + r3 + 2 r6 = − 2 r4 dt So that

27


d [ NO2 ] dt t=0

Updated 01/31/05

= − 2 k4 [O ][ NO2 ]

[ NO2 ] = [ NO2 ]o

The above can be rearranged as 1 d [ NO2 ] d ln [ NO2 ] = = − 2 k4 [O ] dt [ NO2 ] dt Substituting in the expression for [O ] and rearranging yields k [ M ] k7 [ M ][ NO ] k2 [ M ][O2 ] − 2k1 =1+ 5 + + d l n [ NO2 ] k4 k4 [ NO2 ] k4 [ NO2 ] dt

(**)

In the absence of any reactions but 1 and 4 (i.e k2 = k5 = k6 = k7 = k8 = k9 = 0) all oxygen atoms would end up as molecular oxygen giving an overall quantum yield of 2, i.e two molecules of N O 2 disappear for each photon absorbed. In that case − d ln [ NO2 ] = 2 k1 dt

which is observed at very low pressure experiments when [ M ] ≈ 0 . However, that is not the case at atmospheric pressure when other interactions indicated by the mechanism above occur. Now, by assuming that the measurable stable nitrogen species are [N O ] and [N O 2] only, the mass balance on nitrogen requires that

[ NO ] = [ NO ]o + [ NO2 ]o − [ NO2 ] Substitution into eq (**) and integration yields t

[ NO2 ]

o

[ NO2 ]0

−2k1 ∫ dt = t

− 2 k1 ∫ dt = o

∫

⎛ k5 [ M ] k7 [ M ][ NO2 ]0 + [ NO ]0 − [ NO2 ] k2 [ M ][O2 ] ⎞ d [ NO2 ] + + ⎜⎜1 + ⎟ k4 k4 [ NO2 ] k4 [ NO2 ] ⎟⎠ [ NO2 ] ⎝

⎛ k5 [ M ] k7 [ M ] ( NO2 )o + [ NO ]o − [ NO2 ] k2 [ M ][O2 ] ⎞ 1 + d [ NO2 ] ⎟ ∫ ⎜⎜1+ k4 + k4 [ NO2 ] k4 [ NO2 ] ⎟⎠ [ NO2 ] [ NO2 ]o ⎝ NO2

The result of this integration can be presented as:

28


[NO 2 ]0 ⎛ [NO 2 ]0 ⎞ + a 2 ⎜⎜ − 1⎟⎟ (NO 2 ) ⎝ ( NO 2 ) ⎠ ⎛ NO 2 ]0 − [NO 2 ] ⎞ [ ⎟ + ⎜⎜ a 2 [NO ]0 a 3 [(O 2 )0 ] ⎟ NO NO [ ] [ ] 2 2 0 ⎝ ⎠ = (1 + a1 − a 2 ) ln

2k 1 t

where a1 =

Updated 01/31/05

k5 [ M ] k [M ] k [M ] ; a2 = 7 ; a3 = 2 k4 k4 k4

Further simplification is possible by conducting the experiment with [ NO ]o = [O2 ]o = 0 i.e in a closed system containing only N 2 and N O 2 initially. Then 2k1t = (1 + a1 − a2 ) l n

[ NO2 ]o [ NO2 ]

⎛ [ NO2 ]o ⎞ + a2 ⎜⎜ − 1⎟⎟ ⎝ [ NO2 ] ⎠

Monitoring the concentration [N O 2] as a function of time and plotting the right hand side as the ordinate versus time, t , as abscissa allows the determination of k 1 . Agreement between data and experiment is indicated in Figure 4 for two different light intensities. FIGURE 4:

Nitrogen dioxide concentration as a function of time in a system initially comprising 5 ppm NO2 in N2. Experimental data and the predictions of the mechanism in the text are shown for two light intensities (Holmes et al, 1973).

5.5.4 Ozone Formation - CO Role The key reactions in O 3 formation via N O 2 are reactions 1-3 of the above proposed mechanism in section. In addition we will now consider 6 more mechanistic steps that can contribute to ozone formation. They are (for values of the rate constants at room temperature see Table 2):

29


Updated 01/31/05

k4 O3 + hν ⎯⎯ → O* + O2

(4)

k5 O* + M ⎯⎯ →O+ M k6 * O + H 2O ⎯⎯ → 2 OH * k7 a CO + OH * ⎯⎯→ CO2 + H * k7 b H * + O2 + M ⎯⎯ → HO *2 + M k8 HO *2 + NO ⎯⎯ → NO2 + OH * k9 OH * + NO2 ⎯⎯ → HNO3

TABLE 2:

(5) (6) (7a) (7b) (8) (9)

Atmospheric chemistry of CO and NOx

Note that O * is O (' D ) is the excited singlet oxygen atom for which direct transition to O ground state oxygen atom is forbidden . Hence, O * must react with another species present. The proposed mechanism clearly indicates the importance of the hydroxyl radical O H * and of the hydroperoxyl radical HO2* . The excited hydrogen atom H *, formed by (7a), is so short lived and disappears by (7b) that the two steps must be combined by PSSA to yield:. ⎡⎣ H * ⎤⎦ =

k7 a [CO ] ⎡⎣OH * ⎤⎦ k7 b [O2 ][ M ]

Now we can combine steps (7a) and (7b) into step (7) k7 ⎯⎯ → CO + OH HO *2 + CO2 (O2 ) *

(7)

where O 2 is indicated as being incorporated to satisfy the mass balance but does not affect the rate of step 7 which is second order overall, i.e first order w.r.t. C O and O H * each. The value of k 1 is indicated in Figure 5.

30


FIGURE 5:

Updated 01/31/05

Comparison of calculated diurnal variation of NO2, photolysis rate with experimental measurements of Zafonte et al. (1977) in Los Angeles. Reprinted with permission from Environmental Science and Technology. Copyright (1977). American Chemical Society.

