Appendix

Chapter 6

Appendix

6.1 REACTION SCHEME Example 23: Complex reaction scheme A further example in Section 2.1.1.2 with three steps of reaction jc,:

2A-^B

x^\

B + C-^D + E

^3:

D + A->F + G

results in the reaction scheme

x^ Xl 3

A -2

B 1

0 -1

-1 0

C 0 -1

D 0 I

0

-1

E F G 0 0 0 1 0 0 0 1 1

x^ X2 X"!

The transposed matrices of the stoichiometric coefficients v°' is derived from this reaction scheme as f-2 v°' =

0 \^-i

1 0 -1-1 0 0

0 0 0 OA 1 1 0 0 -1 0 1 l y

Using eq. (2.4) one finds

Appendix

474

/ -2

0

- n/^v

1 Ac 0 = lAd = 0 Ae 0 0 A/ ^A^> V 0

-1 -1 1 1 0

0 0 \^3j -1 0 1

(Aa^

lAb

Ar=v'*r

Ch.6

0

^

1

According to the advice given in Section 2.1.1.1 one can easily find the changes in concentrations like Aa = -2A:, -x^, Ad=^X2"'X^,

Afc = jc, - jCj, A^ = ;c2,

Ac = —jCj,

Af^Ag^x^^

Just three concentrations are necessary to observe the progress of the reactions (Aa, Ab, Ad), By insertion, the three degrees of advancement can be calculated as Aa = -2JCI - JC3

jc, =l/3(-Aa + Aft + Arf)

Aft = X, - X2

X2=l/3(-Aa~2Afc + Arf)

Arf = JC2 - JC3

jC3 = l/3(-Aa-2Afe-2Arf)

which results in matrix notation /^v ^ Ai yX^j

The changes in concentration amount to Ac = -^2 = Ae=

1/3 Aa + 2l3Ab-1/3 Ad

;c2 = -1/3 Aa - 2/3 Afc +1/3Arf = -Ac

A/ = Ag

=-y3Aa-2f3Ab-2/3Ad

Ch. 6

Reaction Scheme

475

The more formal approach is given using eqs. (2.4) to (2.8). In principle the matrix (-2 1 V

=

0 -1 -1 1 1 0 0

0 0 0 0 lO

-n 0 0 -1 0 1

1J

can be reduced, since only the reactants A, B, and D have to be monitored:

V

=

f-2

0

-0

1 0

-1 1

0 -1

Thus one finds according to eq. (2.8) with the inverted matrix, which can be easily done using a commercial mathematical tool program package like MAPLE, using the following program lines: > with(linalg): > A:= arrayC [[-2,0,-l],[l,-l,0],[0,l,-l]]); -2 A:= I 0

0 -1 1

> inverse(A);

- ^ Vi

-1 0 -1

Appendix

476

Ch.6

Since no linear dependencies between the steps of the reaction exist, the symbol (*) is unnecessary. The new matrix p can be calculated according to the definition given in Section 2.1.1.

f-K K K ( ^ ° 0 ^ \-'A -% % 0 1 0 % -'A [-% -% %] 'A 0 0 1 -% -% K -y, -% -%\ ij [-'A -% -%\

(-2 0 -1> 1 -1 0 0 - 1 0 1-1 v'v-'=p = 0 0 1 0j 0 0 1

^0 0

This matrix can be inserted into eq. (2.8) and one obtains ( 1 ^b Ac Ad Ae A/

0

0 1

0 ^(Aa^ 0 Ab

A A -A 0

0

yAdj

1

-A -A A -A -A -A V-A -A -A)

finally the same relationships as given above for the different changes in concentration: Ac = -X2 = Ae =

l/3Aa + 2/3 A& - 1/3A^

^2 = -1/3 Aa - 2/3 AZ> + l/3Af/ = -Ac

A/ = Ag = -1/3 Aa - 2/3 A6 - 2/3Arf

Ch. 6

Simple Photochemical Reactions

477

6.2 SIMPLE PHOTOCHEMICAL REACTIONS 6.2.7 Photoreaction via the triplet state Example 2,6: Simple photoisomerisation In Section 2.1.3.3 the photoreaction of an isomerisation from the singlet state for A—!2^B was demonstrated. Here a pathway via the triplet state is considered. It is given by the following scheme A + hv-

xs

A -1 1 1 0 1

<

0

-» A'

A'

- ^ A + Av

A'

2t-^A

A'

*i_ •^A"

x; Xj

A"

- ^ A

A"

•^3

<

A' 1 -1 -1 -1 0 0

A" 0 0 0 1 -1 -1

B 1 xi 0 JA 0 ^JQ' 0 k^d 0 k^d 1 k^a" 1 k^a"

According to the procedure explained in Section 2.1.1.1 rows and columns are filled. All the excited molecules are supposed to form A' first. Therefore both steps A + hv

>A* and

A*

>A'

can be combined to give the single step A + /iv-

^A',

Therefore as in the case of the singlet pathway, one partial reaction can be omitted in the scheme. Application of the Bodenstein hypothesis to the intermediates A' and A" gives

Appendix

478

Ch.6

a = -/A+(/:2+*3+*5K'» b = k^a'\ Using

the following relationships are obtained: k^a'

a =-

*4/A

a =

Thus by insertion fe =

^4^6^A

(/:2+/:3+^4)(/:5+A:5)'

and the quantum yield depends on other photophysical processes than in the case of a pathway via the singlet state: ^4^6

^ =

(/:2+/^3+*4)(*5+*6)

6.2,2 Complex uniform photoreactions Example 2.10 A photochemical with a superimposed backward thermal reaction is the most simple type possible. The following reaction scheme represents the mechanism (see Section 2.1.4.3):

A + Zzv->A

->B

A B * -1 1 ^1 1 -1

•*
Kb

Simple Photochemical Reactions

Ch. 6

479

A forms by absorption of light reactant B, which reacts backward to A in a slow dark reaction. The next scheme gives the elementary reactions A >A'

A + hv

JC,

A'

>A

X2

A'

>K'

^3

A"

>A

^4

A"

>B

B

Xs ^6

>A

-1 1

A' 1 -1

0 1 0 1

-1 0 0 0

A" B 0 0 0 0 1 0 -1 0 1 -I 0 -1

-^* /A Atjfl'

k^a' k,a" k,a"

Kb

As in Section 2.1.3.3 explained by use of the Bodenstein hypothesis a* = a' = 0 can be assumed, therefore one finds /.

a =• k^ + ^5

These equations can be used to calculate a and b as explained in Example 2.6 and introduced in Example 2.4 a = ~/A + k2a' ^-k^d '-\-k^b b--k^d'-k^b and ~a =fe=

^3^5 ^A

-hb^^xlK-Kb

This reaction turns out to be uniform since during the reaction

480

Appendix

Ch.6

a+b-ttn is permanently valid. Example 2.11 In case of a reversible photoisomerisation without a thermal backreaction one can define A to be the trans and B the cis form A Xx - 1 1 Xl

A—!21-^B B—!^A

B 1 -1

Xk

^f/A

^\h

Because of the photochemical back-reaction, a photostationary state is reached. If A " = B" = T is the common intermediate the scheme will be

A + Ziv

>A'

A'

^A

A'

>T

B + hv

>B'

B'

>B

B'

>T

T

>A

T

>B

A x{ - 1 1 x\ 0 X', A 0 A 0 A 0 x'n 1 A 0

A' 1 -1 -1 0 0 0 0 0

T 0 0 1 0 0 1 -1 -1

B

B'

0 0 0 -1 1 0 0 1

0 0 0 1 -1 -1 0 0

A /A

k^a' k^a'

h k^V

Kb' k-jt

kgt

Therefore with the usual assumptions, the concentrations of the intermediates can be caculated

a'=-A. k2 +K3

b'=-

/ p

k,a' + kj}' .^'^y

ks+K

and the change in the concentration of A and B is given by

Simple Photochemical Reactions

Ch. 6

d = ~ / ^ 4- k2a' + A:^^ = (*2+*3)(*7+^8) —

481

(*5+*6)(*7+*8)

s,^^

= -K/A+^2/B

and

In consequence the partial photochemical quantum yields are represented by

^r=-(A:2+*3)(*7+*8)' 3 8

^6^7

^ B = .

