No a catalyst can not change equilibrium. However an adsorbent or a membrane may be able to selectively remove some of the reactor product and allow the reaction to continue to equilibrium on one side of the membrane and "beyond equilibrium" on the other side.

Hi,

I think you're answering your own question: a catalyst does not change the thermodynamics. The chemical equilibrium is a thermodynamic concept. So - the catalyst does not change chemical equilibrium.

Indeed, this is even stated in the (to my knowledge) first formal definitition of a catalyst, by Wilhelm Ostwald: "Ein Katalysator ist ein Stoff, der die Geschwindigkeit einer chemischen Reaktion erhöht, ohne selbst dabei verbraucht zu werden und ohne die endgültige Lage des thermodynamischen Gleichgewichts dieser Reaktion zu verändern." (Translated into English, this goes something like "A catalyst is a material that increases the speed of a chemical reaction, without being consumed, and without changing the final position of the thermodynamic equilibrium of that reaction.")

Hope this helps!

Ricarda

We should say NO - no influence.

But it should be mentioned that influences on the kinetic of the reaction can lead to events when it seems that the equilibrium is influenced (but it is only not achievable in a acceptable time period or it's alternated).

If you have a elementar reaction A to C and you have a catalyst which reduce the activation energy of this process the equilibrium between A and C cannot be changed.

If the catalyst opens up different reaction path the situation can be different. Let as assume that the activated complex AC* have a high activation energy whereas the catalyst leads without energy to a thermodynamically more stable product than A. The reaction from C to B either needs a very high activation energy or is not possible. In this case the catalyst seems to influence the equilibrium between A and B. (It is the same concept as Rick Manner indicated)

This kind of catalyst is also called inhibitor (if A is the starting material). Wedler is a good textbook concerning such questions.

I thank you all for your valuable informations

Michael Schmitt's comments remind me of an unexpeced phenomenon that I have seen at H2 plants that operate at high temperature (1550 F or greater) with feeds that already contain a large amount of H2 (50% or greater). These H2 plants appear to do “better” than equilibrium (as if they were actually 10-20 F higher temperature). I have seen this phenomenon about 10 times a commercial H2 plant but I haven’t been able to come up with an entirely acceptable explanation. I have two theories to explain the phenomenon:

1) CH4+ H2O CO + 3H2 Equilibrium constant is actually higher than reported in the literature as determine by experiments done 50+ years ago and accepted ever since or:

2) There is a phenomena that allows H2 concentration in the bulk phase to be greater than the concentration at equilibrium a the catalyst surface. I have speculated that H2 diffuses into the bulk phase more rapidly than CO, CO2 and CH4 and H2O and as a result the bulk phase becomes richer in H2 than the equilibrium achieved at the catalyst surface.

There are probably other explanations at least as good as the two that I have come up with but I have seen this phenomenon enough times that it no longer surprises me. I tend to believe explanation 1 because it is simpler (Occam’s Razor).

In most cases --- but NOT in all cases ---- a catalyst does not change the position of equilibrium.There are some cases where a catalyst DOES change the position of equilibrium. Examples are cited in C. R. Noller, Chemistry of Organic Compounds; W. B. Saunders: Philadelphia, PA, USA, 1951, pp. 166-167; and in C. R. Noller, Chemistry of Organic Compounds, Second Edition; W. B. Saunders: Philadelphia, PA, USA, 1957, pp.166-167. This is sometimes (but not by Noller) referred to as epicatalysis. [This example of catalysis not only accelerating the of attainment of equilibrium but also altering the position of equilibrium, i.e., of epicatalysis (but without being dubbed "epicatalysis"by Noller) is unfortunately omitted in the Third Edition of this book (W. B. Saunders, Philadelphia, PA, USA, 1965).] In most such cases the catalyst itself is altered, and the work required to restore the catalyst to its initial state is sufficient to prevent a violation of the Second Law of Thermodynamics. Thus a catalyst can remain unaltered and NOT require work to be restored to its initial state only if it does NOT alter the position of equilibrium. Recently there has been work on epicatalysis that poses Second-Law paradoxes: see for example publications by Daniel P. Sheehan. Epicatalysis should not be confused with enhanced one-way catalysis, which catalyzes reactions solely or at the very least preferentially only in one diirection, but always towards equilibrium, without altering the position of equilibrium. See R. T. Ross, Chemical rectifiers: One-way catalysis. J. Chem. Phys. 72, 4031--4038 (1980).

The catalyst does not affect the equilibrium state of a given reversible chemical reaction, but only the rate of achieving this state. Can there be exceptions to this rule, in my opinion not. Regards,

The above comments are 100% correct if and only if the catalytic and noncatalytic reactions are truly the same, which is not in some cases. I quoted the below paragraph from (Introduction to Shape-Selective Catalysis, 2000 ACS):

"All chemical reactions are subject to thermodynamic limitations such as equilibrium composition. However, for a reaction expressed by (A+B=C+D), shape-selective catalysis can often lead to a product mix that differs from the equilibrium predicted composition. For example, the yield of a given product may be higher than that predicted by the equilibrium composition.