The application of PSSA to the 4 intermediates O, O * , OH * , HO2* yields RO = r1 − r2 + r5 = k 1 [ NO2 ] − k2 [O ][O2 ][ M ] + k5 [ M ] ⎡⎣O* ⎤⎦ = 0 RO* = r4 − r5 − r6 = k4 [O3 ] − k5 [ M ] ⎡⎣O* ⎤⎦ − k6 [ H 2O ] ⎡⎣O* ⎤⎦ = 0 ROH * = 2 r6 − r7 + r8 − r9 = 0 RHO * = r7 − r8 = 0 2

Summing the last two rates yields: ROH * + RHO* = 2 r6 − r9 = 2 k6 [ H 2O ] ⎡⎣O* ⎤⎦ − k9 ⎡⎣OH * ⎤⎦ [ NO2 ] = 0 2

In addition, ozone can be treated as an active intermediate: RO3 = r2 − r3 − r4 = k2 [O ][O2 ][ M ] − k3 [O3 ][ NO ] − k4 [O3 ] = 0

We solve now for the concentrations of all the active intermediates

[O3 ] =

k2 [O ][O2 ][ M ] k3 [ NO ] + k4

⎡⎣OH ⎤⎦ = *

2 k6 [ H 2O ] ⎡⎣O* ⎤⎦ k9 [ NO2 ]

31


Updated 01/31/05

k4 [O3 ] ⎡⎣O * ⎤⎦ = k5 [ M ] + k 6 [ H 2 O ] k1 [ NO2 ] + k5 [ M ] ⎡⎣O* ⎤⎦

[O ] =

k2 [O2 ][ M ]

Hence by substituting [O * ] into the expression for [O ] we get k 4 k5 [ M ]

⎧

[O ] = ⎪⎨k1 [ NO2 ] + ⎪⎩

k5 [ M ] + k 6 [ H 2 O ]

⎫

[O3 ]⎪⎬ / k2 [O2 ][ M ] ⎪⎭

Substitution of [O ] into the expression for [O 3 ] yields

[O3 ] =

k1 [ NO2 ] +

k4 k5 [ M ][O3 ] k5 [ M ] + k 6 [ H 2 O ]

k3 [ NO ] + k4

Solving for [O 3 ] we get

[O3 ] = where a =

k1 [ NO2 ]

k3 [ NO ] + k4 a

1 k [M ] 1+ 5 k6 [ H 2O ]

which is constant at constant relative humidity. Now we also get k [O ] k1k4 [ NO2 ] a ⎡⎣O* ⎤⎦ = 4 3 a = k6 [ H 2O ] k6 [ H 2O ] ( k3 [ NO ] + k4 a ) 2 a k4 [O3 ] 2 a k1k4 ⎡⎣OH * ⎤⎦ = = k9 [ NO2 ] k9 ( k3 [ NO ] + k4 a )

etc. The dynamics of the system is now described by: d [ NO2 ] = − r1 + r3 + r8 − r9 dt

32


d [ NO ] dt d [CO ] dt

Updated 01/31/05

= r1 − r3 − r8 = − r7

Upon substitution of the rate forms, and elimination of the concentrations of the intermediates, we get: d [ NO2 ] dt d [ NO ] dt d [CO ] dt

=

=

k1k4 a ( 2 k7 [CO ] / k9 − 3[ NO2 ]) k3 [ NO ] + k4 a

k1k4 a ([ NO2 ] − 2 k7 [CO ] / k9 )

=−

k3 [ NO ] + k4 a

2 k1k4 k7 a [CO ] / k9 k3 [ NO ] + k4 a

with initial conditions at

[ NO2 ] = [ NO2 ]o [ NO ] = [ NO ]o [CO ] = [CO ]o

t=0

The above equations certainly can be integrated on any computer. Before doing so an approximate analysis is in order. Examination of Table 2 reveals k3 [ NO ] >> k4 a for all levels of [N O ] concentrations of interest. Then

[O3 ] =

k1 [ NO2 ]

k3 [ NO ] + k4 a

=

k1 [ NO2 ] k3 [ NO ]

as found before for the simple mechanism consisting of only 3 steps. This part of the cycle can be represented as follows: NO2

hv ,O2 ⎯⎯⎯ →

←⎯ ⎯

NO + O3

Photolysis generates O and N O 2 which with O 2 results immediately in N O + O 3. Ozone reacts with N O to regenerate N O 2. When C O is present the above cycle is modified as N O can be converted back to N O 2 via the hydroperoxyl radical i.e step 8

33


NO2

Updated 01/31/05

, O2 ⎯hv ⎯⎯ →

← ⎯⎯

NO +O3

HO2* But the HO2* radical gets converted to OH *, which with CO regenerates the HO *2 radical. Hence, one has two cycles - the fast NO2 , NO, O3 cycle and a slower CO, OH * , HO2* cycle ,O2 ⎯hv ⎯⎯ → NO2 NO +O3 ←⎯ ⎯ −O2

CO+OH * ⎯ ⎯→ HO2*

The CO/NO x reaction mechanism is a chain reaction with OH * as the chain carrier. The chain length Lc of such a reaction is defined as the number of propagation steps occurring for each termination step. In our mechanism Lc =

r8 r7 k7 [CO ] = = r9 r9 k9 [ NO2 ]

The role of the CO slow cycle is to slowly change (i.e increase) the [NO2] concentration and hence increase [O3 ] concentration which is still given by its pseudo steady state value. Because of the extreme rapidity of NO 2/O3 cycle it is the external independent paths that affect [NO2] which determine maximum [O3] levels. For completion of the ozone story one would need to include reactions of organic compounds and the effect of particulate matter (aerosol) as potential catalyst. This is beyond the scope of illustration of the PSSA assumption. The same rules as illustrated here can be applied.