"'

(*5+*6)(*7+*8)

because of the law of conservation of mass fl + Z? = AQ .

During the total reaction this photoreaction will be uniform. 623 Complex non-uniform photoreactions A typical example is a photoreaction with a superimposed thermal reaction (see Section 2.1.4.4), in which the reaction scheme is given as A—^B

A -1 * 1 ^2 * 0 •

B—5^A B—^C

B 1 -1 -1

C 0
It is assumed that intermediates have very short lifetimes and the Bodenstein hypothesis is valid. The scheme reduces because of the linear dependency of the first two steps according to Appendix 6.3.1 to A x^ - 1

B 1

0

-1

2

C Xk 0
* Xt ^^^ .^1 ^2 =

X

Appendix

482

Ch.6

Stoichiometric summation of the degree of advancement over the column of the reactant gives

Zi = i , - ^ 2 =•*!* "^2

--^a =^l^^A - ^ 2 ^ B - " * 3 ^

C -- Xy "~ .^3 ^ K-yU»

6.3 SIMPLE THERMAL REACTIONS 6.5.7 Reactions including steps of back-reactions Example 2.7: Consecutive reaction: linear dependency For the consecutive reaction with superimposed back-reaction A;f*B->C the following reaction scheme is valid:

A->B B->A B-^C

• x^* ^3

A

B

-I 1 0

1 -1 -1

C 0 kiO 0 kjb 1 k,b

This scheme represents the three partial reaction steps. However, the first two rows are proportional to each other (stoichiometric coefficients can be transferred into each other by multiplication by a factor (-1)). For this reason JC,* and x^ are linearly dependent. Therefore the rank of the original matrix v* of the stoichiometric coefficients reduces to two columns. This is valid for all the reactants A, in the first two linear dependent steps 1 and 2: V,. = —v^-

For this reason the relationship between concentrations and the degrees

Simple Thermal Reactions

Ch.6

483

of advancement is reduced from (stoichiometric summation within one column)

to

Only the difference, taken from the degrees of advancement jCj* and x^, arises in all the equations. The x^ itself neither can be measured nor their dependencies on time be calculated independently of each other. For this reason it makes sense to define the following new linear independent degrees of advancement •

•

Xt ^^ X% ~~ Xy Xy

• 5^ X'l

In consequence the reaction scheme reduces to (the new element in the last column is obtained using both the equations above)

^1 2

A -1 0

B 1 -1

C Xk 0 kfO - k^b 1 kjb

As a result the new time dependence of the degree of advancement jc, becomes

according to definition. Example 2.8 A rather complex reaction scheme is chosen as the next example, which exhibits a cycle of reversible reactions.

Appendix

484

B->A B-^C C-^B C-»A A-»C

A B * -1 1 1 -1 Xl * 0 -1 X3 * 0 1

X4• X,

* Xe

1 -1

0 0

C 0 0 1 -1 -1 1

Ch.6

. •

Xk

k^a k^b kjb

k^c k^c k^a

Considering the linear dependencies, the reduced matrix representation is given by

A«=>B BoC C<=>A

9

xr9 X2 0

xf

A -1 0 1

•9

B 1 -I

C 0 1

k^a — kjb

0

-1

k^c - k^a

k^b - k^c

To demonstrate the reduction, special symbols are used in the following. According to the explanation given above, the six linear dependent steps are first reduced to three steps: Xy

SX,

• -^2'

X'y

— X"!

' •^4»

X-x — Xs — X^,,

But this scheme can be reduced further, since there is a linear dependence according to

by comparing all three columns. For this reason the mechanism contains just two linear independent partial steps of the reaction. The rank s of the stoichiometric matrix v becomes 2. This reduction in the scheme can be done in this example by cancellation of the third row. Its elements have to be subtracted from the first two elements of the last column:

Simple Thermal Reactions

Ch. 6 X-i

— fCeC

K>f.G> •

The two linear independent degrees of advancement are given by Xt = Xt ""• x-y anQ Xy = Xy ""• jc-j and the following representation

X, Jfj

A B C Xk -1 1 0 k^a — ^2^ ~ ^5^ + k^a 0 -1 1 kjb — k^c - k^c + k^a

is found. The reduced stoichiometric matrix is symbolised by v. In consequence the matrix P can be substituted by P = v°-v-'.

Accordingly the reduced degree of advancement x has to be taken. One finds according to eqs. (2.4) and (2.9) Aa°=v°'x*,

Aa°=v°x

and

Aa° = p A a .

Using the equations and the procedure demonstrated in Example 2.2 the following steps of calculus have to be done: ^Aa^ r-1

Ab yAcj

1

,0 1

Aa = -;c,,

fAa\ Ab = v v vAcy

0 r.. \ -1 \^2J

Ab = Xf-X2,

Aa^ Ab,

Ac = X2

-1 0^ -1 0 VAa 1 -1 -1 -l)[Ab 0 1

485

Appendix

486

(^a\ (I AZ? = 0 {^c) [-1

Ch.6

0^ Aa 1 Ab -1

Ac = -Aa-Ab. This final equation correlates with the three equations above. 6.3,2 Parallel reactions Example 2.9 The rank of the matrix v is equal to the number of the degrees of advancement in the following mechanisms. Therefore one would expect that the conditions of equal number of r and rank of the matrix stated at the beginning of Section 2.1.4 is sufficient for the optimal reduced matrix. .* A B C D * -1 -1 1 0 k^ab * -1 -1 0 1 kjab X2

AH-B->C

A + B->D

One realises that in both the reactions the same dependence on the concentrations exists. If the concentrations a and b are chosen to be independent, eq. (2.6) will become explicitly either

-r\(-^ - XVA-]

rAa\ (-\ Ab Ac

-1

-1

1

0

yAd)

[o

vo 11 XAd)

ij

or

fAa\ (\ Ab Ac

0^i^

1 0 \Ad -1 -1

[Ad] [0

ij

By this means one obtains the non-trivial balance-equations: Ab = Aa

and Ac = -Aa - Ad.

Simple Thermal Reactions

Ch. 6

487

Both the degrees of advancement show a linear dependence for kinetical reasons. A comparison of the rates in the last column of the scheme makes it evident that CUi'y

fC'j

dx\

k^

Taking into account that the degrees of advancement are zero at r = 0 (a general rule in kinetics), integration of the above equation gives *

*

where % is defined as

The combination of these equations with eq. (2.4) results in Aa,. = v„.x; + v^,xl = (v^,. 4- xv2i)x, It turns out that this reaction system can be described by a single degree of advancement and the reaction scheme finds the following form:

A Jt,

B

c

D

Xk

-O + X) - 0 + X) 1 X kyob

The reduced scheme contains the same information as any extended one. Considering a as the independent concentration, the changes in the different concentrations are determined by

(^"] (-(i+x)] .Ab -a+x) 1

Ac

[Ad)

K

X

J

+ X))(Aa)

488

Appendix

Ch. 6

and the relationships between the concentrations can be calculated as A6 = Aa,

Ac =

^—Aa,

Arf =

—Aa.

