Chemical Physics 19 (1977) 429-441 0 North-Holland Publishing Company
DISSOCIATIONAND RECOMBINATIONOF A DIATOMICMOLECULE1NVOLVING TWOSTATESWHICHONLY INTERACTVIA THE ATOMICSPECIES I. A method of solution
W.L. HOGARTH* and D.L.S. McELWAIN Department of Mathematics, Universityof Newcastle, New South Wales,2308. Australia Received 4 July 1976
The master equation which describes the relaxation of a diatomic molecule with two states, diluted in some background gas from a nonequilibrium to a final equilibrium population distribution at constant temperature is developed. The only path from one state to the other is assumed to be via the atomic species. An approximate solution is obtained using singular perturbation theory based on the assumption that the internal relaxation time in each state is much shorter than the time for dissociation from either state. This assumption leads to a master equation involving two small parameters, one for each state. The approsimatc solution is obtained by considering two cases involving these small parameters. The existence of rate constants and the validity of the Rate-Quotient Law is also discussed.
1. Introduction
In this paper we are concerned with the dissociation and recombination of a dilute concentration of a diatomic molecule AZ,which has two states, in some backgroundinert gas M. Weconsider that the only path from one state to the other is via the atomic species. At time f = 0, we assumethat the translational degreesof freedom of the molecules A, and atoms A are Maxwell-Boltzmann correspondingto a temperature T, but that both the internal degreesof the molecules A, and the concentrations of AZand A do not take on their equilibriumvaluescorrespondingto this temperature. We assumethat the conditions are dilute enough t&atequilibration takes place via the following processesonly (l)A,(i) + Mf (‘)A#) f M ,
(la>
(l)AZ(i)tM~AtAtM,
(lb)
@)A2(j)+ M* @)A2(Z) + M,
(2a)
and (2b) *Presentaddress: School of Australian Environmental Studies, Griffith University, Nathan, Queensland, 4111, Australia.
where the prefutes (1) and (2) denote the first and second states, respectively.We use the subscriptsi and k to denote the energy levelsin the first state and j and I the energy levelsin the second state. These subscripts take finite values.The atom A is assumedto have no internal states. We feel that this system models such situations as the dissociation-recombination processesin hydrogen where para-hydrogenand ortho-hydrogen are regarded as the two states and in which direct para-ortho conversion is not allowed [I] and [2]. The model may also be used to describe the dissociationof diatomic molecule with two electronic states [3]. The approach adopted by Brau [3] is to extend his work on the dissociationof a molecule with one state to include the effect of the second state on such quantities as the rate constant for dissociationand the incubation time. Our approach is to treat each state sepa-rately and this method givesrise to two smallparameters, one for the first state and one for the second state, whereas Brau’smethod retains only one small parameter even when dealingwith a molecule with two states. His method does not cover the case in which the two states have almostequaldissociation energieslike the ortho-para hydrogen problem or the case of C2 where the first excited electronic state is
1V.L. Hogarth, D.L.S. McElwainlA two-stute &atomic molede.
430
I
only 610 cmdl
above the ground state. In the followingsections,we set up the master and conservationequations for reactions(1) and (2) and then approximatethe solution usingthe boundary layer method [4-71.
@a)
2. Masterequation derivation WI
The molecularconcentrations of (*)A*and c2)A2 are givenby I
(34 and “‘lv,, (t) = C
(2)nj(l!J
9
W
withi= 1,2 ,... andj= I,2 ,... + We adopt the matrix notation developedby Brau [3]. Much of the work of this and the next section follows along the lines of that paper. The distribution functions for (‘)A2 and f2)A2are denoted by I(‘)X(f)> and I(2)X(t)),respectively,with elementsbeing givenby
i
where (1)rzi(f)and (%&r) are the number of molecules per unit volume, at time t, in levelsi and j of (‘)A, and t2)A2,respectively.The relationshipbetween the total molecularconcentration A$~and the concentrations (l)A$~and f2)Nhiis NM = “)IvFn+ “‘N*t .
(4)
Since the reactions (1) and (2) conservethe total number of atoms either in the form of moleculesor as free atoms, the concentration of which we denote by NA, then the quantitiesNA, cl)NIIIand t2)Nhl must satisfy the followingrelation NA + 2(‘Qvfil+ r(%V, = c,
(5a)
where C is a positivecanstant. Using(3) it can be seen that NA+2c(t)ni+2c(21ni=c. i
i
Fb)
The probabilitiesof reactions (la) and (2a) occurring in the forward and reversedirections are denoted by %‘fk, “‘P,i and (*)PjI,(2)P,Prespectively.Similarly “‘Pit, “‘P,i and (‘)I’j,, (z)P,,,denote the probabilities of the dissociationand recombination reactions occurringin (1b) and (2b), respectively.We assume that these probabilitiesare independent of the concentrationsNA, “‘ni and %ri and that we study the relaxation isothermallyso that the probabilitiesare independent of time. The concentrationsNA, (l)ni and (2)nican then be determined from the conservation equation (Sb) and the non-linearmaster equations
‘U&.(t) = W,&)/(‘)y. 1’
(74
and ‘“‘Xj(t)
=
(‘)tlj(f)/(*)lj.