34

THEORY OF ELEMENTARY REACTIONS (CHE 505) M.P. Dudukovic Chemical Reaction Engineering Laboratory (CREL), Washington University, St. Louis, MO


6. 6.1

Updated 01/31/05

THEORY OF ELEMENTARY REACTIONS Collision Theory

Collision theory is the extension of the kinetic theory of gases in predicting reaction rates of a bimolecular gas phase reaction of the type: A + A → P A + B → P

(1) (2)

It is assumed that the reaction rate expressed in terms of disappearance of molecules of A is given by: ⎛ molecules ⎞ ⎟ − ?r A ⎜ ⎝ cm 3 s ⎠

=2 f Z

AA

for reaction(1)

A factor two appears on the right hand side of eq (1) because for each collision two molecules of A disappear. ⎛ molecules ⎞ ⎟=fZ − ?r A = − r? B ⎜ A B for reaction (2) ⎝ cm 3 s ⎠ where

ZA B - is the total number of collisions between molecules A and B in 1 cm3 of reaction mixture per second (frequency of collision per 1 cm3) Z A A - is the total number of collisions among the molecules A in 1 cm3 per second f - the fraction of molecules that possess the required excess energy for reaction i.e the fraction of molecules that upon collision will react. The kinetic theory of gases postulates that the collision frequency Z A B is proportional to the size of the "target", to the relative mean velocity u A B ,, and to the number concentrations of both molecular species. The size of the "target" is interpreted as the maximum area perpendicular to the trajectory of the moving A (or B ) molecule which when it contains a B (or A ) molecule will lead to an inevitable collision of the two. (See Figure 1).

1


FIGURE 1:

Updated 01/31/05

Frequency of collisions: molecule A is moving with speed u relative to B. In

unit time the sphere of radius rA + rB has swept out a volume π (rA + rB ) u and has 2

encountered π (rA + rB ) uN B molecules of B. 2

r B B

u rB r A A

Thus

" t arg et area" = π d 2A B = π σ 2A B where d A B =

d

A

+d 2

B

= dAB is the radius of collision area

d A, d B can be interpreted as equivalent molecular diameter of molecule A and B , respectively, if the molecules are viewed as hard spheres according to the classical collision theory. d A, d B can also be viewed as diameters in space of the sphere within which the attraction forces of A or B would force a collision. The mean relative velocity is given by the kinetic theory as ⎛ 8k T ⎞ B ⎟ u AB = ⎜ ⎝ π µ AB ⎠

1/2

(3)

where k B = 1.38062 x 10-23JK-1 is the Boltzmann's constant.

µ

AB

=

m Am B − reduced molecular mass m A+mB

g ⎛ ⎞ being the molecular mass of A , m the molecular mass of B . with m A ⎝ B molecule ⎠ The number concentrations are:

2

(4)


N

A

Updated 01/31/05

⎛ ⎜ number of molecules of A ⎞ 3 ⎝ ⎠ cm

⎛ number of molecules of B ⎞ ; NB⎜ 3 ⎝ ⎠ cm

The fraction of molecules having sufficient energy levels to react are given by the Boltzmann factor f = e − E m /k

where E

m

B

T

is the difference in energy of an "excited" and "base" molecule.

The rate then becomes:

⎛ molecules⎞ − r? A ⎜ = πσ ⎝ cm 3 s ⎠

2 AB

8k B T

πµ

e − E m/ k B T N A N

B

(5)

AB

⎛ mol ⎞ ⎛ mol ⎞ We want to express the rate in ⎜ and the concentrations in ⎜ . Then: ⎝ lit s ⎠ ⎝ lit ⎠ Na C =N 10 3 A

A

;

Na C =N 10 3 B

B

⎛ moles ⎞ ⎛ molecules ⎞ Na −r A ⎜ = − r? A ⎜ x 3 ⎝ lit s ⎠ 10 ⎝ cm 3 s ⎠ Also k

B

Na= R ;

⎛ M + M B⎞ ⎟ µ AB N a = M AB = ⎜ A ⎝ M AMB ⎠

−1

where N a is the Avogardro's constant

molecules mole M A ,M B − molecular weights N a = 6.02217 x 10 23

The expression for the rate finally becomes ⎛ mole ⎞ 8π R T − E /RT N a 2 = σ AB e − r A⎜ CACB ⎝ lit s ⎠ 10 3 M AB 14444 4244444 3

(6)

k = k o T 1/2 e − E/ RT

(7a)

k

Thus

3


ko=σ

Updated 01/31/05

8π R M AB

2 AB

Na 10 3

(7b)

Collision theory predicts the dependence of the rate constant on temperature of the type T and allows actual prediction of the values for the frequency factor from tabulated data.

1/2

e

-E/RT

For reaction of type (1)

Z AA

1 = πσ 2

2 A

⎛ 8k B T ⎞ ⎜ ⎟ ⎝ π µ AA ⎠

1/ 2

N 2A

(8)

1 is there since all the molecules are the same and otherwise the collisions would have been 2 counted twice. This factor is offset by the 2 in the rate expression which simply indicates that in every collision two molecules of A react.

Factor

mA 2 1/ 2 ⎛ molecules⎞ ⎛ 2 ⎜ 16 π k B T ⎞ ⎜ ⎟ e − E m / k B T N 2A =σ A − r? A ⎝ cm 3 s ⎠ ⎝ ma ⎠ ⎛ moles⎞ π R N a 1/ 2 − E / R T 2 = 4σ 2A e CA − r A⎜ T M A 10 3 ⎝ lit s ⎠ 14442444 3 k o coll 14444 4244444 3

Now µ

AA

=

(9a)

(9b)

k

Some predictions of the collision theory will be compared later to transition state theory. Note: T 1/2 dependence is almost entirely masked by the much stronger e - E / R T dependence. Thus Arrhenius form is a good approximation. The relationship to Arrhenius parameters is: k

o Arr

=ko

coll

(e T )1/2

E Arr = E coll +

(10a)

1 RT 2

(10b)

Original comparison of the collision theory prediction with experimental values for the reaction 2 H I → H 2 + I 2 resulted at 556K in k predicted = 3.50 x 10-7 (L/mol s) as opposed to kexp = 3.52 x 10-7 (l/mol s). This proved to be an unfortunate coincidence as later it became evident that predictions based on collision theory can lead to gross discrepancies with data. For example, for more complex gas molecules predicted pre-exponential factors are often order of magnitude higher than experimental values and severe problems arise for reactions of ions or dipolar substances. The modifications to the collision theory recognize that the collision cross-sectional area is not a 4