A and B have the same stoichiometric coefficients in the reduced scheme. For C and D the equation above can be used to form the ratios: kx

I "^1 +"^2 1

Aa

*. J

/

Ad _k2 Aa kf

I Kt T Ky

k,

{k,+k,) I

K'2

)

{k^-hk^)

Even though some arbitrariness exists in the defmition of linear independent degrees of advancement, the measured value is not influenced at all. For example, the above given degree of advancement x can be substituted by v = ^(i + ;f), if all other stoichiometric coefficients are divided by (1 +;f) in consequence. The first definition x correlates to the overall equation ^ i ^ ^ ( A + B)-^C + ^ D , ^1

k^

the latter (jc*') to A H - B - > — L c + -^2_j) K^ ~r K2

Kt "1 Ky

6.4 EXAMPLES FOR THE DETERMINATION OF CONCENTRATIONS 6.4.1 Second order reactions Two possible cases have to be distinguished for second order reactions (see Section 2.1.7.3). In the first initially only a single compound is present:

Examples for the Determination of Concentrations

Ch. 6

489

B -2

2>4-»B

ka^

In this case the rate is determined by jc =fo= — = ka^. 2 The dimension of the rate constant k is [1 mol~* S"']. By integration (jc = 0, a = a„, and t = 0) one obtains either 1 1 ^ ,

1

= 2Jtr

a

=»

Gn

X^-laAt

- =

a

^—

an

or by rearrangement (2.24)

a = -1 + la^kt

With bQ=0 one finds by substitution into the above b and integration

,.,^_V^

(1 + la^kty

The half-time is reciprocally proportional to the initial concentration (4/2=^o/2) la^k Frequently the concentrations are related to the initial concentration to obtain relative and comparable values. These are called general coordinates. Transformation to these coordinates yields a = —, an

i8 = — ,

^ = —, an

x^lajiU

Appendix

490

Ch.6

and in consequence a =

21 + r

1 1-fr

1

^

1

1

On the other hand a reaction with two different initial components as

A + B-^C

X

A -1

B -1

C JC 1 kab

takes place. The differential equation of the rate has to be solved by a seperation of the variables and by formation of partial fractions. JC = - a = - ^ = c = kab = k{aQ - x){bQ - jc). Integration at / = 0, jc = 0, and a = a^ yields In-rr— = In—- = -(6o - a^Jki [bQ - x)aQ baQ or using exponential notation: X = C

a^b^jl ~exp(~(fcQ ~ ao)kt)) bo - a^expi-ibQ - aQ)kt)

^ UQ (bQ - UQ ) exp(-(fco - gp )kt)

fc =

(2.25)

feo~aoexp(-(&o~ao)^0'

It is common use to define the half-time with respect to the component with the lower concentration. Thus, if a^ < b^ tm^

1 bo-^o

-In 2^^ ^oy

Ch. 6

Examples for the Determination of Concentrations

491

By transformation to the general coordinates a

^

b

c

^

y = —,

X

M = -^,

? = —,

r = a^/rr

an

one obtains ^ ^ ^ M(l-expKM~l)r)) M - exp(-(M - l)r) ' i8 =

u(u -1) u - exp(-(M - l)r)

^ ^ (M-l)expKM-l)r) w - exp(-(M - l)r)

^1/2 =

1

, x^ 7ln(2

^ ),

da -u. 6,4.2 At the end of the reaction In the following, the equations for the concentrations at the end of the reaction are derived for a variety of mechanisms in addition to the text in Section 2.1.8. Example 2A3: Bimolecular reaction (different reactants) For

A + B"->C

X

A -1

B -1

C Xk 1 kab

the use of the schematic representation gives according to Example 2.12 the following relationships by taking a^ - a^Q = Aa, and combining it with the stoichiometric coefficient and the degree of advancement of the considered reaction (row in the reaction scheme): a = aQ — X,

b = bQ- X,

and at the end of the reaction

c=x ,

492

Appendix

Ch. 6

Thus

Using the relationAa, = v,.;c and a^-^a^.x-^x^ whether we are looking at looking at a or 6 either

in depending on

or Coo = ^ c o =

^0

is obtained. To avoid negative concentrations the following restriction is physically necessary: Coo =JCoo=Min(ao,fco)-

Example 2.14: Consecutive reaction For the mechanism

A- • B B- C

JC, JCj

A -1 0

B 1 -1

C Xk 0 k^a 1 k^b

one finds a = —k^a^

b = k^a - ^j^*

c = k2b.

At the end of the reaction all starting material as well as the intermediate B have disappeared: a^^O

atid

b^^Oi

Examples for the Determination of Concentrations

Ch. 6

493

therefore with c^ = AQ and considering only the initial and end products one has to use the first and the third column of the scheme above

Ac = c^ - Co = do - 0 = X2«., a^ = ^2oo or •^loo — •^2o« ~~ ^oo ~" ^ 0 '

Example 2.15: Complex consecutive reaction For the mechanism

B + C-^D

JC,

A -1

^2

0

B 1 -1

C 0 -1

D ^k 0 A:, a 1 k2bc

the rate law is given by d^-k^a,

b = k^a'-k2bc,

c = -d = '-k2bc.

Therefore

is valid. In consequence one finds at the end of the reaction either boo = ^loo"" -^loo = 0

or c«. = Co - X2^ = 0 ^ 2 o o — <^o •

Thus this example gives ^loo = «o

^^d

X200 = Min(a^, Co ).

Appendix

494

Ch. 6

Example 2.16: Complex consecutive reaction including back'reaction For the mechanism A + B-^C C-^A + B C + D-^E

X2

A -1 1

Xj

0

h

B -1 1 0

C 1 -1

D 0 0

E ^k 0 k^ab 0 he

-1

-1

1 k^cd

the rate law is given by a = b ==-k^ab-h k2C, c — k^ab- k2C - k^cd, -d = e = k^cd. In the case where d^ > MinCa^.^^^, the result is ^oo — -^loo "" - ^ a o o " " -^Soo"~ ^

and either

doo^dQ

or

-aQ>0

is obtained. In consequence the final concentration of compound E is given by ^oo =A:3^=Min(ao,feo). Taking the other assumption d^>Min(a^,b^\ one finds the final concentration of D to be zero:

Ch. 6

Consecutive Reactions with Superimposed Back-Reactions

495

In consequence the final concentration of E is determined by the initial concentration of D: •^300 =

^00 =

" 0 •

Therefore an equilibrium is obtained between the remaining amounts of reactants A, B and C according to c^ =

koodoo-

6.5 CONSECUTIVE REACTIONS WITH SUPERIMPOSED BACK-REACTIONS Using the definitions given in Table 2.2 the following equations derived in Section 2.2.1.1 are taken for the calculations in the next sections: Degrees of advancement for all steps of a reaction In matrix notation

^t = kj^aj. {k = 1,2,..., r)

Differential equations with absorption coefficients Non-reduced concentrations

v°*i* = ^^ = v°*Kna* s Kca*" a° = aj + pAa

And its time dependence

^« = KtCaJ + pAa) = ao + K^pAa

Including the defmitions

i j = K^aJ

^* = Kja"

K'C = VKS

Aa = a ^ = K^a

SQ

Reduced time dependence of degree of advancement Changes of selected concentrations Reduced with respect to mass balance

Aa = vx v i = ^ = v Kni = v Kft(a° + BAa)

And reactions steps

K^ = V K Q P

Changes in concentrations

^ = ^0 + K^^Aa ao = v KoSo

Appendix

496

Ch.6

x = v"^ao+v"'^KcVx x = xo + K-x Jacobi matrix

K = v"*KcV

Example 2.17 In addition to the derivation given in Section 22.1 A aspects of Section 2.2.1.2 are included here for the reaction scheme for the reaction A<->B->C A -1 * 1 X2 *

0

B 1 -I -1

• 4>

C 0 kiO 0 k^b 1 k^b

Using eq. (2.40)

we obtain K=

(-{k^+k^) k,

k

and its inverse K-' =

•^/'' "IT

to find a.

