Catalysis: Incompatible with Thermodynamics

Here is a catalyst accelerating the recombination reaction (2H - H2) and SUPPRESSING the reverse dissociation reaction (H2 - 2H):

https://images.nature.com/m685/natur...mms3500-f1.jpg

The publication in Natu

https://www.nature.com/articles/ncomms3500
Yu Hang Li et al. Unidirectional suppression of hydrogen oxidation on oxidized platinum clusters

If a catalyst affects the forward and the reverse reaction DIFFERENTLY, as in the above example, then the second law of thermodynamics is false:

"In the presence of a catalyst, both the forward and reverse reaction rates will speed up EQUALLY... [...] If the addition of catalysts could possibly alter the equilibrium state of the reaction, this would violate the second rule of thermodynamics..."
https://www.boundless.com/chemistry/...lyst-447-3459/

The catalysts rhenium and tungsten also affect the dissociation reaction (H2 - 2H) and the recombination reaction (2H - H2) DIFFERENTLY:

"A small, closed, high temperature cavity contained two metal catalysts (rhenium and tungsten), which were known to dissociate molecular hydrogen (H2) to different degrees (Figure 1). (Rhenium dissociates hydrogen molecules into atoms better than tungsten does; conversely, tungsten recombines hydrogen atoms back into hydrogen molecules better than rhenium.) Because the dissociation reaction (H2 - 2H) is endothermic (absorbs heat), and the recombination reaction (2H - H2) is exothermic (liberates heat), when hydrogen was introduced into the cavity, the rhenium surfaces cooled (up to more than 125 K) relative to the tungsten (Figure 2). Because the hydrogen-metal reactions were ongoing in the sealed cavity, the rhenium stayed cooler than the tungsten indefinitely. This permanent temperature difference - this steady-state nonequilibrium - is expressly forbidden by the second law, not just because the system won't settle down to a single-temperature equilibrium, but because this steady-state temperature difference can, in principle, be used to drive a heat engine (or produce electricity) solely by converting heat back into work, which is a violation of one of the most fundamental statements of the second law (Kelvin-Planck formulation)." http://microver.se/sse-pdf/edgescience_24.pdf

Perpetual flow of the dimer A2 and the monomer A between two catalytic surfaces, S1 and S2:

http://upload.wikimedia.org/wikipedi...SLTD-Fig1c.jpg

See the explanation he https://en.wikipedia.org/wiki/Duncan%27s_Paradox

One of the absurd implications of the second law of thermodynamics is that, if a catalyst increases the rate of the forward reaction by a factor of, say, 745492, it obligatorily increases the rate of the reverse reaction by the same factor, 745492, despite the fact that the two reactions - forward and reverse - may be entirely different (e.g. the diffusion factor is crucial for one but not important for the other) and accordingly require entirely different catalytic mechanisms. The absurd implication is usually referred to as "Catalysts do not shift chemical equilibrium":

"A catalyst reduces the time taken to reach equilibrium, but does not change the position of the equilibrium. This is because the catalyst increases the rates of the forward and reverse reactions BY THE SAME AMOUNT."
http://www.bbc.co.uk/bitesize/higher...um/revision/2/

Scientists should have exposed the absurdity of this implication of the second law of thermodynamics long ago. How can the catalyst increase the rates of the forward and reverse reactions BY THE SAME AMOUNT if these two reactions are entirely different? Consider the dissociation-association reaction

A - B + C

which is in equilibrium. We add a catalyst, e.g. a macroscopic catalytic surface, and it starts splitting A - the rate of the forward (dissociation) reaction increases by a factor of 745492. If the second law of thermodynamics is obeyed, the catalyst must increase the rate of the reverse (association) reaction by exactly the same factor, 745492. But this is obviously absurd! In the reverse reaction the catalyst's function is entirely different - the catalyst must first get together B and C and then join them to form A. It is nonsense to expect the process involving

getting-together-B-and-C

to have exactly the same rate increase, by a factor of 745492, as the process consisting of

splitting-A.

The catalyst may be able to increase the rates of both - forward and reverse - reactions, this is realistic, but not BY THE SAME AMOUNT. The second law of thermodynamics is obviously false.

The false second law of thermodynamics has imposed the wrong and confusing "free energy" interpretation of metabolism:

"Metabolite flow tends to be unidirectional. Living cells exist in a dynamic steady state in which average concentrations of metabolic intermediates remain relatively constant over time. I.e. nutrients go in, they move about getting converted and reconverted etc. and then wastes are excreted. The unidirectional flow of metabolites through a pathway with a large overall negative change in free energy is analogous to the flow of water through a pipe in which one end is lower than the other. Bends or kinks represent individual enzymatic steps. Despite these, the flow is unidirectional which corresponds to the overall change in free energy in the pathway."
https://www.scribd.com/doc/61362780/Enzyme-Activity

The unidirectionality is not determined by free energy changes - it is due to the property of some enzymes to catalyze only the forward reaction, not the reverse. There are countless hints at this in the literature. Just an example:

"It seems exceedingly unlikely, therefore, that the final phosphorylation reaction is irreversible by reason of the endergonic character of the reverse reaction. Since the phosphorylating enzyme system is certainly capable of great activity in the forward direction, we are in the awkward position to postulate a unidirectional catalysis of a thermodynamically reversible reaction." Current Topics in Bioenergetics, Volume 1, Editors: D. R. Sanadi, p.. 108 https://www.elsevier.com/books/curre...-1-4831-9969-6

Pentcho Valev

One can easily see why catalysts are unable to accelerate the forward and reverse reactions EQUALLY on the example of diffusion-controlled reactions:

"Diffusion-controlled (or diffusion-limited) reactions are reactions that occur so quickly that the reaction rate is the rate of transport of the reactants through the reaction medium (usually a solution). As quickly as the reactants encounter each other, they react. The process of chemical reaction can be considered as involving the diffusion of reactants until they encounter each other in the right stoichiometry and form an activated complex which can form the product species. The observed rate of chemical reactions is, generally speaking, the rate of the slowest or "rate determining" step. In diffusion controlled reactions the formation of products from the activated complex is much faster than the diffusion of reactants and thus the rate is governed by collision frequency." https://en.wikipedia.org/wiki/Diffus...olled_reaction

Can a catalyst accelerate a diffusion-controlled reaction? Obviously not - catalysts do not accelerate diffusion. On the other hand, nothing prevents the catalyst from accelerating the reverse reaction. Consider the association-dissociation reaction

A + B - C

which is in equilibrium. Let us assume that the forward reaction

A + B - C

is diffusion-controlled. The reverse reaction

C - A + B

is not diffusion-controlled, obviously.

We add a catalyst which starts splitting C so efficiently that the rate of the reverse (dissociation) reaction increases by a factor of, say, 745492. If the second law of thermodynamics is obeyed, the catalyst must increase the rate of the forward (association) reaction by exactly the same factor, 745492, which is impossible - the catalyst cannot accelerate the forward (diffusion-controlled) reaction at all! The second law of thermodynamics is obviously false.

Pentcho Valev