Updated 01/31/05

constant, but a function of energy, introduce a steric factor and acknowledge that molecules do not travel at the same speed but with a Maxwell-Boltzmann distribution of velocity! If one considers just the translational energy, ε t , one can argue that the rate constant can be predicted from k (T ) = (

8

∞

3)

πµ (k T )

1/ 2

∫ε

t

σ (ε t ) e −

ε t / kT

d εt

(11)

o

Similar but more complex expressions can be derived accounting for rotational and vibrational energies. Modern analytical tools allow measurement of effective reaction cross-section when reactant molecules are in prescribed vibrational and rotational states. When such information is available collision theory in principle leads to the 'a priori' prediction of the rate constant. Example 1. Using collision theory estimate the specific rate constant for the decomposition of H I at cal 321oC, σ H I = 3.5 A, E = 44000 . Experimentally the rate constant is found to be mol ⎛ lit ⎞ . Collision theory prediction is: k = 2.0 x 10 −6 ⎜ ⎝ mol s ⎠ π RT N a − E / R T k = 4σ 2H I e M H I 10 3 k=4 x

(

3. 5 x 10

π 8. 314 x 10 7 x (273 + 321)

)

−8 2

σ H I = 3.51 x 10

−8

cm ;

−

e

44000 1.987 x 594

127. 9 R = 8.314 x 10+7 erg / mol K

x

6. 023 x 10 23 103

T = 273 + 321 = 594 K , M H I = 127.9 N a = 6.023 x 10 23 molecules / mol k = 6.63

x 10 −6

lit ← mol s

collision theory estimate

lit ← exp erimental value mol s Collision theory usually gives the upper bound of the rate constant. ________________________________________________________________________ k = 2.0

6.2

x 10 −6

Classical Transition State Theory (CTST)

This is the most promising of the rate theories and deals again with elementary reactions. However, even an elementary reaction is not viewed any more to occur exactly in one step (this shows how flexible the definition of elementary reactions has become since they are supposed to occur in one step). According to transition state theory every elementary reaction proceeds through an activated complex - a transition state.

(reac tants ) ⇔ (transition state) ⇔ (products ) A + B ⇔Z* ⇔ Q + P

5

(12)


Updated 01/31/05

The transition state is a little more than the fraction of excited (energized) reactant molecules as in Arrhenius or collision theory. Its structure is neither that of the reactants nor of products, it is someplace in between and its concentration is always orders of magnitude lower than that of reactants or products. The transition state can either be formed starting solely with reactant molecules from the left or with product molecules from the right. The energy picture is as shown in Figure 2. Energy Diagram (Simplified for the Classical Transition State Theory)

FIGURE 2:

* #

Z Z Energy Level Eof*

Ef Eob* Eoz −∆H

Eb

Eor

Eop

Base Level

Reaction Coordinate

Eor - energy level of reactants Eop - energy level of products Eoz - energy level of transition state Eof* - activation barrier for forward reaction Eob* - activation barrier for reverse reaction ∆ Ho - heat of reaction E *o f

=E

E *o b

= E b = E oz − E

∆H

=E

∆H =E

f

= E oz − E oR

op f

−E

oR

oP

(

) (

= E oz − E b − E

oz

−E

f

)

−Eb

When the overall system, i.e reactants and products, is in equilibrium, clearly the net rate of reaction is zero, i.e r = rf - rb = 0

However, even in overall equilibrium a certain number of reactant molecules gets transformed per unit 6


Updated 01/31/05

time into the transition state, and the same number gets transformed from transition state to reactants as required by the principle of microscopic balancing. At the same time a certain rate of exchange exists between transition state and products which is balanced by the exchange between products and transition state. That exchange rate at overall equilibrium is proportional to ν C z where ν is the frequency of ⎛1 occurrence of exchange ⎝ ⎞⎠ and C z the number concentration of the transition state. s The three fundamental assumptions of transition state theory are: 1.

The rate of reaction is given by r = νC z

(13)

even when the system is removed from equilibrium i.e when there are no products, or products are being removed, or they are present in amount less than required by equilibrium. This is equivalent to assuming that the equilibrium between the transition states and reactants is always established. 2.

The frequency ν involved is a universal frequency, it does not depend on the nature of the molecular system and is given by

ν=

k BT

(14)

hp

where k b = 1.38062 x 10-23 JK-1 Boltzmann's constant h p = 6.6262 x 10-34 Js Planck's constant T (K) - absolute temperature

3.

The reaction system is "symmetric" with respect to the transition state and the above is also true when starting from products to make reactants (from right to left). A more rigorous treatment is outlined in appropriate textbooks on kinetics (e.g. see Laidler, Chemical Kinetics). This is a very bold assumption asserting that the equilibrium rate of exchange is equal to the rate for the system out of equilibrium. If we refer to Figure 3, the above assumptions indicate that once reactant molecules have reached the col of the activated complex (transition state) from the left there is no going back and they get converted to products. Similarly, the product molecules that reach the col from the right cannot turn back and do get transformed to reactants.

7


Updated 01/31/05

FIGURE 3: Profile through the minimum-energy path, showing two-dividing surfaces at the col separated by a small distance δ . Parellel dividing surfaces

δ

[X ] m l

[X ] m r

Reactants Products

Distance along minimum-energy

Let us look at the forward reaction (from left to right) r

f

= νC z

Reactants and transition state are at equilibrium K*=

az = a AaB γ

γ zCz =Kγ Kc A γ BC A C B

C z = K * K −1 γ CACB r

f

=

k BT K * γ hpγ z

k BT hp

ν=

Aγ BCACB

=

(15)

k BT * K aAaB hpγ z

Now we can define a rate constant for the forward reaction that is a function of temperature only k T k fi (T ) = B K * . This yields hp rf=

k fi (T )

γ

aAaB

and

K =e *

−

∆G * RT

=e

−

∆H * RT

z

8

e

* ∆S R

(16)


Updated 01/31/05

Similarly for the reverse reaction r b = ν C z , products is always established. K≠ =