A Xx - 1 0 X2

B 1 -1

C Xk 0 k^a — k^b 1 k^b

Consecutive Reactions with Superimposed Back-Reactions

Ch. 6

497

Example 2.19 For the most simple consecutive reaction A -^ B -> C one finds for the steps given in detail in Example 2.17 the reaction scheme

JC,

A -1

B 1

X2

0

-1

C ^k 0 kfO 1 ^2^

r-1 o\ V

1 0

=

-1 V = 1 0

-1 Ij

X = X* = K„a° =

(-ki a = A:,

n.

fc.

0

V-*2

0

Ky

0 -k^

0

0^ -1 1;

OVa^ 0 b

0^ -1

" 1 V .<^J

with

(-k^

0

0\

0

itj

0,

K^ =

r-it, it,

0 -k^

^ 0

A:2

a = -i:,* = —A:,a(?)

-1

0

1

-1

v-' =

0

(a\ h\

k^ ojtc^

^a^ (-1 1

V=

fki '^o ~ '^o ~

-1

0

-1

-1

OYa'i 0 Oy

v-/

0

0

k2 0

498

Appendix

r-1 o\ P=

1

-1

0

1

-1 -1

^a^ 0

a=

^1

0 -1J=

^1 0

Ch. 6

0^

0

1

-1

-1

0^ 1 tA6

v^y

^"=^o+K;pAa =

'-*.«o' *,ao

^

+

-ife,Aa

^

/:, Aa - ^jA^

I 0 J ^^

k^ti^b

J

a = -^j^o - /:, Aa + k2^b b = ^j^o + /:, Aa - /rjAfc c = ^2 Aft /" ' A

Jt,

V^ay

0

x = Kor =

^a^ ON 6 it^ 0) ^<^> 0

i , = /:,a(0 i j = *2^(0

vx = a = V K„a = v K^ (al + pAa] with

-k^Aa \

Kc=vKoP = *i

0 y

Finally the rate matrix is obtained

r-i oYfe, 0 1,-1 -\ik, -k, and

-1

0

^-)k,

0 ^

-1

-1

V ^2

~*^2y

- -

Consecutive Reactions with Superimposed Back-Reactions

Ch.6

499

to find

0 ;

VM1-M2;

1^200;

\a^

Example 2.20 For the consecutive reaction A->B=?^ C one finds for the reaction scheme A -1 * 0 •^2

B 1 -1

0

1

f-l V

=

K;=

C 0 1 -1

k^b kjC

0 -1 1

0^ 1 -Ij

fkt

0

0^

0 {0

Jkj 0 0 k^

1 0

a =

Kj

0

B 1

X2

0

-1

-1

0^

1

-1

0

1

V

=

Kn =

fe,

r^*\

(L

0

y^ij

0 {0

k^ 0 ky)

X = KQ • a =

f-k.

Xl

A -1

k^a

0 -Jt AC2 AC')

0

-)t.37\^J

^k kfO fCyly ~~ IC-yC

0

0^fa^ 0 b

f-k,

OYa\ ^3

C 0 I

with

Kc =

0 "~fC'>

0

0 ^ AC-i

3 /

Appendix

500

(a'' f-l b = 1

0 -1

.0

1

<^J

V =

p=

oVx;] M,

iA

*.

0 -k.

*3

-V {_XjJ [o

h

-hj

1

1

*

X2

=

0^

r

i^j

(-.*-,)• - ( : ; - . ] -1 1 0

0 -1 -1 -1

0 -1

1

n a

Ch.6

b

0

c

0

1

l-i

-1

o\

0

1

-1

-1

c = - A a - A£> -*,Aa a°=ao+KJ;pAa =

^, Aa - ^jA^ — ^jAa — /^j Afe 0

\^

k2Ab + kyAa + kjAb

j

a = —^jflo ~ kiAa b = kfOo + (*, - kj)Aa -{k^ - k^)Ab c = kjAa + (ATJ + kj)Ab ^^ ^ x = Koa° = K^2j

(k^

0

0

0

*2

-ifcj

^2 = Jkj^CO - ^3^(0

Ch. 6

Consecutive Reactions with Superimposed Back-Reactions

501

^ = vKoa°=vKo(ao+PAa) -k^Aa (L - L )Aa - {k. + k. )Ab

-*i«o

with

Kc = vKoP =

r -k,

0

Finally the rate matrix is obtained

{-\

o y -*i

v-i

0

o^_

-k, *2

0 -(^2+*3)

and

r -V*.

0 1

K '= I

fCi yiCj

~r fC-j y

^

fCj "i rC-j i

to find

x=

and

0

2

1 "" v 2

3/

2

^loo

/r^a, 2"0 V ^ 2 "*• ^3 y

Example 2.21 For the complex reaction with two equilibria A ^F^ B :?^ C according to one obtains the reaction schemes and the matrices B

C

. •

1

0

k^a

0

-1 -1

0 1

0

1

-1

A * -1 •

2 X-y

* JC4

1

^2^

X,

-1

B 1

k^b

X2

0

-1

k^c

A

C ^k 0 A:,a - ^2^ 1 kjb — k^c

Appendix

502

V

r-1 o^

r-1 1 0 0^ = 1 -1 -1 1 0 0 1 -1 0

.2

0

h

0^ 0 0

1,0

0

kj

(k

K

• •

0

T^*

V

and

o

Kft =

=

1 0

^1

~1 1

"1

-^2

-*4J 0

0^ ^a^ 0 0

0

k^j v^v

0 0

-•O

X =Ko-a =

^0

-K

Ch. 6

*2

a =

0 with

0

-*, Kc =

fCj

^^^2 ' '^3)

0

^a^

r-1 1

V'-V

^4

-it.4 /

1 -1 0

0 -1 1

p'1

/ -k. 0> x'2 1 = I ^1 —

-1;

•^3

V 0

0 Ya^ v^2

A:3

'^3)

^4

- A : 4 v^y

Consecutive Reactions witli Superimposed Baclc-Reactions

Ch. 6 V=

-1 1

0 -1

.-I

-1 -1 -1 1

P= 1 0

=(::

-.)

O

0^

0 -1

0 -1

(I 0 + 0

0\ Aa 1 Ab

"0

a = \b =

.0,

1 -1

,-1 -ij

c = -Aa-Afe

f-k,a^ a'=aoH-K^pAa = 0

^

-k^Aa-hk2Ab ^ + (it, - k^ )Aa - (k2 + /:3 + A:4 )Ab k2Ab + ^4 Aa + ^4 Afe

^1^0 —ktAa-^-k-yAb '^1 a = Lan k^a^ + (/:, - k^)Aa - {k^ + ^3 + A:4)Afc

c = k^Aa-{'{k2 •^k^)Ab *,

x = Kor = V-^2,

0

-Jk2

0

Jt3 -A:4

^a^ ft

X,

=kya{t)-k2b{t)

X2 = A:3fc(0 - ^4^(0

a = vKor=vKo(a^+PAa) a b with

-"*i«o

fc, a^ j

'k.Aa-\'k^Ab v(^i - ^4 )A^ - (^2 + ^3 + ^4 )Afe

503

Appendix

504

Ch. 6

~ * i

Kc=vKoP =

Finally the rate matrix is obtained J-l [-1

OY -*, -ij^.(^k^+k2+k,)

*, Y-1 (k^+k,+k,)Xl

0 -1

'-(k,+k^) -(kj+k^) and AC^/C^

1V 3

K-' =

4 -^

^^2 4

1 >^ 3

4 •'

^2 4

k^ +k2 \^

k^ (k-^ + ^4 ) •*• ^2^4

^1 V^3 "*" ^4 ) "^ ^2^4 )

to find x=

^1^0 "I M^i+*2)-^l+*2-^2

0 J y k^x^-(k^-^k^)x2

and f

k^(k^+k^)aQ ^ k^(k^-hk^) + k2k^ kik^ciQ

[ ACj yK^ I K^ ) "T ^^2^4

Example 2.18 The method discussed is applied to the chemical mechanism of a cyclic reaction according t o A ^ B ^ C ^ A given in Example 2.8 with the reaction schemes

Consecutive Reactions with Superimposed Baclc-Reactions

Ch. 6

1

• •

A

B

c

-1

I

Xy

1

-1

0 0

k-^b

*

0

-1

1

kjb

Xt 4>

X"!

k^a

JC4

0

1

-1

k^c

*

1

0

-1

k^c

-1

0

1

k^a

•^5

xl

A

B

C

^i

-1

1

0

k^a — ^2^ - k^c + k^a

X2

0

-1

1

k^b — k^c — k^c + k^a

Xk

an the following matrices:

V

=

K* =

r-i

1

1

u

-1

-1

0

1

ffc.

