Yes. You have written the kinetic expression in a form that will guarantee an approach to equilibrium as t approaches infinity because the term in parenthesis goes to 0 as residence time approaches infinity. The amount of time it take to approach infinity could vary greatly depending on the value of k2.

There are commercial units (Tatoray) that perform the opposite of this reaction to produce para-Xylene. Transalkylation of toluene + TriMethyl Benzene to form mixed Xylenes (O-X, M-X and PX) is typically modeled as an equilibrium limited reaction at commercial (Tatoray) plants. Your expression is incomplete because it does not include the all of the isomers of Xylene and trimethyl benzene.

A technique using approach to equilibrium is probably a more reasonable way to model this type of system. If you are analyzing kinetic data for this reaction system you should consider all of the Isomers of Xylene and trimethyl benzene. I am not sure why anybody would want to intentionally convert Xylene into TMB and Benzene. Para-Xylene is valuable because very desirable building block molecule for the production of polyesters..

Dear Tarlan, after long direct discussion with Rick I deleted all my comments and he deleted his replies to my answers. We did not want to confuse you much. My opinion is that the cited reaction rate law is not correct. In addition I don't understand what are the Kb and Pb.

The kinetic equations of the "forward" reaction and of the "backward" reaction are not the same. If they were the same, then for example for reaction A B: rA = kfcA - kbcA = 0. Reaction rate constants can have the same value, but they usually do not. In fact, the form of the kinetic equation and the value of its parameters depends on the results of a suitably planned experiment.

You also ask if the same kinetic equation as for the 2X T + TMB reaction can be used to describe the T + TMB 2X reaction? If in the kinetic studies, T and TMB (in different proportions) were added to substrate X (at different concentrations), then it is possible. In another situation, I would suggest additional verification. There is, however, no a problem with an equilibrium constant.

In the kinetic equation above (L-H type), the kinetic term is consistent with the stoichiometry. In the soprption term, however, there is the partial pressure of the reagent (B), which is not in the stoichiometric equation In this form, the equation is not useful for describing the above reaction. Regards,

Dear All,

Thanks all for comments and recommendations. Unfortunately, I am confused a little bit. Let me clarify it again. Question:

If rate (in picture) is consumption rate of xylene for reaction 2X -> T+ TMB. can I write it for production rate of xylene for reaction T+TMB-> 2X.

Second question is about units. rate has unit " mol/ g(cat) * hour * bar^2 " question: is it ok to have bar^2 on rate? will it show pressure or reaction ?

Thanks.

1) Yes a reversible reaction rate equation for a reaction that is equilibrium limited can be used in both direction if it is formulated properly. However many rate equations re only valid over the narrow region where thet wer derived so you may need to adjust the values of the consyants or the expression itself to fit data in a region far away from where it ws developed.

2) The rate constant should have units that cancel the bar^2 (and the gcat) to leave the the product rate i mol/hr (or similar units)

3) If you are still trying to design a Totoray at conditions similar to commercial units you don't actually need to use a rate expression you can assume equilibrium or a modest approach to eqilibrium for you work.

We can continue this discussion in more detail here or move it to email or RG messaging if you prefer

Yurii V Geletii

Sir, B stands for Benzene , therefore, they are constant Kb and partial pressure Pb of benzene in mixture. The reason is doe to fact that another reaction takes place at the same time. 2T--> B+ X.

It's easy to guess that it's about benzene. Only that to determine its current concentration an additional stoichiometric equation and kinetic equation are necessary.

For a reversible reaction, the kinetic equation should apply in both directions. But one need to know how experiments were carried out. For example, has the inhibitory or enhancening effect of products been studied? Regards,

I agree with Rick that "reversible reaction rate equation for a reaction that is equilibrium limited can be used in both direction if it is formulated properly. " The reaction 2*T = B + X is reversible and the reaction rate constants of this reaction (not equilibrium constant Kb) should be present in the rate law. Again, your reaction rate law is not correct.

Yurii V Geletii,

But k2 is rate constant for reaction. isn't it?

One more thing, if we say " yes this is correct one and we can consider it for reverse direction" in this case , should I recalculate Kex value again? since it is from deltaG= -RTlnK. and deltaG is not same for each direction ( I mean , sign will change, as a result , K value itself will change)

Regards

The reaction rate law should include the rate constants for 2*T = B + X .

Your overall process includes two reactions, not one. Both reactions should be included in derivation of the rate law, totally 4 rate constants and initial concentrations of all reactants. This is the classical kinetic approach.

Minor comment. " I mean , sign will change, as a result , K value itself will change) ". The sign of deltaG depends on which one of two reversible reaction your call "forward" and "reverse . " Keep in mind that K = kf/kr

In usual the rates of direction and reversible reactions is different. You can use the references that is existed about your reaction.

Thanks all comments

The rates of individual reactions (forward and backward) are of course different (omitting state of equlibrium). It should remain unchanged the form of a kinetic equation, but one must remember to change the values of reaction rate constants (one on the other) or about changing the rate constant and equilibrium constant values. The problem may also occur if in the kinetic equation there will be pressure activities instead of partial pressures.

Yes, the above kinetic equation does not make sense unless we know the value of equlibrium partial pressure of benzene and the value of the constant KB has been adapted to this situation. Regards,