*
O’b)
Here, (1)~~and (2)~~are the elements of a normalized Boltzmanndistribution such that
and c
‘2’ri =p2 ,
i
@b)
where the partition functions p 1 and pa, which are assumed to be of order 1 in magnitude,sum to 1. The molecular concentrations of (‘)A2 and f2)A2 can be written as ‘%v&) = (1)(V$)X(r))
Pa)
and %VM(r)= (11(z)l+2)X(r)),
WI
with the diagonalmatrices (l)I’ and (%? defined by (l)r = ‘l’+k
Wd
and (2)p= “‘rjsjl ,
Wb)
where sik and ai1are Kronecker deltas. Weintroduce a scaled atom concentration y(r), where
W.L.Hogarth,D.LS. McElwain/A two-state diatomic molecule. I
y(r)
=N*(r)/Kff.
01)
Kes is the overall equilibriumconstant and is related to the equilibriumconstants for reactions (1) and (Z), namely fl)Kcq and (2)Keqby the followingexpressions (l)K,, = K,JPl
431
The elements tl)Lik and t2)LiIare givenby (UL, = CUri = (ULki and
(124
and
’
(2lKeq = k&2
Wb)
-
At equilibriumwe have the fnllowingrelationships 72 =j$
(13a)
-(20b) where we have used the principle of detailed balance. Since (‘)I, and 12)Lare symmetric then L is also symmetric.
and I(‘)%=&,II),
(135)
PNI =is,ll> )
(13c)
where the elements of the vector II>are all unity and a bar denotes an equilibriumquantity. If we form the overall distribution vector IX(t)) as IX(t)>= PX@), ‘Q(r))
(14)
3. Dissociationwithout recombination With the aim of developinga method to deal with the full non-Iinearmaster equation (18), it is useful to look first at the simpler case where the atom concentration is small enough so that recombination can be neglected. If we let y = 0, our master equation (I 8) reduces to [L - r(d/dt)]lX) = 0 .
and the matrix F as
(21)
SinceL has the form in (19) then the eigenvaluesof
r=
(Or
0
0
car
(
1
(15)
then we can write the total molecular concentration for the system as &t(r) =
.
(16)
It should be noted that the vector 11)wiiI be of the length appropriate to the expression in which it is 1
found. Thus it is of different lengths in (13b), (13~) and (16).
In terms of the scaled atom concentration y(r) and the distribution function IX(r)) the conservationequation {5) can be written as
L will be the eigenvaluesof (‘IL and t2)L denoted by
1; and $,*, respectively.We can obtain the eigenvectors of L in terms of the eigenvectorsof (l)L and (l)L in the followingway. If the eigenvectorsof (l)L and c2)L are givenby l(1)xE)and lt2~$), respectively,then the eigenvectorsof L will be
f2)x,) = IO,q*)Y
f 2(llI-lX(t)>=
c.
07)
In the same variablesthe master equation (6) becomes r(dlX(0/dr) = L (IX(t), -y2(t)l I)),
(18)
P = 0, Wx;prp)x;)=spO ,
L=((Z(2yL) .
-I-1; (l)r)((l)x*)
p = 0,i)
p, 0 = 0, I,
... ,
(234
.._ ,
C-1
and we define a quantity IV; by A$ = (1i”)rlt’)x*)P’
with the relaxation matrix L having the form
(22b)
-
We require that Xl, I($;) and $z, I(“)$ satisfy the following equations (W
Gy(r) eq
(22a)
I”)X,) = I”‘$, 0)
y=O,l....;
(23~)
and similarly
(19)
((2)~ t 9; (+)1(2)xu*) =0, Y=0, 1, ._.,
(244
W.L. Hogurth, D.L.S. McEiwain/A two-state diatomic molecule. I
432
((qolWl(2)~~ = svo,
v,Q=O,l,...,
(24b)
Ev = ((2’X~l’*)rl’2)~(0)>,
with Mz defined by p=O,l,....
,
J%f;= (ll”‘rl’2’$)
(24~)
The matrices (*IL and (*)L.are negative definite so we can order the eigenvalues as folkX!s o
<...
(25)
and
u=O,l,....
(31)
if we take the time t sufficiently long so that (XT -h$t 9 1 and (J/T - Jl:)t S 1 then (28) and (30) reduce to P)X(t)> = &rlfl+$> exp(-XG t) ,
(324
WVb,(t) = iQr\it exp(-X5 r) ,
(32b)
and
o<$;
(26)
The solution of (21) in terms of I($+,) and l(2)xV) is
t C 6Vl(2)~,) exp(-$,*f) Y
,
(27a)
and hence N&)
where
= C ii,NE exp (-X;t) + C Zv,M3exp ” P
(-$zt) , (2-1
where the coefficients ZP and gV are determined from the initial conditions by the formulae ZP = P)x,lrlx(o),
(1=0,1,...