0

0^

0

Ky

0

0

0 0

0

k.

0

0

k,

^,*^6

0

oj

0

0 1

1 -1^ 0 0

-1-1

and

V

0

l<

ij

KQ

/ r-1 0^ = 1 -1 0 1

-k,

=

-(*4+*5)

0 /'„•^

X =

KQ

\^2J

a =

ki Kg

0

K2

0 0

0

•a =

-ikt+k^)

505

0

0

k,

0

0

k.

V*6

0

k^ -ik2+k^) K3

^a^

v-y

ki

Ya\

k^

lb

—{k^+k^^Kcj

Appendix

506

Ch.6

with -(k^+k,)

k,

K^ =

-(k^+ks))

fa\

f-i

1 1

y'-j

-1 0

0 1

0

-1 1 -1

1 0

0

-1

1

a = -x; +xl+ xl - i ; = -(*, + k^)a{t) + k^m + kjciO b = Xi -xl-xl+xl=

kia(t) -(*2 + kjMO + k^c(,t)

c = i j - ^4 - i j + ig = k^a{t) + k^bit)-{k^ + k^)c{t)

v=

r-1

V

1 -1

0

P = 1 -1 -1 -1 0

^a^ a = b

-1 -1

0

1

=

n 0

-1

J

1 0 + 0 -1 vOy

0 -1

1 -1

0\rAa\ 1 -1

{Ab)

c = -Aa - Ab (-(k,+k,)aA ^° = ao+K^pAa =

-(iki + Jkj + itg )Aa+(jfej - *5 )Ab \ (ki-k^)Aa-(k2+kj+k^)Ab [ik^+k^+k^ )Aa + (ikj +^4+^5 )Ab)

Consecutive Reactions with Superimposed Baclc-Reactions

Ch.6

^ = ^j^o + (A:, - k^ )Aa - (A:2 + ^3 + ^4 )Afe

/^. ^

x = Koa^ =

(ki -\-k^

k2

k^

\-^2j

^a^ V^y

•^2 = ^6^(0 + ^ 3 ^ 0 - (^3 + k^)c{t)

^ = vKor=vKo(a^4-PAa)

with ^-(k^+k^-^k^)

k^-k^

Finally the rate matrix is obtained

(-1

-iX

*,-*4

~(ifc, + ^ 2 + ^ 6 ) ^3 "~ ^6

-(*2+*3+*4)Al *2~"*5

'\

~V^3 + ^4 "^ ^5 ))

and I

k'^ -\r k^ '\- k^

^2 "*'^5

1

itr

J

kr

K-' =

^3 "" ^6

V with

/:r

-ij

507

508

Appendix

Ch. 6

to find

x=

f{k^ 4-fcg)aQ "^ r~(/ci + /:2 "^ ^6 )-^i "^ (^2 "" ^5 )-^2 V

6

/

V ^3

^6 /-^l "" v^3 •*" ^4 •*" ^5 )-^2

and

Jtr [/:t/:3-h/:6(^3-^^2)K jfcr

V-^2oo

Example 2.22 An application of Table 2.6 is the simple consecutive reaction A —> B —> C. According to Section 222A it has the eigenvalues: r, = - / : ,

and

h ~ ""^2

looking at the elements on the diagonal of the K-matrix. Therefore one finds according to Table 2.6 for the concentrations of the three reactants the following relationships: For A: No compound A is present at the end of the reaction, therefore a^ = 0 . Furthermore taking just the first step of the reaction in the initial phase as a boundary condition one can use a^ = —k^a^. Both are introduced into the elements given in Table 2.6 for .y = 2:

^

r,-r2

-/:, + ^2

_^io"-n(^»o~^»oo)_ r2-r,

-/:2 + *,

= ^n

=0

Thus using eq. (2.57) and substituting the calculated elements one finds

Ch. 6

Consecutive Reactions with Superimposed Back-Reactions

509

a = aoexp(-/:,/). For B: Assuming the boundary conditions b^^^Q, b^-Q 60 = -^0 == ^1^0 ^"^ ^^" derive accordingly

"~fC|

fC|

I Ky

_ fc,ao + ^1 (0 - 0) _

I Ky

fcjao

Ky

and

"" fCi

__ k^a^

and obtain for the time-dependence of the intermediate, inserting the coefficients above into eq. (2.57) ka fc = —^(exp(~/:,0~exp(-A:20) • Ky

Kt

For C: Finally considering the concentration of compound C at certain conditions of time one can set CQ = 0, C^ = a^ and CQ = 0 and obtains for the elements p^f^

i\>y

Ky

Kit

"^ rCi

The result is according to eq. (2.57):

c = ao + -. ^^_l e x pexp(-fe,r) ( - ) f c , 0 + - '_° ^ expC-fe^f) Ky Kt Ky ^ K-t

Ky K,y

IKt

c= 1

1

c = 7 - ^ [ * 2 ( l - exp(-^,0) - *, (1 - exp(-*2r))] ^2

^I

Appendix

510

Ch.6

6.6 RELATIONSHIP BETWEEN QUANTUM YIELD AND PHOTOMECHANISM 6.6.1 Details for simple uniform photoreactions 6.6.1.1 Photoaddition Without desensitisation by an additional partner via singlet: A ^A'

A + Av A'

>A -»C

A' + B

A

-1 1

X',

0

<

A' 1 -1 -1

B

C

K

0 0 -1

0

/A

0 k^a' 1 k^a'b

via triplet

>A'

A + AvA'

>A

A'

>A"

A"

>A

A" + B-

•4C

A x[ - 1 x\ 1 A 0 A 1 A 0

A' A" B 0 1 0 -1 0 0 -1 1 0 0 -1 0 0 -1 -1

c

A

0 /A 0 k^a' 0 k^a' 0 k^a" 1 k^a"b

With desensitisation by an additional partner via singlet

A + hv A'

>A' >A

A' + B B' A' + C

»B' + A ^B >D

A' 1

B

B'

x{

A -1

0

0

A

1

-1

X^

1

-1

0 -1

0 1

A A

0 0

0 -1

1

-1

0

0

C 0

D

A

0

IA

0

0

k^a'

0 0 -1

0 0

k^a'b k^b'

1

k^a'c

Ch. 6

Relationship between Quantum Yield and PhotoMechanism

511

via triplet

A + hv

>A'

x;

A 1 A' 1 A"1 B -1 1 0 0 1 -1 0 0 0 -1 1 0

B" 0 0

D

0 0 0

0 0 0 0

k^a' k,a"

0

k^a"b

A'

>A

4

A'

>A"

X-i

A"

>A

<

1

0

-1

0

0 0

4 4

1

0

-1

-1

1

0 0

0

0

0

1

-1

0

0

0

-1

0

0

-1

A" + B B"

>B" + A >B

X',

A" + C

6.6.1.2

>D

Sensitisedphotoreactions

Example 3.8: Physically sensitisedphotoreaction The physically sensitised photoreaction B-

kv

^C

with

^A'

A + /iv A'

>A ->B' + A

A' + B B'

>B

B'

>C

A x[ - 1 A 1 A 1 A 0 A 0

A' B 1 0 -1 0 - 1 -1 0 1 0 0

B' 0 0 1 -1 -1

c

A

0 IK 0 k-jO.' 0 k^a'b 0 k^b' 1 ksb'

yields a = 0,

-^b-c-ip^Ip^,

kjk^b
{k^+k,b){k,+ksy

1 ^*

c

'A /Cj^'