γ zCz γ Qγ PCQC P

Cz =

aQap

γ

K =e

K≠

z −

≠

and the equilibrium between transition states and

∆G ≠ RT

−

∆ H≠

= e

∆ S≠

RT

e

R

The reverse rate is now expressed by rb =

k b i (T )

γ

a Qa

P

where k

(T ) =

bi

z

k BT K hp

≠

(17)

In the above ∆ G * , ∆ H *, ∆ S * and ∆ G ≠, ∆ H ≠, ∆ S ≠ are the Gibbs free energy, heat of reaction, and change in entropy for formation of the transition state starting from reactants and products, respectively. The net rate of reaction then is r = rf − rb =

1

γz

[k

fi

aA aB − kb i aQa P

]

or H * AS* ∆H≠ ∆S ≠ ⎤ − − ≠ kB T * ko T ⎡ − ∆RT R RT R r= K aA aB − K aQ aP = e e a a − e e a a A B Q P⎥ hpγ z hpγ z ⎢⎣ ⎦

[

]

(18)

At equilibrium r = 0 so that the equilibrium constant K for the reaction is recovered: ≠

*

∆H −∆ H − aQ aP RT K= =e e a A aB

*

∆S − ∆S R

≠

=e

−

*

≠

∆G −∆ G RT

Therefore, the following equations hold:

∆ H * − ∆ H ≠ = ∆ H ° − heat of reaction ∆ S* − ∆ S ≠ = ∆ S°

− entropy due to reaction

∆G* −∆G ≠ = ∆G°

− Gibbs free energy change due to reaction

∆ H* = E f

∆ H≠ = Eb

Let us call k f (T ) = k r=

1

λz

[k

fi

f

i

k b (T ) = k

a A a B − k f bi aQ a P

bi

rate constants which are functions of T only.

]

(19)

9


Updated 01/31/05

We usually express the rate in the form: r = k f C AC B - k

b

(20)

CQC P

This gives the relationship: k

f

=k

γ f

Aγ B

γ

i

γ

kb = kbi

z

Qγ P

γ

(21)

z

This is a very important result of the transition state theory. It tells us that whenever we write and evaluate a rate expression with its driving force being expressed in molar concentrations the rate constants, k f , k b , etc. that appear in such expressions are not only functions of temperature but also could be functions of pressure and concentration. This is obvious from the above since k f i , k b i are functions of temperature only but γ A , γ B, γ Q, γ concentration.

P ,γ z

can also be functions of pressure or

The above relationship between rate constants k f , k b in nonideal systems for rates whose driving force is as usual expressed in concentrations, and the ideal rate constants k f i , k b i which would be observed in a system with all activity coefficients of unity, is frequently used to correct the constants for pressure effects in gas reactions or for concentration effects in ionic solutions. Relationship to Arrhenius parameters k

k

o Arr

⎛ k T⎞ ° = ⎜ e B ⎟ e ∆ S /R ⎝ γ zh p ⎠

; E

o Arr

Arr

e − E Arr / R T is:

=∆H*

+ RT

(22)

Note: Sometimes transition state theory is interpreted by asserting that the rate is proportional to the product of the universal frequency and the activity of the transition state (rather than to its concentration). This would lead essentially to the same expressions as shown above except that γ z would not appear, i.e wherever it appears it would be replaced by unity. ________________________________________________________________________ Example.2 Estimate the rate constant for decomposition of methyl aride CH3N3 by transition state theory at 500K, given ∆ H * = 42 ,500 cal / mol , ∆ S * = 8.2 cal / mol K. k T k= B e hp

−

∆H *

∆ S*

RT

e

R

1.38062 x 10 − 23 x 500 − 1 .987 x 500 k= e e 6.6262 x 10 − 34 42500

8 .2 1.987

k = 1.705 x 10 −4 s −1

10


Updated 01/31/05

When we use transition state theory to calculate the rate constant for reaction of the type A = B there is no ambiguity as to the units of the rate constant. However, when we deal with bimolecular reactions of the type A + B ⇔ P + Q and }ν k BT * kf = Kd hp

}ν k T ; kh = B K hp

≠ d

⎛1 the units of the rate constants are not governed by the kinetic factor ν ⎝ ⎞⎠ but rather by the s

thermodynamic factor K *d and K ≠d , i.e by the choice of standard states for calculation of ∆ G * and ∆ G ≠ upon which the calculation of these parameters K *d ,K ≠d are based. If concentration is chosen for the driving force in the rate expression then d = c and Kd* = Kc* = C*z / CACB , if pressure is chosen then d = p and K *p = Pz* / pA pB . Similarly,

K ≠c = C *z / C p C q , and K ≠p = Pz* / Pp Pq . It is useful therefore to review the use of standard states and relationships between thermodynamic quantities at this point. For a general reaction s

∑

ν jA j =0

j=1

The thermodynamic quantities K, ∆ H °, ∆ G ° , ∆ S °,C °p are generally tabulated for standard states of 1 atm and ideal gas. Frequently we want to transform these to standard states expressed in concentration mol units, generally 1 (1 M ) . lit ________________________________________________________________________ Example. 3 S

When using transition state theory for a bimolar reaction for which

∑υ

j

= −1, since two moles of

j=1

reactant give one mole of transition complex, and when the quantities K *, ∆ H *, ∆ S * are based on standard states at 1 atm of ideal gases , then from transition theory

∆ H *p = ∆ Hc* − RT . k

p

k T = B e hp

∆S R

* p

−

e

∆H RT

* p

(atm

, s −1 )

−1

Compared with Arrhenius equation

11


k

p f Arr

=k

Updated 01/31/05

' po

e

(atm −1,s −1)

− E 'p /R T

yields the frequency factor k 'p o = e

k BT e hp

∆ S *p RT

(atm −1,s −1 )

and activation energy E 'p = ∆ H * + R T = ∆ U * Compared with Arrhenius equation in concentration units k c =k

⎛ lit e − E c /R T ⎝ , s −1 ⎞⎠ mol

co

it gives the frequency factor:

k

co

= k 'p o (R' T) e = e 2 k T = e2 B e hp

since

k BT R' T e hp

∆ S *p + Rln R'T

= e2

R

∆ S *p

k

R

BT

hp

e

∆ S *c / R

∑ ν j = − 1 going from two reactant molecules to one molecule of the transition state