,
1(2)X(r)) = i$ I(2Jxi) exp (-$z r) , rr>N,,(f) = @2~ exp(-$:r)
v=O,l,...
.
(274
Since we are considering dissociation without recombination, then reactions (1) and (2) are independent of one another. This enables us to write PX(r)) = C iiJ1)~~> exp (-Xit) P and
(28a)
)
p=O,l,....
(29)
Similarly, 1(*)X(f)>= C 6Vl(2)x~)exp (-Q) Y and t2)N,,(t) =C V
ZJt4,* exp(-@)
,,bs cl)NM
(34a)
and d(2)iVbI/dr = -(2)7cD ohs t2&VM,
WI
where fl)/c, Obsand’@& ohs are the observed dissociation rate konstants give; by (I)k, &s = a;, ,
(35a)
and (28b)
ZP = Px;lt’)rl(‘)X(oy
(33b)
Thus we see that reactions (1) and (2) proceed with an initial transient phase during which the initial distributions decay to new distributions characterized by I%$) and I(“)$) and the rate of this decay depends on hi for state one and I/I: for state two. These new distributions then decay at much slower rates which depend on A; for the t1)A2 reaction and rji for the t2)A2 reaction. We believe that the initial transient corresponds to internal relaxation and the slower time phase to a “steady state” during which dissociation takes place. Following the initial transient, the dissociation rates will approach the values d(%$, /dr = -(I&
1
.
(27~)
and& = P)x”lrlx(o))
(33a)
(3Oa)
(2)kD ohs = l/J; . I
(35’4
We now compare the observed dissociation rate constants with the equilibrium dissociation rate constants for t1)A2 and t2)A2. The equilibrium dissociation rate constants are given by the expressions (l)kD,eq = c
‘I+ (t)J’Jic= _(I lilt)
(36s)
i
and ,
(30’9
Q)k
D,eq
= c “‘Tj “‘Plc = -(l j(%jl) j
.
WV
W.L.Hogatth. D.L.S. Mc.Hwain/Atwo-state &atomic molecule. I
433
Wenote that cllrll>=txl~~~rll>+~Il”~rll~= 1 )
(37)
where (Il(‘)rll)=p~
(38a)
and (11(2)rll>=p~ .
(38b)
If the assumptionthat the observedand equilibrium dissociationrate constants are of the same order of magnitudeis also made (see Brau [3]), then (44) implies $ =%),
If we express the vector 11)as
p>O
(45a)
U>O,
(45b)
and Il)=cN;l(~)~;) P
(39
then it can be shown that (lI(‘)Lll>= -c
h;NG2
(4Oa)
and
(ilWli)=Cfp
=pl .
fl
(4Ob)
M3 =Oco),
where we also assumethat all the it, p> 0 are of the same order of magnitudeand a similarrelationship holds for all the $J,*,u> 0. However,in order to fulfil conditions (40b) and (42b), we require flo =pi’2 +6(G)
(46a)
and
Similarly,with
M*=p;n SO($).
lU=C~l’2’X~
(41)
V
Wb)
These conditions are consistent with (44).
we get
4. Dissociationand recombination (1I”‘_LI1)= - c $,*M,*2 Y and
(42a)
(11(2V11)=C@=p2.
(42b)
Y
By assumingthat AZ4 h;, $6 @XT,J/z < $i and q < JIi then we can introduce two parameters c = (x;/x;)llt
(43a)
and P = (51;/51;Y’2,
Mb)
such that ES 1 and p < 1. Physicallythis meanswe have assumedthat the internal relaxation time for both (l)A2 and f2)A, is much shorter than the dissociation time for either (‘)A, or (2)A2.In terms of E and p we can now show
We now turn to consider the full non-linearmaster equation (18) involvingboth dissociationand recombination. FollowingBrau [3], we split the distribution function IX(t)) into two parts as follows IX(t)) =v2(r)ll> + IZ(f)) ,
(47)
where it can be seen from (13) that l.?(f)) tends towards IO>as the system tends towards equilibrium, the elements of 10)all being zero. lZ(t)) is essentially a measure of how far the bound states and the atom concentration are from being in equilibriumwith one another. In terms of IZ(t))andy(f), wecan write the master equation (18) and conservationequation (17) as (L - r(a/dr))lZ> = (dr2/dr)rll>
(48)
and
(l)kD,eq~(l)$obs =G2 i- (l/e2) & and
@;/A;) Ni2
(44aj
respectively,where a tilde over a quantity denotes its initial value.