0 Kb" l \ h,a"c

Appendix

512

Ch.6

If Bodenstein's hypothesis is applied to A' and B' and /TJ ^O 10* s"* and k-^ toto 10*0 10*0moh* moh*s~* s~*the thebb concentration concentratio has to amount to 10^ 1 mol"* at k^ least to supply a remarkable turnover. Example 39: Physical sensitisation using the triplet state Using the same mechanism with the exception of an energy transfer from the triplet state of B, one finds the modified scheme

A + hv

>A'

A'

>A

A'

>A"

A"

>A

A'' + B

>B''-f A

B''

>B

B"

>C

A x{ - 1 1 ^2 0 ^3 1 X\ x's 1 0 < 0 X'-,

A' A" B 1 0 0 -1 0 0 -1 1 0 0 -1 0 0 -1 -1 0 0 1 0 0 0

B" 0 0 0 0 1 -1 -1

C 0 0 0 0

K /A

k^a' k^a' k^a" k,a"b

0 0 1

Kb" k^b"

with the rate laws a = 0,

^br:zc^ip^l^,

Jkjfc

(p^ ^ Ky

I K'i KA

t ^s

k-, 6

7

ithin the quantum yield the factor k^b k,-¥k^b

b

ii+* *,

determines the minimal necessary concentration of B to allow energy transfer. Concentration b must reach the order of magnitude of k^/k^. If ^4= 10^ s"* and k^- 10*^ 1 mol"* s~*, the concentration of component B must become approximately 10~^ mol 1"*.

Relationship between Quantum Yield and PliotoMechanism

Ch. 6

513

Example 3.10: Physically sensitised photoreaction with another partner Another possible scheme for the physically sensitised reaction B + C-

hv

^D

takes into account a bimolecular step of reaction and is given by the following partial steps of reaction:

-»A'

A + kv A'

<

>A

X-i

^B' + A

A' + B B'

X
x;

>B

B' + C-

4

->D

A -1 1 1 0 0

A' B B' C 1 0 0 0 -1 0 0 0 -1 -1 1 0 0 0 -1 0 0 0 -1 -1

D K 0 /A 0 k^a' 0 k^a'b 0 k^b' 1 k^b'c

One finds
k^kAbc V3/V41 (k2 + k^b)[k^ 4- k^c)

A measurable turnover is obtained, if b-^

and *3

h ^5

Assuming the bimolecular rate constants to be diffusion controlled and having a value of approximately 10^^ 1 mol~^ s-^ and taking the average lifetime of A' and B' to be 10-^ s, then the minimal concentration has to be &-'C~10"^moll-'.

Appendix

514

Ch. 6

Example 3.11: Physically sensitised photoreaction with another partner and via triplet pathway In comparison to the previous example for this mechanism the energy is transferred via the triplet state according to the following scheme

>A'

A + Av

^;

A'

>A

A'

>A'

4 4

A''

>A

<

->B" + A

A" + B B"

<

>B

B" + C

A xf,

A -1

A' A" 1 0

B 0

B" 0

C 0

D 0

xl

1 0 1 1 0 0

-1 -1

0 1

0 0

0 0

0 0

0 0 0 0

-1 -1 0 0

0 -1 1

0 0 0 1 -1 -1

0 0 0 -1

k^a' k^a' k^a"

0 0 ksa"b 0 k,b" 1 k^b"c

0

IK

->D

Thus - c = ~fe = rf = y)^/^

and ,p- =

3 5 7

{k^+k^){k^^•kib){k^+k^c)

are obtained. Turnover becomes measurable, if b~^

and

c--^.

Therefore the minimal concentration has to be fe ~ c ~ 10"* mol r* for bimolecular rate constants of 10'° 1 mol-' s"' (the same as explained in Example 3.1) and for average lifetimes of A" and B" amounting to 10-4 s.

Ch. 6

Relationship between Quantum Yield and PhotoMechanism

515

6.6,2 Details for simple non-uniform photoreactions 6.6.2.1 Physically sensitised parallel photoreaction Example 3.20 B-

hv

D-

^c.

•^1

B -1

C 1

X2

0

0

D 0 -1

•^E

E 0

^^A

1

^2/A

^k

Overall scheme: A x{ - 1 1 ^2 X3 1 0 JC4 JCj 0 JCg 1 x^ 0 4 0

->A'

A + hv A'

>A ^B' + A

A' + BB'

>B

B'

>C

A' + D

>D' + A

D'

>D

D'

>E

A' B 1 0 -1 0 - 1 -1 1 0 0 0 -1 0 0 0 0 0

B' 0 0 1 -1 -1 0 0 0

c

0 0 0 0 1 0 0 -1 0 1 0 0

6.6.2.2 Physically sensitised-consecutive photoreaction Example 3.21

C—^D

B_^C,

Xl

B -1

C 1

D 0

H

0

-1

1
H


D 0 0 0 0

D' E 0 0 0 0 0 0 1 -1 -1

Xk

u

0 ^2^' 0 k^a'b 0 k^b' 0 k,b' 0 k^a'd 0 k^d' 1 hd'

S16

Appendix

Ch.6

The overall scheme: A + hv A'

>A' >A

A' + B

x; X2

»B' + A

^3'

B'

>B

x'.

B'

>C

^5

A' + C

>C' + A

C

>C

C

>D

A -1 1 1 0 0

*;

1 0

•^8

0

^6

A' B 1 0 -1 0 -1 -1 0 1

B' 0 0 1 -1

C 0 0 0 0

0 -1

0 0

-1 0

0 0

0 0

0 0

1 -1 1 0

c0 0 0 0 0

D 0 0 0 0

1

0 0

-1 -1

0 1

xi /A

kjtt'

k^a'b k^b' ksb'

hb'c kyC' kgC

6.7 EXAMPLE OF A PHOTOKINETIC EVALUATION As a typical example of a photokinetic examination the consecutive photoreaction of rmn^-stilbene to phenanthrene is taken. Referring to the different theoretical and experimental sections, some essential equations are reviewed, the relevant figures are redrawn and the procedure of the evaluation is explained stepwise. Computer programs for the evaluation exist. They use measured signals to construct absorbance diagrams of different orders to extract the number of linearly independent reactions and apply Runge-Kutta algorithms as well as exponential curve fitting to determine photochemical quantum yields. Since a general overall program for all possible mechanisms is rather difficult to supply, the given example is supposed to allow the reader to modify the parts where the specific mechanism is treated. Graphical presentation, user interface, and part of the algorithms are independent of the mechanism. The following procedure is proposed. (1) The stirred solution is intermittently irradiated - best in a combined irradiation and measurement device. At least 30 spectra are taken for a simple photoreaction. In the case of more complicated mechanism preferably 25-40 spectra should be recorded per linearly independent step of reaction. Care has to be taken that the total wavelength range is recorded in which during reaction progress spectral changes occur, especially in the UV. (2) Next this reaction spectrum is plotted. Wavelengths are selected at

