When a chemical thermodynamic analysis is associated aside with kinetics it helps to express the rates in terms of activities. Such arises when doing non-equilibrium thermodynamics: it establishes more clearly the association of rate and driving force (chemical potential difference).  It would only be an error to use the same rate coefficients as when concentrations are used.

You raise the interesting question. My brief answer is to use activities. All reactions are reversible. "Irreversible"  means that the equilibrium is significantly shifted to one side. The transition state theory in chem. kinetics is actually based on thermodynamic. I also give an example which clarifies the answer. Consider equilibrium: 

A -> B with k(f)

B -> A with k(b)

In equilibrium k(f){A} = k(b){B} or  k(f)[A]= k(b)[B] ?? what is right?

The equilibrium constant K = {A}/{B} or K = [A]/[B]?

Evidently that K = {A}/{B} and d[A]/dt = -k(f){A}

Wow, I'll be interested to follow this discussion! When a God of chemical engineering like Danckwerts says something, we should listen. But my instinct is to agree with the two answers above, so why did Danckwerts say what he said? I have no idea... To be continued, I hope :-).

Thanks so far, I've thought some more about this. I agree with Yurii, but equilibrium can also be ensured by modifying the rate constants appropriately. 

Thinking about activity coefficients (g), they essentially cause an increase/decrease in chemical potential energy (u):

u = u0 + RT ln {A}

   = u0 + RT ln (g) + RT ln [A]

   = u0 + /\u + RT ln [A]

Where /\u = RT ln(g), so g = exp( /\u /RT)

Now the rate equation kind of accounts for energy with the rate constant, and collision frequency in the concentration term (please help me out here). If I use activities and the Arrhenius equation:

r = k{A}

   = k0 * exp(-E/RT) * g * [A]

   = k0 * exp(-E/RT) * exp( /\u /RT) * [A]

   = k0 * exp(-(E - /\u )/RT) * [A]

So by using activities, you essentially reduce the activation energy required by the same amount as the chemical potential energy is changed by the activity coefficient. Does this make sense?

Let’s make things simple. First, we have to define the reaction rate. We are counting the number of reacted species A in the react A -> B. Therefore the rate = d[A]/dt. The reaction rate law can be written as d[A]/dt = -k’[A] or d[A]/dt= -k{A}. The second expression is theoretically correct and suggest that k is constant. However, you can’t integrate this equation. Therefore, we should write d[A]/dt = -k g [A], where g is the activity coefficient of A and k’ = kg. Finally you have d[A]/dt = -kg[A] and d[A]/[A] = -kg dt.

As a result, experimentally you are dealing with kg, which is dependent on experimental conditions (ionic strength, solvent, etc.). Therefore, Dankwerts’ statement is intrinsically contradictive. The expression d[A]/dt= -k{A} is correct but practically useless. d[A]/dt= -k{A} can’t be replaced by d[A]/dt = -k[A] as claimed by Dankwerts.

Again, thank you for the input. After even further reading, I've come upon the Bronsted-Bjerrum equation, which uses transition state theory:

[A]+[B] [AB]* -> [P]

Where [AB]* is the activated complex, in quasi-equlibrium with reactants [A] and [B] so that K* = {AB}*/{A}{B}.. The reaction rate is:

r = k*[AB]* = k*K*(gA)(gB)/(gAB*)[A][B] = k (gA)(gB)/(gAB*) [A][B]

In this equation, "k" is dependent on temperature but independent of composition, and gA, gB and gAB* are the activity coefficients of reactants A, B, and the activated complex. The difference between the Bronsted-Bjerrum equation and the standard kinetic expression with activities instead of rates is the inclusion of the activity coefficient of the activated complex.

For elementary reactions, is this appropriate?

This problem discused  very intensive in non ideal plasma. I can send your some papers if you send your E-mail. My E-mail : [email protected].

Sorry for english.

Alexander Khomkin

Dear Tobias,

If you want to link kinetics and thermodynamics, you should write

[A]+[B] [AB]* [P]

The classic thermodynamics is applicable only for equilibria.

Regardless of theory, you should consider the reaction

[A]+[B] [P].

If you write [A]+[B] -> [P], then delta(G) of the reaction is equal to infinity. The “thermodynamics of irreversible systems” deals with kinetically stable systems and by definition includes kinetics. The term “activity” belongs to the classic thermodynamics

Dear Yannick,

I never heard about T. De Donder theory. It’s related to the “thermodynamics of irreversible systems.”

The form suggested by classic transition-state theory (going back to Eyring, 1935, J. Chem. Phys. 3, 107-115) indicates that one should use activities, not concentrations. That is what we do (largely) in my field of aqueous geochemistry. In simulating the interaction of aqueous solution with minerals, it is important that the rate equations be consistent with equilibrium as an end point (that is, the rate goes to zero as equilibrium is approached). That is doubtless important in other areas as well. Nevertheless, rate laws written in terms of concentration are common. By the way, de Donder seems to have been the purveyor of the affinity concept, although the notion of activities predates that. Since Tobias Louw brought up the activity coefficient of the activated complex, it is common belief (as stated in textbooks and such) that the rate of the decomposition of the activated complex to products is proportional to the concentration, not the activity of the the activated complex. This particular issue is considered "settled science" owing to analysis of experiments involving aqueous ions, particularly ones of high charge number. A specific reference to this topic is given by Laidler, 1965, Chemical Kinetics, 2nd edition, McGraw -Hill, New York. See pages 221-223, especially Fig. 50. This is regarded as the "proof" in favor of concentration. I can tell you that the impression of good fit is in part due to not showing all the actual data points given by the cited sources of the data. In defense of Laidler and many others who present similar plots, this is likely done only to achieve clarity for pedagogical purposes. A key point is that the same slopes result from either theory (concentration or activity). Significant difference in slope would only result from reactions involving highly charged species of the same charge sign. So perhaps the analysis in favor of concentration is wrong. Perhaps the people in the plasma world would weigh in? 