IV.1,.Hogarth, D.L.S. McElwainlA two-state diatomic molecule. i
434
We expand IZ(t), in terms of the eigenvectors of the matrix L as follows IZ(t)> = c n,(t)i’?tJ P with
f c b,(t)Px,) ”
,
a,(t) = w&lrlz(t))
60)
(5la)
and b Y(f) = Px,irlZ(r))
.
(5lb)
By taking account of the relationships established in section 3 we are able to write a; =eQu ?
jq =,y
$ $&)
x;=a,
N; = Eh;, ,
(52a)
cr>O,
(S2b)
and $; = P’$+J 7
hli = ,;I2 + p2& ,
!g=$,,
M, = pMv,
(SC) v>O,
(52d)
where the quantities A,, N,, for p > 0 and I$“, &I, for v> 0 are of order 1 in magnitude. Using the expression for IZ(r)) in (50) and the relationships (52) it is possible to show that the coefficients a,(t) and b,(t) must satisfy the master equations (d/dt) [no + p 112 1 J’2 1 = -c2 [X,a, + Nu(dy+!t)j
,(53a)
[(dldr) + +J uP = -eNP(dyZ/dr) ,
(53b)
(d/dr) [bu +p$“$]
p > 0,
= -~“[~~‘O~+M~(dy?ldt)l,
[(d/d?) f $,I b, = -pM,(dy*/dr)
,
v > 0,
5. Method of solution Equations (53) are symmetric in E and p so we assume, without loss of generality, that p < E. We shall only seek the solution of these equations for values of p such that 6(e2) < p < O(E).Since E < 1 we only carry out the perturbation analysis to include first order terms. Consequently, a term of order ep will be order e2 or smaller and will be neglected. Our approach to approximating the solution of these equations involves the consideration of two cases, the first where p =6($) and the second where 0(&-Q GO(e). 5.1. clsc 1: p =0(G)
5.1.1. t Time scale Since p = ai then p-* > e-2 so that if we consider a time scale of order less than es2 then we can neglect the right-hand sides of (53a), (53~) and (53d). If we also neglect terms of order e2 and p in the conservation equation (53e) then the system of equations we solve on this time scaie is (d/dr) (au +p;“y2)
= 0,
Pa)
[(dldt) + ?.,]a, = -eN,(dy*/dt) (dldt) (b. +
p;“v*)= 0 , v>o,
K$y
f 2y’ + 2pf’G0 t 2pj”b0+2e
X0 N&
(53d) = K$
+ 2j72+2p;‘2Zo t 2p;&-. t 2~ Pqo A’,&
(54e) We approximate the solution of these equations using the standard perturbation approach, with P(t)=a(0)(t)tfa(‘)(t)tO(E2) P P
9 pB0
a
0
WI
and (53~)
K,$y t 2~2 t 2(p~12,eZN0)a~t2(p:lr+p2~~)bo
eq
(54b) (54c)
Kdldt) + $,I b, = 0,
and the conservation equation
=1(1/2~t2p2t2(p~‘2t,2N
, Y> 0 ,
b u(r) = b”‘(r) t-&(t)(r)+6 V V
)a0
(e*)
,
vao,
>
W4 Wb)
and + Z(ppfl’ +p2Mo)b0
y(r) = y(O)(r) + ey(l)(r) t 6 (e2) _ The initial conditions are
(W
(55c)
435
WL. Hogarth, R.L.S. McEhain/A two-state diatondc nrolecule. I
b(‘)(O) =I5 6ro’ V Y
vao,
Wb)
and y(‘)(O)= j%ro -
(564
The approximation to the solution up to and including
rived from the expansionsin (57) by consideringthe limit in which t becomes large.The conditions are -1/2- c r?u=ii,-4epl 112Kcq y
p>o
$=O,
N z Ir P’
Wa)
cc>O,
(59b)
first order terms in e is i, = go -4~p:“K;;l*j~
“0 = Z,, - 2~pp:‘~yQ, Qp= zp exp (-1, r) ,
(574
D>O,
(57b)
b,=&+ep;‘2$Q, b, =&exp(-$vr),
(57c) v>o,
(574
and y=Y+eQ,
(57e)
6” =o,
(59c)
v>o,
(594
and P=Pf&K;;‘2
c N li . &I>0 p fi
We)
If we carry out standard perturbation on eqs. (58b) and (58d) up to and includingorder E then we obtain ap=O,
where
c N n” K>O Ir P’
Ir>O
GOa)
u>o,
GOb)
and b,=O,
The expression fory shows that there is little change in the atom concentration during this time phase. However,both aB, y > 0, and b,, v > 0, decay rapidly and this indicates that the internal distributions of (‘)A, and t2)A, are changingrapidly to new distributions._Therate of this decay in (*)A, depends on A,,,cc> 0 and in t2)A1on I,$, v > 0. 5.1.2. e2t Time scale Let us now consider the system of equations on a time scale of order em2. We do this by introducing a time variableT= e2t explicitly into eqs. (53). The equations obtained are (d/dr) (a, + p f12y2)= -Xunu - e2No(dv2/dr) , (58a) e2(du,/dr) + Su, = -e3rV,(dy2/dr),
p > 0,
and
t2e
c Npap+2p c N,b,=C.