Ch. 6

Example of a Photokinetic Evaluation

517

which the change in absorbance is typical. The present program on the Internet allows bars to move across the reaction spectrum realising these effects. (3) Absorbance-time curves are plotted. (4) Rank analysis allows the determination of the number of linear independent steps of reaction. Graphical approaches are preferable since deviations of a correct determination of this number become better visible than fuzzy results which make the decision delicate. Absorbance diagrams are constructed at different orders. (5) According to these results and additional information, a mechanism is proposed, which has to be proven next. (6) The rate laws according to the mechanism are set up. (7) Since photokinetic equations cannot be solved in a closed form, equations are preferably rearranged due to allow formal (numerical) integration of the integrals. It is best to choose a matrix representation. The elements of the Jacobi matrix are obtained by (a) Runge-Kutta iteration (b) parametrisation via curve fitting (c) transformation of the time axis to reduce complexity and parametrisation. (8) Only rather simple mechanisms give the quantum yields directly. Furthermore according to Lambert-Beer the absorbances have to be converted to concentrations. Complex Q-matrices are obtained (see Section 5.4). In addition in most cases only products of absorbance coefficient times quantum yield are obtained or sums and differences of these products. Therefore additional information is necessary. (9) It is best to find a way to construct concentration diagrams either by synthesising an intermediate or by using an additional method such as HPLC. (10) For all the results it has to be proven that the mechanism is fulfilled (a) not only with respect to the number of linear independent steps, (b) but also considering backward reactions and superimposed dark reactions. This is rather generalised advice to treat photokinetic data. Some of these points are applied to the stilbene photoreaction as an example. The mechanism of a consecutive photoreaction is obvious. The procedure outlined above is followed in the next paragraphs step by step using a program which can be obtained from the authors on disk by re-

Appendix

518

Ch.6

quest. In this program Runge-Kutta iteration and curve fitting is demonstrated in source code to allow the user to set up his own program lines for other mechanisms. The data measured can also be transfered to EXCEL (Microsoft) or ORIGIN (Microcal). The first is used for the graphics, the second can be used for curve fitting or even formal integration. 6.7.1 Measurements Usually absorbance measurements are taken using the equipment demonstrated in Section 4.2.2.3. Since these are not selective to the different components, in addition other methods are used such as chromatography (see Section 4.5.3). 6.7.1.1 Absorbance First a reaction spectrum is taken according to the statements above.

C3 JO

X)

<

Fig.4-16

Wavelength [nm]

Ch. 6

6JA2

Example of a Photokinetic Evaluation

519

Chromatography

Fig. 5-55

6.6 7.2 7.8 8.4 9

\U]

Loss in concentration is corrected for each injection according to the next graph. Fig. 4-30

50 injection

Appendix

520

Ch.6

6J.2 Graphics 6.7.2.1 Absorbance-time curves Using the reaction spectrum at selected wavelengths absorbance-time curves are plotted first. c

320 nm

Xi

- • — 3 1 3 nin

o JO

<

- • — 3 0 6 nm

0.6

— A — 292 nm — T — 2 7 3 nm —•—250nm

0.4

0.2

0.0 Fig.4-17

200

100

300 Irradiation time [s]

6.7.2.2 Absorbance diagrams Next absorbance diagrams are plotted in various combinations. A typical graph demonstrates curves, which means the reaction contains more than one linear independent step of reaction.

Fig.4-18

1.0

1.2

1.4

1.6

Absorbance at 230 nm

Example of a Photokinetic Evaluation

Ch. 6

521

Furthermore absorbance difference diagrams can be plotted as demonstrated in Fig. 5.7. Whereas the additional information of this type of diagram is small, absorbance difference quotient diagrams can be used to decide whether the number of linear independent reactions amounts to 2 or more as given in Fig. 5.11.

Fig. 5-7

Flg.3-n

Taking the data as used for Fig. 4.18 the following linear relationship is obtained demonstrating that the reaction is represented by two linear independent steps. Therefore any of the mechanisms given in Table 2.5 can be used. On the other hand chemical information requires a consecutive reac-

Appendix

522

Ch. 6

tion pathway. Furthermore one finds starting at cw-stilbene the same reactants. Therefore the first step trans- to cw-stilbene must be an equilibrium. 6.7,3 Evaluation of absorbance data 6.7. i. 1 Set-up of differential equations Accordingly the mechanism A;=^B—!^C can be assumed. As done for a similar reaction in Table 2.1 the schemes of the reaction can be set up as

A -1 * 1 * 0

B 1 -1 -1

. 4>

C 0 'PUK 0 ¥>X 1
Xi

—" Xt

Xy

X'^ -~" Xi

Xi Xy

A -1 0

B 1 -1

C Xk 0
with the differential equations in analogy to Section 3.2.2 as dx. "•1 dt

^

at

•o[l-exp(-£„')]

— =_ i,A =1000

= x, = 1000^«^/o[l-exp(-£;)] t,„

or using decadic absorbance units i , = 1000/o(^^'Aa -
Ch. 6

Example of a Photokinetic Evaluation

523

According to Section 5.4:

+lOOOIo
\dE, = \(mi;f^z^\{E,-E^)F{t)dt

(A = 1.2....,5)

or {(lOOO/o)-' F(0-' dE, = J^z,,\E,dt -\Y,z,,E^dt

= J^z,,\E,dt -

z,,\dt.

where the Z;^J^coefficients are found according to Example 5.22 as

ZM

= [ ( e , 2 ^ ? "-G„^2)('i2

-^22KB

-Gll^f^'A'ia]

^21 =[(022^3 -G21^2)(ni --^2lK -G21^lVAr,,]

^22 = [((322^3 - 021^2 )(n2 ' ^22^8 "

QIX^X^'K^,^]

^20 ~ V"~^21^1oo "• ^22^2oo j

By formal integration one can obtain these Z;^^ elements. To calculate the quantum yields, however, the absorption coefficients of all the reactants at all the wavelengths of measurement as well as at the wavelength of irradia-

524

Appendix

Ch. 6

tion have to be known. Looking at the equations one can understand that the correctness will be rather small because of the many factors in each zelement. For some mechanisms a more simple approach is possible: As explained, the Jacobi matrix and the Z-matrix are similar matrices. Therefore the trace and determinants are equal and one obtains

D = ipx^\(p^e'^ = ^11^22 - ^12^21

For some mechanisms two such equations can be used to calculate the quantum yields, if the absorption coefficients are known. For the transstilbene photoreaction, however, the elements are too complex. 6.7A Evaluation of concentration data 6.7A. I Absorbance triangle The absorbance diagram of the stilbene photoreaction is shown schematically below. Since the first step is an equilibrium, the absorbance line from c/5'-stilbene to phenanthrene does not form a tangent to the measured absorbances. The point B is determined from the absorption coefficients of the c/5"-stilbene at both the wavelengths 250 and 292 nm and the initial concentration aQ. Point C is obtained by the absorbance coefficients of phenanthrene and the initial concentration of rran^-stilbene. The lines ^A, K^ and ^c correspond to the calibration curves for the three reactants absorbance/ concentration. The triangle can be considered as a vector space, given by u, v, , Vj. Using these vectors the concentrations of all the reactants can be obtained for each reaction time according to ii = Eo-E = v,+V2 = eAao-[eAao+(eB~^A)b + ( E c - 8 A ) c ] = (eB-eA)b + (ec-eA)c Using the distances

Example of a Photokinetic Evaluation

Ch. 6 1.0

u

j^__

,

j_ _ -

r

-]-

,

,

r

J

1 - .,„.„j

,-. ,

525 -^,..„.-„ ,

j^.

\

0.9 0.8

e

0.7


0.6 0.5 0.4

f

' ^i\

1

U

.

IM

0.3

<

0.2

1

H

'^«

V' " \*

^

A

^ ^ ^ N . . ^ ^ ^

jl

\

Si

O c« Xi

-J

""A^

a

cs '*-* a o o ced

\

1 ^^ k't^.o J ^

l»

^'''*^^^ J

' h " ^

,Kx:cj

'^j^E(t)

,,,.—-^-r-^"^^"'^'*'^ c 0.1

Ec'Ecao

k VS»o '^' r [ ;

- ' 1 '

0.0

-

' .

^^1—-

— ^ ' T ' ^ " ^ ^ '

\ T

1

,

0.2

0.4

"

'

-_ t.