The activation energy is E c = E 'p + RT = ∆ H *p + 2 R T = ∆ U * + R T ________________________________________________________________________

Example. 4. The homogeneous dimerization of butadiene (CH2=CH-CH=CH2) has been studied extensively. An experimental rate constant based on disappearance of butadiene was found as: ⎛ cm 3 ⎞ 9 − 23960/ RT ⎜ ⎟ k = 9.2 x 10 e 1424 3 ⎝ mol s ⎠ k o

a)

Use collision theory to predict a value of k o at 600 K and compare with experimental value of 9.2 x 109. Assume effective collision diameter 5 x 10-8 cm. ⎛ cm 3 ⎞ ⎜ ⎟ = 4σ ko ⎝ mol s⎠

(

π RT

2 B

k o = 4 x 5 x 10 −8

M

)

2

Na e

B

π x 8.314 x 10 7 x 600 54

12

x 6.023 x 10 23 e


Updated 01/31/05

⎛ cm 3 ⎞ ⎟ ← collision theory prediction k o = 8.81 x 10 14 ⎜ ⎝ gmol s ⎠ ⎛ cm ⎞ k o = 9.2 x 10 9 ⎜ ← exp erimental value ⎝ gmol s ⎠ b) Use transition-state theory to predict k o at 600 K and compare to experimental value. Assume the following form of the transition state (a hint from a friendly chemist!) ' ' CH2 -CH=CH-CH2-CH2-CH-CH=CH2 * ∆ S *p k T k T k o = e 2 B R' T e = e 2 B e ∆ Sc /R hp hp R

∆ S *c = ∆ S *p + R ln R' T We have to calculate the change of entropy for the formation of transition state from reactants. To do this the group contribution method and appropriate tables are used. [D.A. Hougen and K.M. Watson "Chemical Process Principles" Vol. II: pp 759-764 Wiley, NY, 1947; or S.W. Benson "Thermochemical KInetics, 2nd ed. - Methods for the Estimation of Thermochemical Data and Rate Parameters", Wiley, NY 1976]. Using Benson's Tables (attached) we get for CH2=CH-CH=CH2 cal mol K 500 26.64

S ° = 2 x 27.61 + 2 x 7.97 = 71.16 T 300 Cp 18.52

400 22.78

600 30.00

For CH2-CH=CH-CH2-CH2-CH-CH=CH2 using the contributions of CH2 to be between those for -CH3 and -CH2 radical we get So = 110.995 cal/mol K. T 300 Cp 40.0

400 49.42

500 57.82

600 65.02

Then

∆Cp

T

∆

S *T

=

∆ S *o

+

∫ T

∆ S *T = ∆ S *o +

T

dT

o

⎡ 1 ∆ C p (T o ) ∆ C p (T 1 ) ∆ C p (T 2 ) 1 ∆ C p (T 3 )⎤ + + + ∆T ⎢2 ⎥ 2 To T1 T2 T3 ⎣ ⎦

13


Updated 01/31/05

⎡ 1 40.00 − 2 x 18.52 + 49.42 − 2 x 22.78 + ⎤ 2 300 400 ⎥ x 100 ∆ S *T = (110.995− 2 x 71.16) + ⎢ ⎢ 57.82 − 2 x 26.64 1 65.02 − 2 x 30.00 ⎥ + ⎣ ⎦ 600 500 2 cal ∆ S *T = − 31.325 + 2.785 = − 28.54 mol K 28. 54 + 1.987 l n (0 .0821 x 600 )

k

co

k

co

k

1.38062 x 10 23 x 600 − 1.987 =e e x 10 3 6.6262 x 10 34 ⎛ cm 3 ⎞ ⎟ ← transition state theory prediction = 2.63 x 10 12 ⎜ ⎝ gmol s ⎠ 2

co

⎛ cm 3 ⎞ ⎟ ← exp erimental value = 9.2 x 10 9 ⎜ ⎝ gmol s ⎠

A closer prediction but still an upper bound. Remember the structure of the transition state was only hypothesized and all thermodynamic quantities only estimated. Table 1: Some useful orders of magnitude Quantity mean molecular velocity universal frequency

Expression

⎛ 8k T ⎞ B ⎟ ⎜ ⎝ π µ A B⎠ k BT hp

Order of Magnitude

Units

5 x 104

(cm/s)

1013

(1/s)

1/2

collision frequency gas-gas

π σ 2AB u A B

10-10

(cm3/s - molecule)

collision frequency gas-solid

u 4

104

(cm/s)

It is instructive to gain an insight into the order of magnitude of some important quantities by examining Table 1 above, and to learn a little more about the comparison in prediction of the TST and collision theory from Table 2. ___________________________________________________________________________________

Table 2: Comparison of Theories Reaction

Logarithm of kco Observed

NO + O 3 → NO 2 + O 2 NO 2 + O 3 → NO 3 + O 2 NO 2 + F 2 → NO 2 F + F NO 2 + CO → NO + CO 2 F 2 + C l O 2 → FC l O 2 + F 2C lO → C l 2 + O 2

11.9 12.8 12.2 13.1 10.5 10.8

Transition State Collision Theory Theory 11.6 13.7 11.1 13.8 11.1 13.8 12.8 13.6 10.9 13.7 10.0 13.4

Transition theory seems to give consistently better predictions. 14


Updated 01/31/05

6.3. Transition State Theory explained by Statistical Mechanics

In this section, we will investigate the relationship between molecular level concepts and thermodynamic properties. In order to be able to start looking at molecular level, we will introduce partition functions that describe a specific system using statistical mechanics. Let’s start with a few definitions. An ensemble is a collection of subsystems that make up the thermodynamic state. Different ensembles are obtained depending on the intensive (e.g. temperature (T)) or extensive (e.g. volume (V)) variables that define the thermodynamic state. Figure 4 shows an example of an ensemble that has 16 states. If there is no material exchange between the states, composition (N) is constant for each state. In a microcanonical ensemble, V, energy (E) and N are kept constant. 1 NVE 5 NVE 9 NVE 13 NVE

