fYou have the reaction A = 2B

The problem is not well posed. You are considering not one reaction but two different reactions, a first order one and a second order one. Your equation has not an equilibrium solution (from thermodynamic point of view), but you are finding a quasi-steady-state solution when A > B is the formation of the B species and 2B>2A is the distruction.

The detailed balance principle is valid for each reaction separately, therefore

reaction    rate  dir           eq                 rate rev

A > B           K1d             Keq           K1r=K1d/Keq         

2A > 2B      K2d              Keq^2     K2r=K2d/Keq^2

Both reactions bring to the same equilibrium state.

If in your kinetic scheme K1r

Dear Prof. Gelatii, 

No, it is the reaction A = B, or equivalently, 2A = 2B. This is when one writes it as a chemical equilibrium (and not as a kinetic equation). This chemical reaction may have any of the following three mechanisms (written as kinetic equations):

a) both direct and reverse reactions are of the first order, i.e. A -k1-> B   and   B -k2->A;

b) both direct and reverse are of the second order, i.e. 2A -k1-> 2B   and   2B -k2-> 2A;

c) one is of the first order and the other is of the second order as Eq 2 and 3 above.

Whatever the mechanism, the reaction is not A = 2B. One can check that a) and b) yield equilibrium condition Beq/Aeq = K, where K is either k1/k2 or sqrt(k1/k2). If the two reactions are of different order, we get a different equilibrium condition.

Dear Prof. Colonna,

Yes, you are correct. There must be 2 equilibrium reactions (or 4 kinetic processes, two direct and two reverse). A consequence of the general mechanical principle for reversibility of time is that if A -> B is possible, then B -> A must also be possible (even if 2B->2A is a much faster process). The equilibrium condition Beq/Aeq = K is only valid at infinite dilution. But at infinite dilution, kr1*B will inevitably become larger than kr2*B^2, which indeed means that the problem is not well posed.

However, I disagree that B^2/A = const is only a stationary and not equilibrium state. At conditions at which 2B -> 2A is faster than B -> A, the solution or the gas must be strongly non-ideal. The equilibrium thermodynamics then yield the more general condition gammaB*Beq/gammaA*Aeq = Keq, where gammaA and gammaB are the respective activity coefficients. And then it turns out that the activity coefficients can be related to the mechanism of the reaction, which is, in fact, again a principal question - all equilibrium thermodynamics can be derived from kinetics. 

For the mechanism A -> B & B->A, one can calculate k2 from k1 and K, i.e., equilibrium thermodynamics (standard chemical potentials) can be used to find a kinetic constant. If there are four processes (A -kd1-> B, B-kr1->A, 2A-kd2->2B and 2B-kr2->2A), it turns out that one  can calculate kr1 from the value of K AND kr2 from the series expansion of the activity coefficients (more precisely, from the second virial coefficients of the mixture of A and B), which also means that kd2 and kr2 contain the information for the binary interactions...

So thank you for the answer.

@Radomir

There is something I cannot get from your comment. First of all, why the possibility of having 2B > 2A faster than B > A lead to the conclusion that the gas is not ideal? In my experience it often happens that the direct and reverse processes follow different path also in ideal gas, leading to quasi-steady solution. Moreover in non-ideal gas, not only the activity  of the species, but also the equilibrium constant should change, depending on the composition, because the mixture virial coefficients are a function of the particle density of each species.  

Secondly, I do not agree that thermodynamic equilibrium always comes out from kinetics. Diamond-grafite transition is not observed. Thermodynamic equilibrium (at least for gases), starting from basic principles of statistical physics, tells what is the most probable state for the statistic. If such a state can be reached is a different problem that can depend on the nature of the interaction between particles. If there is no open channel between the actual state and the equilibrium one, the transition could not happen. 

However, I have experience in plasmas, with small molecules and probably with large organic molecules things can be different. Do you have a practical example of the case you are proposing in which both thermodynamic and kinetic equilibrium is B^2/A and not B/A?

Dear Radomir, 

I'm sorry but you are not right. For the reaction mechanism you MUST write a correct elementary reaction. If you write  A B, this means that both forward and reverse reactions are unimolecular. In you analysis you actually suggest the mechanism 

A B

B+ B A, which does not make any sense. 

Dear Gianpiero,

I completely disagree that: "Secondly, I do not agree that thermodynamic equilibrium always comes out from kinetics." For all elementary reaction it must be K = k1/k(-1)

Dear Yurii

In some sense you are right. The problem is what happens if K1=0? or K1~0?

You will never reach the equilibrium configuration.

The stable configuration in normal condition of carbon is graphite. Why diamond does not reach its equilibrium configuration? Why methene-air mixture do not burn if you do not increase the temperature? When I thing that the kinetic and equilibrium can give different results I mean exactly this.