„ 1

0.6

t

1

1—

0.8

i__

_.

i__ __i

1 _

1.0

Absorbance at 250 nm

AB = | E 5 - E o | = (eB-eA)-ao AC = | E c - E o | = (ec-eA)-ao and knowing the initial concentration |Vl|^|(gB"gA)| b AB |(eB-eA)|ao'

N^ICBc-gA)! C AC |(ec-eA)|ao

the concentration are calculated. 6.7.4.2 Data from HPLC According to Section 4.5.3 concentration data can also be obtained by chromatography. 6.7.4.3 Concentration-time curves The result of these "concentration" determinations are given below for the triangle and HPLC measurements

Appendix

526

Ch.6

cis-stilbene

phenanthrene trans-Stilbene

2000

6000

10000

s

Data from triangle Cone.

0.5

,

,

,

.,

,

1

,

1

|i trans-stilbene

[lO'^moI/1] 0.4

\

1 /'''^'•'•'^^^-^ 0.3 r t f

cis-stilbenc

^

^^^ •"*- '

0.2

L

J

H f\o

r / v>

phenanthrene a

0.1

o

1 B

Fig. 5-59

NOg-

-wlt-*''*''"'^'^ -

2000

4000

r-2

6000

8000

/[si

HPLC-data

Ch. 6

Example of a Photokinetic Evaluation

527

6.7.4.4 Photokinetic equations The differential equations using concentrations are given by a = {-ip^E\a{t)+
A,

B

B , |JlOOO/oF(0 = K•a•1000/oF(0•

6,7.4.5 Photochemical quantum yields Formal integration of these equations is possible using ^da = -lOOO/o^jf e ; jaF(r) dt

jdb = lOOOlQ(p^e\jaF{t) dt - 1000/o(^^ + ^3 )^B J^^(0 dt The different coefficients allow us to calculate the partial photochemical quantum yields knowing the absorption coefficients. Using the absorption coefficients for the three reactants at the wavelength of irradiation (313 nm): f^ = 15900, ^B = 1320 lmol-^cm~* the three quantum yields are obtained as ^f = 0.46,

ip^ = 0.28,

ip^ = 0.21

in perfluoro-compounds as solvents, (p^ is obtained from the Element Zj^ and (jof from a formal integration of the third rate law jdc = 1000/0^3 CBJ^^CO dt and finally using this (pf to get (pf from the element Z22 above. Since at maximal concentration of cw-stilbene &max

528

Appendix

Ch. 6

is valid, a possibility exists to check the obtained quantum yields for plausibility. 6.7,5 Computer programs Modem spreadsheets and data evaluation software can be programmed to solve these kinetic problems. Graphs can be easily obtained as absorbance diagrams of different order using EXCEL. Curve fitting programs such as ORIGIN can be used to evaluate differential equations after transformation of the time axis (see Section 3.3.11). Modem software pools supply numerically stable programs to estimate parameters (e.g. Runge-Kutta approaches) as necessary in formal integration. A demo version of the program is available at the URL in www:http://barolo.ipc.uni-tuebingen.de/photokinetics. Updates of this evaluation can be found at this address as well as future approaches for the use of EXCEL in kinetic evaluations. A simple example is given below. Based on the discussion in Section 5.4.4 and using e.g. eq. (5.93) for a simple uniform reaction, it is possible to determine the reaction constants by means of formal integration. The following EXCEL result has been obtained with the Macro for numerical integration, using data from a thermal reaction first:

rearranged for formal integration:

Y

=Zx\+

Zx2^

l\ I

Ch. 6

i u

PQ

Oil

2 Cuii

<

col

o

Q

ol

<

2 13 Q

c:

X)

<

i

^

*-^

529

'-3v2NOOoooooooor-r*t^r^r-r^vo

0000 •-
Example of a Photokinetic Evaluation

^

en m fo 00 en CO en »o m m oo '^ »-^ Tf n

888888888 o o o o o o o o o o o o o o o o o o o o o o o o o

en r^ fnr* ..cn'-«--^r^t^fno>oooor-'QONm _ , . ON _ --H i - i T f t ^ o o g s Q O N t ^ T t o « n Q f n v o o o O N O O N O N O O « oONc nCo«nio»r>»nvovovovovor^r^r--

S S S ? 8 S CS 0 0 § 8 S CS 0 0 sor^t^ooososOO'^cNfSenen

o o d o o o o o o o o o o o o o o o d o o c > o ^ ' - J

voooeninenvo*-^ t^»o»or*-'-'00oooooNO\ONO\ONOO

?8 SSS?

vo es 00 '-^'-^csmenrf'^u-j

to^enON'^voooen r^»o r^vqenvoinpenrf'^t'-j^ncN ^-H^pw^oo^^vo-^w-jr-.—^ "csvd^o6o6oN'-^»r>dvo^* 0'-Jo6oNen^enr^Ti^'^r^c4d 0\rtQ»o—tr^Ttdt^enO TJ•«o^\Ot^t^OOO^O^O'-< ,-H,-^,—icsesr^enen

CS 0 0

i8S

CM 0 0 ON

o o cs 00 v O O O ^ ^ c S c S e n e n O N '-< '-^ •-< '

?8S

8S

^t^ON'^vO"^r-cS'-'OovovooocsoNvo^oOTt»-Hen^rf»OTj-cs OooTt»-Ht--rnON*r>ON'^ooenr-'—iTfoO'—"ent^Qen«nr*0\'-^en ^^raenenTtrtv->w-)VOvor-r-ooooooONONONpOOOp^'-«

§8S

^^esenenTj-Ti-invovor^r-

s o r«i 0 0

Appendix

530

Ch.6

0,0016 Y 0.0012

0,0008

0,0004

t

Y =-0,00094 X +0,00138 ^

+ J

calculated data

1

1

1

0,2

0.4

0,6

0.8

Since Zxi = kE^^^ and 1^2 - ~^ > ^^^ ^^'^^ constant amounts to 9.4 x 1(H s-' and £;,„= 1.468. In the case of a photoreaction the photokinetic factor has to be included. Various evaluations are possible according to Figs. 5.18-5.20, the simplest one being the use of formal integration like (5.93) f,.l-10-^^'"^ dt + 2,, I E, it„) ^=^J: " e(0

dt

or \"EM„)^—^

AE I •''1

•=

dt

Zi,+Zi

dt

I

E (t )

J/,

E(t„) -

dt

with Zxi=lOOOe\(p''IoE^

and

/ ,^Ai Z;^^ =-1000e>'^/o

Ch. 6

Example of a Photokinetic Evaluation

531

or by analogy with Fig. 5.20

In both cases the Macro can be used to calculate the integrals in a more complex way. 'NumlntVl.O 'Numerical Integration using the trapeziod method 'Author: Felix Heise 'The function requires a field with two rows (or two columns) as input parameter. 'The first row (column) should contain the time-values, the second (e.g.) the absorbance-values within the integral. The algorithm of the Macro requires at least three pairs of data Function Numlnt(ParField) Field = ParField If UBound(Field; 1) = 2 And UBound(Field; 2) > 2 Then For Count = 1 To (UBound(Field; 2) - 1) Numint = Numint + (Field(l; Count + 1) - Field(l; Count)) * (Field(2; Count + 1) + Field(2; Count)) / 2 If Field(l; Count) > Field(l; Count + 1) Then Numint = CVErr(xlErrValue) Exit For End If Next Count End If If UBound(Field; 2) = 2 And UBound(Field; 1) > 2 Then For Count = 1 To (UBound(Field; 1) - 1)

532

Appendix

Ch. 6

Numint = Numint + (Field(Count + 1; 1) - Field(Count; 1)) * (Field(Count + 1; 2) + Field(Count; 2)) / 2 If Field(Count; 1) > Field(Count + 1; 1) Then Numint = CVErr(xlErrValue) Exit For End If Next Count End If If UBound(Field; 1) > 2 And UBound(Field; 2) > 2 Then Numint = CVErr(xlErrValue) End Function This macro was used to calculate the results according to the figure on p. 530 using columns 3 and 4 of the table on p. 529.