2 NVE 6 NVE 10 NVE 14 NVE



FIGURE 4. An example of a microcanonical ensemble with 16 states each at constant N, V and E

Microcanonical ensemble is therefore isolated since each state is at the same energy and no energy transfer occurs between the states. In a canonical ensemble, the states are not isolated and T is fixed instead of E. By this way, E may be transferred as heat. If the states are not closed but open to material exchange and chemical potential (µ) , V and T are fixed, the ensemble is defined as grand canonical ensemble. Finally, the ensemble is called isothermal-isobaric for states with constant N, T, and pressure (P). In statistical mechanics, each ensemble can be connected to the classical thermodynamical information using partition functions. The derivation for the connectivity equations are explained in detail in statistical mechanics books (e.g. Donald A. McQuarrie, “Statistical Mechanics”, Harper Collins Publishers, NY, 1976 and M.P. Allen, D.J. Tildesley, “Computer Simulation of Liquids”, Clarendon Press, Oxford, 1997) Table 3 summarizes the connectivity equations between the micro-level partition functions and the macro-level classical thermodynamic equations. Table 3. Connectivity equations Ensemble microcanonical

Thermodynamic Quantity S

Equation

canonical

A

grand canonical

pV

S = k B Q NVE A = − ln Q NVT k BT pV = k B T ln Q µVT

isothermal-isobaric

G

G = − k B T ln Q NPT

In the table Q represents the partition function for the ensemble. (e.g. QNVE represents partition function for the microcanonical ensemble where N,V, and E are constant.) If the molecules that make up the system are independent of each other as in an ideal gas at the canonical ensemble, the partition function, Q can further be divided into individual molecular partition functions as: 15


Q=

Updated 01/31/05

qN N!

(23)

The molecular partition function itself consists of different contributing partition functions of different contributing partition functions arising from different modes of motion, mainly translational, rotational, vibrational, electronical and nuclear. In other words, q = qtranslational q rotational qvibrational q nuclear q electronic

(24)

Let’s focus on the individual contributions: Translational contribution:

⎛ 2πmk T ⎞ B ⎟ =V⎜ ⎜ h 2 ⎟ p ⎝ ⎠

qtranslational

3/ 2

(25)

with m being the molecular mass and h p being the Plank constant = 6.62608 x 10-34 Js Rotational contribution:

q rotational =

1

σ

∑J (2 J + 1)e

−

J ( J +1) hcB k BT

(26)

with J = 0,1,2,….,

σ = symmetry number, 1 for heteroatomic molecules and 2 for homoatomic molecules c = speed of light B = rotational constant for each molecule (cm-1) Rotational contribution is usually approximated by Eq. 27. q rotational ≈

k BT σhcB

(27)

Vibrational contribution:

q vibrational =

e − hν / 2 k B T 1 − e − hν k BT

(28)

1/ 2

⎛k⎞ 1 where ν = ⎜⎜ ⎟⎟ ⎝ µ ⎠ 2π with k being the force constant of the molecule and µ being the reduced mass.

16

(29)


Updated 01/31/05

Nuclear contribution:

q nuclear is taken as the degeneracy of the ground nuclear state. It is usually omitted in the calculations because for most cases, nuclear state is not altered for states of interest. Electronic contribution:

q electronic = f (e)

(30)

The electronic partition function is basically given as a function of degeneracy of the electronic ground state and is equal to 1 for most cases. Rate Constant Related to Molecular Partition Functions

If we look at our reaction Eq (12) and express the equilibrium constant, K using partial pressures, we can see how molecular partition functions correlate with transition state theory.

(reac tants ) ⇔ (transition state) ⇔ (products ) A + B ⇔Z* ⇔ Q + P

(12)

From classical thermodynamics: G = A + pV

(31)

Combining and keeping in mind that nRT = k B NT qN + nRT = −k B T ( N ln q − ln N !) + nRT = −k B TN ln q + k B T ( N ln N − N ) + nRT = N! ⎛q⎞ ⎛q⎞ − k B TN ln⎜ ⎟ = −nRT ln⎜ ⎟ N ⎝ ⎠ ⎝N⎠

G = −k B T ln

(32) (33)

We also know that ∆Gr = ∑ j =1ν j ∆G fi s

(34)

Then combining with Eq. 33 ⎛q ⎞ ∆G fi = Gi + GT0 = − RT ln⎜ i ⎟ + U To ⎝N⎠

(35)

Note that we lost the n factor of Eqn (35) since G is in terms of molar quantity. If we select our reference system as T = 0, then G = U. Combining Eqs (34) and (35) gives

17


Updated 01/31/05

⎛ qj ∆Gr = − RT ∑ j =1 ln⎜⎜ ⎝N We also know that s

νj

⎞ ⎟⎟ +∆E 0 ⎠

(36)

∆Gr = − RT ln K

(37)

Finally we have K in terms of q by ⎛ qj K = ∏ ⎜⎜ j ⎝ N

νj

⎞ − ∆E0 / RT ⎟⎟ e ⎠

(38)

6.4 Some Consequences of TST 6.4.1 Rate Constants in Dilute Strong Electrolytes

Debye-Hüchel theory relates the activity coefficient of dilute strong electrolytes with the ionic strength I of solution and with the charge of the ion in question: ln γ j = − AZ 2j

I

(39)

Where Z j − charge of j ion

γ j - activity coefficient of ion I=

1 2

s

∑C j =1

Z 2j - ionic strength

j

(40)

C j - molar concentration of j ion A - constant (A ≈ 0.51 for water at 25ºC) For reaction of A

ZA

+B

Z B

→Z*

(Z A +Z B)

(41)

the transition state theory predicts: k 1c = k 1

γ Aγ γ *Z

B

(42)

Clearly the transition state must have a charge of Z A + Z B in order to satisfy the law of conservation of charge. Taking logarithms

18


Updated 01/31/05

l nk 1c = l n k 1 + l n γ

− ln γ

* Z

= l nk 1 − A I Z 2A + Z 2B − (Z

A

A

+ ln γ

[

B

+Z

l nk 1c = l n k 1 + 2 A Z A Z B I

B

)2 ]