0-J
V>Cl
(dldr) (a0 + #y’)
= -+,a0 :
(d/dr) (bO +p!12y’) = 0 , and
(58b)
(d/dr)(b0+~;‘2y2)=-~2e-2$obo-~2&,(d~2~d~), (58~) e*(db,/dr) + il;b, = -p e2M,(dv2/dr), v > 0, (58d)
K112y cq t2yZ+2(p :Rt&v&70f2@2
which, of course, is consistent with the initial conditions (59b) arid (59d). Thus, on this time scale, if we are consideringonly terms of order zero and one, we can neglect eqs. (58b) and (58d). The right-handside of (58~) is of order e2 since p = 0 (E’) and thus can be neglected.We also neglect the second order temr $No(dy2/dr) on the right-handside of (58a) as well as the terms ez and p in (58e). Consequently, on this time scale the behaviour of the system can be approximated by the equations
1’2+p*M,-,)b,
We)
= K$
t 2jj2 t 2p;i2Zo t 2p:“&, + 2~ c
CC,,
say.
00
N,&
(61~)
The initial conditions for these equations are givenin (59). Eliminatingou and b0 in (61) we have dy/d~=2p,X,~~~“[(C2/2pl)-(K,1112y/2pl)-y21, (62)
The initial conditions on these equations can be de-
where
W.L.Hogarth,A L.S. McEkvafn/Atwo-statediotomicmolecule.i
436
c2 = K$
+ 2~~92 + 2p:“a,
If
+ 2~ c N ii . p>o rl p
jJ > -(Kl’2/4p1) eq
This equation has two critical points
+A
(69)
then y1 = -(K$?4pl)
+ [($.,/lGp:)
+ (C2/2pl)l 1’2 (63a)
+DJ, coth( 2p1 ADK’!2Ar q
(70)
where
and y2 = -(K$/4pp,)
u=-(K,L~/4p,)+A
- IU‘&ll6~;)
+ (C2/2~l)1 1’2.
WI By consideration of the change of sign in (62) it can be shown that the negative critical point,yg, is unstable and the positive critical point,yl, is stable. It is essential from physical arguments that the positive critical point be stable sincey, the scaled atom concentration, must be non-negative at equilibrium. We consider the solution of eq. (62) as follows. If the fast time scale leads to a value of j in the range 0Gf-G(K$4p,)+A,
(64)
D, = coth-’
[(jj t K;f/4pt)/A].
(71)
In this case using (59c), (59e) and (61b), we obtain b. = go + p;‘2y2 - (Ke,p:‘2116p3 + (K;~p;‘2/2p&A
coth (2p,XoK;tAr
- pti2A2 coth* (2p,X,K;f 7
+ Dl)
AT f D ). E
(72)
Similarly, from (62c), (70) and (72) we obtain no
= -p’DA2 csch2(2pthoK;f 1
A7 + D E’)
(73)
+ (CJ~PJ”~,
(66a)
We see that in both regions u. changes markedly and tends to zero when T becomes large. This indicates that (l)A2 is approaching an equilibrium with the current atom concentration which is also changing significantly as is shown by the expression fory. From eqs. (60b) and (6 1b) we fmd that if we neggtect terms higher than the fast, then
BE = tar-K1 [@ f K$4pp1)/Al,
(66b)
(2)1vhl= p$2(bo t p$2)
then the solution of (62) is y = -(K$4p1)
+A tanh (2plXoK$4r
+ B,), (65)
where A = [(K,$~P;)
. with B, depending on E throughj. To obtain b. we solve (61b), using (59c) and (59e), giving b, = E0 +p;12Y2 - (Keqp;“l’6p;, + (K;_fp;/2/2pl)A
tanh (2plhoK;$4r
- p;/‘A tanh2(2pthoK,‘~Ar +-BE).
f B E)
SimilarIy solving (61c), using (65) and (67), yields an expression for fro, namely
A=‘1
+ BE).
(68)
(76a)
and c2 = Cl - 2t2% M’ Consequently (62) can be written as
rzo= pij2A2 sech2 (2p,h,KgAr
(74)
Hence the concentration of (2)A2 is constant, to our approximation, on this time scale. Let us now recast eq. (62) in terms of the concentrations tl)IVM, t2)NM and N*. We note for terms up to and including first order 2t1)NhI t 2t2)NM + N
(67)
=p;12(Eo t p$2).