@Yannick: I agree we cannot determine Tau without calculations, and using activities with a value for Tau which was experimentally determined for use with concentrations is incorrect. However, Tau will depend on the non-ideality of the solution. I'm looking for an expression which will yield an (experimentally obtained) Tau which is independent of composition.

@Yurii: I am interested in the reversible reaction. In transition state theory, one of the assumptions is that the reaction only proceeds in one direction from the activated state, thus my expression [AB]*->[P]. I also believe that the concept of activities is not exclusive to classical thermodynamics, but it has a solid foundation in statistical mechanics. It can and definitely should be applied to non-equilibrium systems.

@Thomas: I completely agree that this should be "settled science", that's why I'm surprised to find so many different opinions on the subject :). On the point of end-point equilibrium: let's use the reaction [A]+[B] [AB]* [P]. Just substituting activities in place of concentrations would yield the following:

d[A]/dt = kf{A}{B} - kr{P} = kf(gA)(gB)[A][B] - kr(gP)[P]

Where gi are the activity coefficients. On the other hand, if we use transition state theory (TST), we should use:

d[A]/dt = kf(gA)(gB)/(gAB*) [A][B] - kr(gP)/(gAB*) [P]

At equilibrium, the activity coefficient of the activated complex will cancel, so both expressions result in the same end-point equilibrium.

However, (gAB*) will affect the actual rates. If we follow TST, then it does become a question, just as Thomas said, of whether we believe the rate is proportional to the activity or the concentration of the activated state. According to TST, the rate of product formation (rp) is given by:

rp = k[AB]*

In this expression, "k = Kappa*v" where "v" is the frequency of vibration of the activated state along the reaction coordinate and Kappa accounts for the fact that every oscillation doesn't result in product formation. If Kappa is independent of the solution, then the use of concentrations in "rp = k[AB]*" is appropriate, since we are dealing with actual amounts, as opposed to the "apparent" amounts which are represented by activities. So, is there any reason to believe that Kappa will be dependent on the chemical potential energy of the activated state?

Dear Colleagues,

Let’s look at the question from the practical point of view.

No doubt that the rate is measured as d[C]/dt (in concentrations). The reaction rate law must be written in concentrations, otherwise you will be unable to simulate the kinetics. In this case the reaction rate constants are not constant and depend on ionic strength, solvent, etc. Now we want to link the kinetics to the classic thermodynamics. In the classic thermodynamics the deviations from ideal systems are described by introducing the activity coefficients. In classic thermodynamics all reactions are reversible, otherwise you have deltaG = infinity. All thermodynamic equations are written using activities, which are g[C]. The kinetics and thermodynamics are linked to each other through equilibrium constants. Therefore the reaction rate laws expressed in [C] include the activity coefficients, the rate constant k for the ideal system becomes gk.

Thomas correctly wrote that “The form suggested by classic transition-state theory (going back to Eyring, 1935, J. Chem. Phys. 3, 107-115) indicates that one should use activities, not concentrations.” At the same time “It is common belief (as stated in textbooks and such) that the rate of the decomposition of the activated complex to products is proportional to the concentration, not the activity of the activated complex.” I agree with Thomas that “So perhaps the analysis in favor of concentration is wrong.” Here are my thoughts.

I mentioned several times that all reactions should be written as reversible, but if the equilibrium is shifted to the right hand side, we simplify kinetic analysis assuming that the rate of reverse reaction is too slow. Nevertheless, for the accurate analysis all reactions should be written as reversible

The Tobias equation d[A]/dt = kf(gA)(gB)/(gAB*) [A][B] - kr(gP)/(gAB*) [P] is not correct

It should be d[A]/dt = -kf(gA)(gb)[A][B] + kb [AB*] and

d[AB*]/dt = kf(gA)(gb)[A][B] – (kb - kp)[AB*] + k-p(gP)[P],

If the steady-state is applicable, then [AB*] = ( kf(gA)(gb)[A][B] + k-p(gP)[P] )/ (kb -kp) and

d[A]/dt = -kf(gA)(gb)[A][B] + ( kb /(kb -kp) ) ( kf(gA)(gb)[A][B] + k-p(gP)[P] ) = ( kp /(kb-kp) ) ( kf(gA)(gb)[A][B] + k-p(gP)[P] )

Thus, an additional term ( kp /(kb-kp) ) appears in the rate law.

At equilibrium d[A]/dt =0, and [P]/[A][B] = ( kp /(kb-kp) ) ( kf(gA)(gb)/ k-p(gP) If the reaction is thermodynamically very favorable, kb