Another problem is that the equation you write start from an assumption that for a given reaction you can apply the detailed balance principle. That is a "principle" and therefore it is not demonstrated. In binary reactions, the detailed balance principle is derived from the micro-reversibility principle, that is not demonstrated neither. For small molecules, considering spherical symmetry in the  collisions it is considered valid, but for large molecules, where the collision geometry is not isotropic, this principle is not valid (Huang, Statistical Mechanics). We use to say that because an equilibrium state exists, it should be the thermodynamic one and therefore the detailed balance is valid. But as far as i know, there is no general demonstration of its validity.

Dear Gianpiero,

We are discussing the problem which is beyond of the original question. The thermodynamics and kinetics are linked by a simple eq:   K = k1/k(-1). This does not mean that the equilibrium is always achievable (you absolutely right). All organic materials around us (including  human beings) are thermodynamically unstable but stable kinetically, in your words k1 ~0.  In any case (deltaG of the reaction)/RT = -Ln(K equilibrium).  

When you write whatever kinetic scheme,  you MUST have a mass balance. "where the collision geometry is not isotropic." This is described by the entropy of activation. 

Dear Prof. Geletii,

"In you analysis you actually suggest the mechanism

A B

B+ B A"

No. I suggest the mechanism

A -> B

B+B -> A+A     (note the absence of

Dear Gianpiero,

"First of all, why the possibility of having 2B > 2A faster than B > A lead to the conclusion that the gas is not ideal?"

The technical answer: it yields equilibrium condition B^2/A = K, which is possible only if the activity coefficients differ from one (non-ideality).

The physical answer: if both B -> A and 2B -> 2A are possible, then the second reaction involves the interaction potential between two molecules B (the kinetic constant will be higher if two B interact with an attractive potential and will be lower if they repulse). Interaction means non-ideality.

"In my experience it often happens that the direct and reverse processes follow different path also in ideal gas, leading to quasi-steady solution."

I would be really interested to see an example. But for a reversible process AB in ideal gas, Beq/Aeq = K is an exact result. Either the mechanism you speak about must yield that, or the gas is not ideal.  

"Moreover in non-ideal gas, not only the activity of the species, but also the equilibrium constant should change, depending on the composition, because the mixture virial coefficients are a function of the particle density of each species."

No, the equilibrium constant in a gas is independent on composition and don't involve virial coefficients, by definition. Both factors affect the equilibrium composition only through the activity coefficients. The constant of the process A = B is defined as K = exp( mu0[A]/kT - mu0[B]/kT ), where the standard potentials mu0[A] and mu0[b] are characteristics of a single molecule interacting with nothing. However, equilibrium condition reads K = a[B]/a[A], where a is activity and involves non-ideality.

"Secondly, I do not agree that thermodynamic equilibrium always comes out from kinetics. Diamond-grafite transition is not observed... If there is no open channel between the actual state and the equilibrium one, the transition could not happen."

I totally agree with that statement, but in the considered case you have that open channel. You have a path which can bring you from A to B, and a different path which can bring you from B to A, and both processes are of measurable velocity. I think that A and B must then reach thermodynamic equilibrium, whatever the two processes are. But this is an interesting point and I might change my mind. Can you give an example which violates my statement?

"Do you have a practical example of the case you are proposing in which both thermodynamic and kinetic equilibrium is B^2/A and not B/A?"

The process A = B in concentrated solutions will of course deviate from B/A = K, due to activity. I can think of an example which yields nearly B^2/A = K, but it will be perhaps artificial. A better example - I once checked that experimental data for a process of the type A = B can be modeled successfully by a scheme involving two kinetic equations, A  B and 2A 2B, to account for the deviations from ideality and deviation from the simple mass action law with the increase of concentration. The same can be done with bimolecular processes (in dilute system only 2A2B matters, in more concentrated 3A 3B starts to play a role) and others of the same kind.

Dear Radomir,

Now your reaction mechanism is written correctly

A -> B

B+B -> A+A

What is wrong?

"For this mechanism Consider the equilibrium process A B. Let its equilibrium constant be K; the condition for equilibrium is

Beq/Aeq = K, (1)

where Aeq and Beq are the concentrations at equilibrium. Let us further imagine that the forward process is following a first-order kinetic equation:

A -k1-> B, mass action law v1 = k1*A, (2)

NO!

d[A]/dt = -k1[A] + k2[B]^2

but the reverse process follows a second-order mechanism,

2B -k2-> 2A, mass action law v2 = k2*B^2. (3)

NO!

d[B]/dt = -k2[B]^2 + k1[A]

The velocity of the equilibrium reaction is v = v1 - v2, and in equilibrium it must reach zero, so that

k1*Aeq = k2*Beq^2, Yes!!!   k1[A] = k2[B]^2

[A]o = [A] + [B]

i.e., we have the equilibrium condition

Beq^2/Aeq = k1/k2. (4) Yes! k1/k2 = (([A]o -[A])^2}/[A]

Thus, your first statement is wrong:

"Consider the equilibrium process A B. Let its equilibrium constant be K; the condition for equilibrium is

Beq/Aeq = K, (1)" NO!! K = (([A]o -[A])^2}/[A]

Dear Yurii,

“NO!

d[A]/dt = -k1[A] + k2[B]^2”

Yes,precisely! d[A]/dt = -v1 + v2 = -v. It is incorrect to call d[A]/dt “velocity of the direct process”. v1 is the velocity of the direct process.