(43)

This equation gives excellent agreement with experimental data and is very useful for correlating liquid phase reaction data. 3.2.2 Pressure Effects in Gas Phase Reactions

For gases aj= f

j

=φ j P j = γ jC j =

γ jPj Z j RT

Z j − compressibility factor of j

γ

j

− activity coefficient of j

P j − partial pressure of j

(44)

C j − molar concentration of j

φ j − fugacity coefficient of j f

j

− fugacity of j

From the above

γ

j

=φjZ

j

RT

(45)

For a reaction A + B → Z * k c =k

γ

A

γ

γ * Z

B

=k

φ A Z A (RT )φ B Z φZZZ

B

(46)

φ φ Z Z k c = k A B A B RT φZ ZZ φ j → 1, Z j →1 At sufficiently low pressures ⎛ k c ⎞ ⎝ k⎠

(47)

= RT low pressure

At high pressure ⎛k c⎞ ⎝ k⎠

= RT high P

φ Aφ B Z A Z B φZ ZZ

(48)

19


Updated 01/31/05

Thus the ratio ⎛ ⎜ k c, high P ⎞⎟ = φ A φ φZ ⎝ k c,low P ⎠

B

Z

AZ B

(49)

ZZ

In the case of a reaction 2 A → ⎛ ⎜ k c high P ⎞⎟ = φ ⎝ k c low P ⎠ φ

2 A

ZZ

assu min g Z

A

≈ZZ

(50)

Z

The variation of the thermodynamic properties with pressure was calculated for H I decomposition (2 H I → I c + H c ). The above equation agreed excellently with all experimental data up to 250 atm which led to density variations of 300. 6.4.3 Dependence of Rate Constants on Temperature and Pressure In Chain Reactions

So far we have treated only simple reactions (often elementary ones) and their rate constants. Let us take a look at more complex overall reactions - say in free radical polymerization: r

pol

kd f =kp CM C I kt 14243

(51)

k

using Arrhenius form for k and each other constant k p ,k d ,k t we get

⎡ ⎛kd lnk =ln ⎢k p o ⎜ o ⎜ kt ⎢ ⎝ o ⎣

⎛k d ⎞ o k o = k p o⎜ ⎟ k ⎝ to ⎠

Range of values

⎞ ⎟ ⎟ ⎠

1/ 2

1 1 E p+ Ed − Et ⎤ 1 2 2 ⎥ + ln f − RT ⎥ 2 ⎦

1/ 2

E=Ep+

1 (E d − E t ) 2

(52)

(53)

30 < E d < 35 kcal/mol 5 < E p < 10 kcal/mol 2 < E t < 5 kcal/mol 17 < E < 26 kcal E ≈ 20 mol

Polymerization rate increases with temperature. However the degree of polymerization X proportional to the following group that depends on temperature:

20

n

is


Xn

∝

Updated 01/31/05

k

p

(54)

(k d k t )1/ 2

The activation energy for the degree of polymerization is then given by Xn = Xno e

−

E

(55)

xn

RT

1 (E + E t ) 2 d − 15 < E x n < − 6 kcal / mol

Exn = Ep −

(56)

Thus, degree of polymerization decreases rapidly with increasing temperature. The dependence on pressure is

d l nk − ∆ V * = dP RT

(57)

∆ V * - volume of activation, i.e change in volume in going from reactants to transition state. ∆ V *R = − 20
> 20,000 cal) and moderate temperatures (T < 600 K) the value of E is not affected by the choice of units for the driving force i.e C A (mol/lit) or P A (atm), etc.

21


Updated 01/31/05

4.

In case of low E ( E < 10,000 cal) and high and moderate temperatures the value of E is greatly affected by the choice of variables for the driving force i.e C A (mol/lit) or P A (atm) etc.

5.

The reaction rate can increase dramatically with temperature. The larger the E the more rapidly the rate increases with T.

FIGURE 5: Effect of Temperature on the Rate Constant E = 80,000

E = 20,000 cal/gmol

E = 60,000

E = 40,000

10,000

1,000

Say T o= 300 K E = 10,000 cal/gmol kT 100 kT o

50 10 5 1 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

←T

0.8

0.9

⎛ 1 1⎞ 3 1.0 ⎜ − ⎟ × 10 ⎝ To T ⎠

kT = rate constant at temperature T o kTo = rate constant at T = To = 300K = 27 C

The above figure (Figure 5) demonstrates the possible rapid rise in reaction constants and rates if activation energy is sufficiently high. The rate increases 10 times for E = 10,000 cal for a ∆ T of 50K. For the same rise in temperature the rate with E = 20,000 cal will increase 100 times. It takes only 35oK to raise a rate constant 1000 times for E = 40,000. 25oK temperature increase gives a 1000 times larger rate for E = 60,000 cal and a staggering 10,000 times larger rate for E = 80,000 cal! 6.

E f − E b = ∆H o reaction for elementary reactors. For exothermic reactions (∆H ° < 0)

E

< E

and a rise in temperature will promote more the reverse reaction and thus reduce dl n K ∆H = equilibrium conversion. This is obvious also from 2 < 0. dT RT f

b

22


Updated 01/31/05

dlnK < 0 as T ↑ K↓. However, forward rate rises also with T to some extent. dT There always is a T optimal which balances the increased rate with more unfavorable equilibrium. ∆H < O

7.

For endothermic reactions E f > E b , and for irreversible reactions, the higher the temperature the higher the rate and the higher conversion is obtainable.

8.

A temperature excursion of 10-20˚C can cause dramatic increases in the rate and lead to runaways and explosion.

9.

The reaction rate is so much more sensitive to temperature than to concentration. For a first order reaction with E = 20,000 cal a rise of 50˚C in T leads to an increase of 100 times in the rate. To accomplish the same augmentation of the rate by changing concentrations we would have to increase concentration 100 times. For a 2nd order reaction with the same activation energy we would have to increase concentration 10 times. For higher activation energies the difference between temperature and concentration sensitivity of the rate is even more pronounced.

23