OW
437
W.L. Hogarth, D.L.S. McEhvain/A two+tate diatomic molecule. I
d%$Jdt
= (‘)kR obSN; - “‘$, ohs ‘l)NM, , 9
(774
~*(cVQ&-,
+pl
l/2 2 Y I=-X,0,
- ~4N,(dy21dr41, (soa)
where e4(&Jdr4) (l)k
(8’Jb) (d/dQ (b. +p;“y*)
R
(774
,
Thus we see that (62) is, in fact, just the usual phenomenological rate equation for the dissociation or recombination of (‘) as if (*)A were not present. The rate constants 4(l kR,obs and2 (1)kD,cbs satisfy the Rate-
-(w;c~
c4(ti,ldr4)
= -(WOE* + o1 e9 t ... )M,(dy2/dt,),
K’l2y t 2y* t 2@
If we look at the overall reaction, we can write - kD,obs%l
(794
+
eq
t 2b;/2 t
+ 200w1e5 + ... )] b,
t M&;e4
2~ PFO Npal + Z(W,e’ + w1 e3 + -._) zoM,b,=C. (804
Note that since p = 6(e’) then we have expanded p as a function of E as follows
by
D,obs =.(l)kD,obs (L%f~NM’
v>o,
c*No,a,,
but althOUghkqobs iSjUSt (‘)kR,obs, kD,& is @en
k
(80~)
(god)
(‘)k
(7%
+ 200wl~ + ... )Ijlobo
f rCl$,
and
D,obsl(l)kR,obs = “‘%q
= -(w;
+ 2w0wl es f . .. )M,,(dy "/d$),
Quotient Law for such a reaction up to first order in E, namely
mM1dr =‘R,ob&
P>O,
C’b)
D,obs
and (l)k
+ hPar = -E5N,(dy’/dr,),
(7W p = WOE2+ f_$e3 i- ... .
a time and concentration dependent quantity. Here k D obs and kR obs do not satisfy the Rate-Quotient La&. ’ In region (64) there is an increase iny resulting from dissociation of (‘)A2 molecules whereas region (69) corresponds to a decrease iny resulting from recombination of atoms to form (I)A, molecules.
The initial conditions for these equations can be obtamed from our solutions for aP, p > 0, b,: v 2 0 and y on the previous time scale by considering the limit in which T becomes large. The conditions are ai = 0, ai=O,
51.3. e4t i?me scale Since the right-hand side of (58c) is 0(e2) on the time scale 0(e-*), it was neglected in the above analysis. Consequently these terms have not effected the solution up to a time which is as long as E-*. We therefore must consider a longer time scale and since the terms in (53~) whose effect has yet to be investigated are of order e4 then we examine the system of equations on a time scale of 0 (F4). This is done by introducing the variable f4 = e4t explicitly, into eqs. (53). In terms of this variable, the equations become
(81)
(824
b; = ii,, - &;‘*/p,) +(K,,/8pl) b;=O,
(82b)
Y>O,
[(K;@)
+ ~$i?~ v>o,
+ pi’*+, - (K,‘tAAI%
WC) (824
and y” = -(K,l$4@
+ A.
(824
ML. Hogarth, D.L.S. McElwainjA two-rtakdiatomic molecule. I
438
Carryingout standard perturbation up to and includingfirst order in Eon eqs. (SOa),(8Clb)and (80d) yields u. = 0, IIII = 0,
negativecritical pointy4 is unstable.Wewould expect this from physical arguments,since y, the scaledatom concentration, should be non-negative.Althougheq. (85b) can be solvedin the regions
(83a) P>O
(83b)
0
and &=O,
V>O,
(83c)
which is againconsistent with the initial conditions. ConsequentIyon this time scalewe neglect eqs. (gOa), (80b) and (sod). If we only considerzero and first order terms in Ein the two remaining equations then the system of equations reduces to
(d/dr4) (bO+ pi12y2) = -(c$ + 2oowt e)JI&,, (84a) and K$y + 2yl+ Zp:/26u = ct.
(84’4
Eliminatingb0 from these two equationsyields
+2w0wte] [(5/Z) - (K$$2)
-y2]. (8Sa)
we find that the solutions involvey only implicitly and for this reason we do not present them here. Oncey is found, b, can be obtained from (84b). Wehave already pointed out that at the end of the c2t time scale the bound states in (l)A2 are in equihbrium with the instantaneousatom concentration and this is indicated by u,, = 0 up to first order in e on this time scale. Since the atom concentration, represented by y, is changingduring this time scalethen the populations of the bound levelsof (‘)A, are also changing. Eqs. (84) indicate that b,-,wiU change considerablyduring this phase and this impliesthat the @)A, levelsare approachingan equilibriumwith the atom concentration. Ify+ is in the region (87a) theny increasesdue to dissociationof (2)% and ify+ is in the region (87b) (2)A2. Eq. (85b) rewritten in terms of the concentrations NhI and N* becomes
This equation may be rewritten as
@$tldf = kR,,b&$ - kD,obsNM’
(@f + 4~5~) (dyldt)
where
=2pT~~[(C,/2)-(~~~y/2)-y~].