“NO!

d[B]/dt = -k2[B]^2 + k1[A]”

Yes, precisely! d[B]/dt = -v2 + v1 = +v. It is incorrect to call d[B]/dt “velocity of the reverse process”. v2 is the velocity of the reverse process. So we have no disagreements here except for terminology.

“Thus, your first statement is wrong:

...Consider the equilibrium process A B. Let its equilibrium constant be K; the condition for equilibrium is Beq/Aeq = K...

NO!! K = (([A]o -[A])^2}/[A]”

Yes, k1/k2 = B^2/A, according to the kinetic scheme. You disagree that the kinetic scheme A->B & 2B->2A corresponds to the equilibrium process A=B. But it does. The condition for chemical equilibrium between A and B is

 muA = muB

where mu is the respective chemical potential. This follows from the stoichiometry of the process and is exact irrespectively of the kinetic mechanism (except if Prof. Colonna is correct about equilibrium and stationarity). The condition for equilibrium is not muA = 2muB. So

   muA0 + kT ln(a[A]) = muB0 + kT ln(a(B))

where a is activity, which yields

   a(B)/a(A) = K .

If a(B) = B^2 and a(A) = A, or if a(B) = B^(3/2) and a(A) = A^(3/4) or similar, then there is no contradiction between the thermodynamic equilibrium condition and the kinetic stationarity condition. But this is possible only in concentrated solutions. At infinite dilution, a(A) = [A] and a(B) = [B] so [B]/[A] = K, which is undoubtedly correct and which is in a contradiction with kinetics. And the solution is the microreversibility principle / basic mechanical equations independence of direction of time, as discussed in the posts above. If there is a process A->B, then inevitably there must exist a monomolecular process B->A parallel to 2B->2A. At some conditions one can possibly neglect B->A, but at low concentrations B->A inevitably will become faster than 2B->2A. Thus, the thermodynamic equilibrium condition at infinite dilution holds.

Dear Radomir,

There are numerous semantic problems in chemical thermodynamics and kinetics. The IUPAC Gold book definitions are very confusing and often wrong.

First we have to write dawn the overall reaction. Commonly if delta(G) of the reaction is large, we write the reaction as irreversible. If delta(G) of the reaction is small, we write the reaction as reversible. The reaction delta(G) is independent on the reaction pathway. When you apply the kinetics to describe the process, you should write the reaction mechanism. Different reaction mechanisms result in different reaction rate laws, which include certain reaction rate constants. Therefore, the equation for the equilibrium constant via the reaction rate constants depends on the kinetic model. If you assume that

A -> B, k1

B+B -> A+A, k2

you have k1/k2 = {([A]o -[A])^2}/[A]

If you write A -> B, k1

B -> A, k’2

you have k1/k’2 = ([A]o -[A])/[A].

In this scheme k2s are different, the second and first orders.

k2 = k’2 ([A]o -[A]) and you always have a(B)/a(A) = K . I do not see any problem (riddle) in linking kinetics and thermodynamics. These links depend on the reaction mechanism

Just small comments

“d[A]/dt = -v1 + v2 = -v. It is incorrect to call d[A]/dt “velocity of the direct process”. v1 is the velocity of the direct process.”

I do not call d[A]/dt as the rate of the direct process. I do not understand what does “the rate of the direct process” mean. I wrote kinetic equations for consumption/formation of A and B based on the mass action law.

“the forward process is following a first-order kinetic equation”

“the reverse process follows a second-order mechanism”

I guess that it would be correctly to say that the reactions are unimolecular and bimolecular.

Dear Radomir,

in my opinion your assumptions simply violate the detailed balance (see: http://en.wikipedia.org/wiki/Detailed_balance). So no riddle is in your question.

Dear Radomir

Thermodynamics is very clear on this. If you have an equilibrium between a species A and a species B than from the fact that the Gibbs reaction energy vanishes in equilibrium it follows that the ratio of the mole fractions (concentrations) of the species are related by the equilibrium constant; the latter is related to the standard chemical potentials of the species. This is completely independent of the kinetic pathways by which the reactions proceed.

What confuses you is that only in the case where there is a reversible reaction between these species and the forward and backward reactions are both first order then the above equilibrium constant is equal to the ratio of forward and backward reaction rate coefficients.

It so turns out in your case that the backward reaction is much more complex than first order. Then this simple relation does not hold as amply exposed above.

Ger

To elaborate Fabio's argument: If there is a reaction A->B, it is in equilibrium with B->A; If there is a reaction 2B->2A, then there is also 2A->2B. The overall equilibrium then is a combination of the two equilibriums: k1A+k2A^a=k-1B+k-2B^2, where the four k are not independent.