Why is this possible ? Heterogeneous catalysis begins with adsorption on the catalyst surface. The standard statements about equilibrium quoted above are valid only if the catalytic and noncatalytic reactions are truly the same, and yield identical products. Catalytic reactions are as much subject to the laws of thermodynamics as are noncatalytic reactions. However, it is rather difficult to find truly comparable pairs of catalytic and noncatalytic reactions that yield the same products in hydrocarbon processing and chemicals synthesis. The catalytic reactions inside the pore channel of a shapeselective catalyst can be substantially different from those that occur in homogeneous catalytic or noncatalytic reactions involving the same reactants."

Regards,

Thank you dear all for your answers and explanations

First, thermodynamic calculations are subject to error (equation describing the dependence of heat capacity on temperature, standard values ​​of thermodynamic functions, selection of linearly independent reactions (method based on equilibrium constants), selection of reaction mixture components (method based on minimum of Gibbs function), deviation from ideal state etc. Secondly, if side reactions occur as a result of the use of a given catalyst, they must be taken into account in the equilibrium calculations. Thirdly, please give an example of the situation: initial state (inlet): set of substrate composition, set temperature and pressure; in progress: isothermal conditions, isobaric conditions, no phase transitions, no local overheating, the bed has only a catalytic function and is not a hybrid with an admixture of adsorbent selective to the products, no so called enhancement effects;

final state (outlet): conversion is higher than equilibrium conversion. And fourthly, Locally can all happen together with reverse diffusion. Regards,

Strictly speaking, it can not. But a catalyst can produce an APPARENT shift of equilibrium. In some cases, due to the slow kinetic rate, a reaction nearly does not move in spite of that the equilibrium conditions are violated. Such a situation looks as equilibrium. By using a catalyst, one can trigger the reaction movement in a direction defined by the thermodynamic conditions. Such a behavior may yield an illusion that the catalyst shifted the thermodynamic equilibrium. But it is not so.

This is starting to become a philosophical question, isn't it? If catalytic and noncatalytic reactions are not the same (and if side reactions occur, and so on), does this make any kind of statement about the thermodynamics of the noncatalytic reaction (including the equilibrium) itself?

If the rate of reversible chemical reaction is low, the equilibrium (in fact, the state close to equilibrium) will be reached in a longer time. That's all.

"slow kinetic rate" - no words ............ Regards,

A longer time can be 100 years, for example...

Be well

I want to remind everybody that sometimes a catalyst can change the course of a reaction in a way that an impossible reaction becomes possible. Numerous examples can be seen in organometallic chemistry such as ring closing metathesis, acetic acid synthesis, etc. By this I mean that the problem of equilibrium becomes irrelevant.

Indeed. If a reaction does not occur (anymore) in presence of the catalyst, or if it's just very slow compared to the catalytic reaction, its equilibrium is obviously not particularly relevant for what actually happens. Instead, the new reaction path determines the outcome. I'm not trying to argue that. However, the equilibrium of the not-occuring reaction itself has not actually changed. This might be a purely semantical thing, though.

Also, I think ring-closing metathesis is an excellent example of catalysts making new reaction pathways possible - it shows that considerations of thermodynamic equilibrium are not sufficient to predict the actual outcome of a catalytic reaction.

a catalyst will affect both the forward reaction and the reverse reaction in the same way and will not have an effect on the equilibrium constant. The catalyst will speed up both reactions thereby increasing the speed at which equilibrium is reached

A catalyst is a substance that increases the speed of a reaction. Overall, a catalyst is not a reactant and is not used up, but it still affects how fast a reaction proceeds. However, a catalyst does not affect the extent or position of a reaction at equilibrium. It helps a reaction achieve equilibrium faster.

The time of reaching the state close to equilibrium may be shorter in the case of fast reactions and much longer than "100 years" in the case of very slow ones. It is obvious.

We are talking, however, here whether it is possible to obtain a conversion degree greater than equilibrium conversion degree. And this is impossible and has nothing to do with the rate of reaction.

And one more thing - the discussion concerns the chemical equilibrium, not the metastable states accompanying the phase transitions. Regards,

No. This is one of the tests for equilibrium.

No.

Catalyst changes only rates of reactions, but never the equilibrium state.

It seems to me that the discussion is heading to a dead end. The problem is that kinetics has its own laws, just like thermodynamics, and one can not mix these issues. The catalyst has an influence on the rate of the chemical reactions, but it can also affect the mechanism of transformations taking place in its presence. What certain is, however, that both without the presence of a catalyst and in its presence it will not be possible to obtain, for the same chemical reaction, a conversion degree higher than equilibrium conversion. If someone thinks that one can get more than equilibrium conversion, I am asking for an example.

But, really, the discussion should have started with specifying the initial conditions, because otherwise it is difficult to reach a consensus.

Regards,