(8Sb)
This equation has two critical points, namely y3 = -(@4)
+ [(K,J16) + (C,/2)1 1’2,
k ,,@s = !@
+ 4p,N,4K;;),
k R&s = d&@
(86a)
+ 4plN&;)*
(SW
Wb) WC)
Thus we have obtained the usual phenomenological
rate equation as we did on the e’)t time scale. The rate coefficients satisfy the Rate-Quotient Law
and y4 = -(K”2/4) - [(K 116)+ (C,/2)] 1’2. eq eq
(87b)
then y is decreasing due to recombination involving
(K”2 eq + 4~ty) (dy/dt4) =26&J;
y+ > -(K1’2/4) + [(K 116)+ (C1/2)] U2’ eq eq
(86b)
By consideration of the changeof sign of the derivativeofy at the critical points it is possibleto show that the positive critical pointyj is stable and that the
kD,obslkR,obs = %q
(89)
and we note their dependenceon the atom concentration, NA. In summary,if p =6 (e2) there are three time phases
W.L.Hogarth, D.L.S. McElwainlA two-state diatomic molecule. I
and
in the problem which can be represented by the time variablesr, e2t and e4t. We shall further discussthe physical interpretations of these phasesin section 6.
y=y+&tpH,
5.2. Case2: C?(Z) < p < 0(e)
where
5.2.1. Fast time scale Weagainconsider the e s. (53) and first investigate a time scaleless than0 (e- 9 ). As before we neglect the right-handsidesof (53a) and (53~). However,in this case, we must retain the right-handside of (53d). If we neglect terms of 6 (e2) in (53e), then the relaxation is describedby the followingset of equations on this time scale (d/at) (no + ~;‘~y~) = 0f
(904
[(d/dt) + A,] p; = 4’(dy2/dr),
P>O,
(d/dr) (bu +~;‘~y~) = 0,
(9W
(9W
[@ldt) + $,I b, = --pM,(dy2/W,
u> 0,
(9Od)
and K,1/92y + 2y2 + 2~;‘~ao f 2p;‘2b,,
+2e
EN;; 00
=C3,
+2p CM& fi p 00
(904
say.
Weapproximate the solution of these equations up to and includingfirst order in Eand p by standard perturbation technique, first expandingin E and then in p. This process yields
439
w-l
(9W Wenote that these results are similarto those given on the t time scale for the case where p = S(E’), with aO, b, and y changingonly slightly and 4, p > 0, b,, v > 0, decaying rapidly. 12.2. Slow time scale Wenow return to eqs. (53) and consider time as long 2s 6 (C2). Under this condition we can neglect eqs. (53b) and (53d) since aP and b, willbe zero to first order in e and p for n > 0 and V> 0. This is similar to the ~?r time scale in the previous case where we neglected eqs. (58b) and (58d). In the remainingthree equations we againneglect the second order terms e2No(dy2/dr) in (53a) and p2M (dy2/dt) in (53~) but here we must include the term p9 Gobo. If we also disregard the effects of the second order terms in Eand p in the conservation equation (53e) then on this time scale we wish to examine the followingsystem of equations = -e2h 0a0’ (d/dt*) (a0 t p’12y2) 1
(924
(d/dt*) (ba t ~“~y~) = -p2$ 0 b0’ 2
(92b)
and Ki(12yf 2y2 + 2p:12ao -t 2pp:/*b,,= C,,
ww
a,--=Zo- 2pi’2ji(eG + pH),
(914
aK=ZP exp(-$r),
(91b)
where t* is used to denote that the time is longer than that represented by the fast time scale. The initial conditions to be imposed on these equations can be deduced from (91) by consideringthe limit in which c becomes large.These conditions are
(914
(93a)
(91.d)
W)
b, =&,-
P>O,
2p;j2F(Ec tpH),
b, = 5, exp(-$J),
u>o,
W.L.Hogarth,D.L.S. McElwainjAtwo-state diatomic molecule. I
440
That eq. (94a) can be used to follow the relaxation to equilibriumcan be seen’fromthe followingargument. From eqs. (53) we obtain, for times t 2 6(eS2)
and y+q
+d+pJ,
cP = -eN,&‘(dy2/dr)
where
J=2K,-,1j2
&fg.
v>o “”
By systematicallyeliminatingthe variablesno and b, from eqs. (92) it is possibleto obtain the following equation for y (d2y/dt*2; + [4(pIe2X0+ p,p2$o)K;,‘/2y + e2Xot p2Jlo] (dy/dt*) - &rrt12e2h0p2$o(C~- Kity - 2y2) = 0, (94a) with 6, and a0 then beinggivenby b. = [Z@dy/dt*) - ezXo(C3- K$y - 2y2)] /2p”2do2$u - e2X0 ) 2
Wb)
and a0 = (C3 -
K$y - 2y2 - 2#2bo)/2#2.
(94c)
To solve(94a) we not only require an initialcondition for y but also an initial condition for dy/dt*. To obtain this we first differentiate (92c) and use (92a) and (92b). The resultingequation is Kg2(dy/dt*) = 2pf/2e2hoao + 2p;“p’$,,b,,.
(95)
Sincethe initial valuesof a0 and b, are givenby (93a) and (93b) then the initial condition for dy/dt* can be found by substitutingthese expressionsin (95). The initial condition is thus
+6@),
P> 0,
(97)
and a simiiarexpressionfor b,, u > 0. Thus as long as eq. (94h) yields ay(t) which approachesits equilibrium valuein a regularway the contribution of a,,, j.r> 0 and b,, v > 0 is negligibleand eq. (94a) remainsvalid. Thus, in consideringtime as long asO(~-~) we find that the problem of solvinga system of first order differential equations and one nonlinear equation canessen tially be reduced to solvingone second order differential equation (94a). Weshah, in a succeedingpaper to be called part II 181,solveeq. (94a) numericallyusing the Runge-Kutta method to obtain valuesofy and dy/dt* at different times. The quantitiesa0 and b0 can then be determined usingeqs. (94c) and (94b), respectively. Duringthis time phasey, go and b, are,allchanging markedly.Wesee that unlike the case where p = 15@), (‘)A2 is unable to establishan equilibriumwith the atom concentration before (2)A2beginsto either dissociate or recombine. In summary,we believethat if6 (e2) < p d 6(e), then relaxation takes place in two time phases.The first, a time phase of order lessthan ce2, ‘involvesinternal relaxation, and the second, a slowertime phase at least of order e02, during which the concentrations of tl$ and t2)A2approach their fmal equilibrium values.In this case, it is not possibleto describethe reaction in terms of a phenomenologicalrate equation and thus there are no well-defmedconstants.
6. Discussionand comments Before discussingthe physicalaspectsof the results obtained by the boundary layer approachwe make some commentson the perturbation procedure used in this paper. In neither of the casesconsidered,p = O(&) or O(e2) < p G 6(e), da we givea composite expansionfor the solution to cover the whole time range.The reason for this is that we were unable to obtain an explicit expressionfor y on severalof the time phases. In the first case,wherep 4 (e2),onewouldex-
ML. Hogarth, D.L.S. Mc&lwain/A two-state diatomic molecule. I
perturbation e2N (dy2/dr) in (58b) on the time scaleO(E-~) to have
pect the
441
secondstate then the first state dissociateson this time phase. The rate constants for the reaction as a whole can be related to the rate constants for the reaction of the second state. Henceover the whale time rangethe rate constants for the overallreaction willchangequite markedly. For the case0 (e2) < p G U(e) there are two time phases.Duringthe secondphase,one of the states beginsto dissociateor recombine,but before it can establish an equilibriumwith the atom concentrationthe secondstate commencesto dissociateor recombine. The rate constantsfor this caseare not welldefined.
dy/dr = X K-“2(C, - K’12y- 2~~). 0 eq eq This is completelyequivalentto the casepreviously consideredby the authors [6] for the dissociationof a diatomicmoleculewith only one state. Physicallyour resultsare interpreted as showing that there is a period of internal relaxation for both of the states with very little dissociationor recombination, This occurswitbin the time phaselessthan 0(eS2) For p = C,(e2) the solution displaysthree time phases.The second phasecorrespondsto the dissociation or recombinationof the first state whilethere is little reaction for the secondstate. The first state sets up an equilibriumwith the atom concentration. During this phase the rate constantsfor the system as a whole can be tepresentedin terms of the rate constants for the reaction of the first state. The third time phase involvesthe dissociationor recombinationof the second state in order to reach an equilibriumwith the atom concentration. If dissociationoccurs then the fmt state recombines,but this is very rapid on this time scaleand the first state maintainsits equilibriumwith the atom concentration.Similarly,if recombinationoccursin the
Acknowledgement One of the authors (W.L.H.)wishesto expresshis appreciationfor the support givenby a Commonwealth PostgraduateResearchAward.Wewould also like to thank the referee for someuseful commentson section5.
References [l] J.E. Dove and D.G. Jones,Chem. Phys. Letters 17 (1972) 134. [2] RE. Roberts, J. Chem. Phys. 54 (1971) 1422. [3] CA. Brau. I. Chem. Phys.47 (1967) 1153. [4] J.D. Cole, Perturbation methods in applied mathematics (Blaisdell,\Valtman, hlassachusetts, 1968). [S] AH. Nayfeh, Perturbation methods Wiley-Interscience, New York, 1973). [6] W.L. Hogarth and D.L.S. McElwain,J. Chem. Phys. 63 (1975) 2502. [ 71 1V.L.Hogarth and D.L.S. McElwain,Proc. Roy. Sot. 345 (1975) 251. [S] W.L. Hogarth, Chem. Phys. 19 (1977) 443.