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… catalysis, the acceleration of chemical reactions by substances not consumed in the reactions themselves—substances known as catalysts.

(Source)

Now as I’ve understood, to keep a reaction going, you must keep the solution heated and sometimes add more of the catalyst. Why then, is the definition that the catalyst is not consumed? Something of the catalyst must surely be consumed.

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  • $\begingroup$ "Catalysts generally react with one or more reactants to form intermediates that subsequently give the final reaction product, in the process regenerating the catalyst." From Wikipedia article on catalysis. Syrely, the Encyclopædia Britannica article you quote also mentions this, but I don't have access to the full text. $\endgroup$
    – Wildcat
    Jul 6 '15 at 16:14
  • $\begingroup$ Read specifically this section of Wiki article. $\endgroup$
    – Wildcat
    Jul 6 '15 at 16:15
  • $\begingroup$ @GeoffHutchison, usually I wait for a response from OP on some hints provided by me in comments to be sure I got the question right. :D $\endgroup$
    – Wildcat
    Jul 6 '15 at 16:39
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    $\begingroup$ Even if a catalyst isn't consumed by the reaction it's intended to facilitate, it may be consumed by other not-necessarily-desirable reactions. $\endgroup$
    – supercat
    Jul 6 '15 at 17:51
  • $\begingroup$ @supercat, sure. The excerpt from Wikipedia in my answer mentioned this in its very last sentence. $\endgroup$
    – Wildcat
    Jul 6 '15 at 18:07
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In short, a catalyst does participate in a chemical reaction the rate of which it increases, it just do so in a somewhat tricky way so that it does not appear in the overall reaction equation.

Wikipedia article on catalysis provides a concise explanation of why this is the case:

Catalysts generally react with one or more reactants to form intermediates that subsequently give the final reaction product, in the process regenerating the catalyst. The following is a typical reaction scheme, where $\ce{C}$ represents the catalyst, $\ce{X}$ and $\ce{Y}$ are reactants, and $\ce{Z}$ is the product of the reaction of $\ce{X}$ and $\ce{Y}$: $$ \ce{X + C → XC} \tag{1} $$ $$ \ce{Y + XC → XYC} \tag{2} $$ $$ \ce{XYC → CZ} \tag{3} $$ $$ \ce{CZ → C + Z} \tag{4} $$ Although the catalyst is consumed by reaction 1, it is subsequently produced by reaction 4, so it does not occur in the overall reaction equation: $$ \ce{X + Y → Z} \, . $$ As a catalyst is regenerated in a reaction, often only small amounts are needed to increase the rate of the reaction. In practice, however, catalysts are sometimes consumed in secondary processes.

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    $\begingroup$ There is also the possibility of the catalyst being poisoned, although this might be what you meant by "secondary processes" $\endgroup$
    – thomij
    Jul 6 '15 at 17:56
  • $\begingroup$ Thus, the rate equation for this reaction might look something like $$\text{rate}=\dfrac{\Delta [\ce{Z}]}{\Delta T}=k[\ce{X}][\ce{Y}][\ce{C}]$$ Adding more catalyst will increase the rate. $\endgroup$
    – Ben Norris
    Jul 6 '15 at 18:55
  • $\begingroup$ Yes @thomij, and a good example is the poisoning of the $\ce{Pt-Ni}$ catalysts in catalytic converters due to an illegal existence of TEL in the fuel. $\endgroup$
    – M.A.R.
    Jul 6 '15 at 22:41
  • $\begingroup$ Thanks for all your answers. I understand about its regeneration process. But still, this bothers me: "In practice, however, catalysts are sometimes consumed in secondary processes." So, when we realise that our solution needs more of the catalyst, we are actually just helping these "unwanted" secondary processes? $\endgroup$
    – Sork
    Jul 7 '15 at 16:55
  • $\begingroup$ @Sork, I would rather say that we are helping our process of interest and not "unwanted" secondary processes. $\endgroup$
    – Wildcat
    Jul 11 '15 at 12:12
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Think of it like an assembly line. A worker in an assembly line takes two parts and puts them together. The worker isn't expended in the process.

Similarly a catalyst helps molecules react. But the product is just a combination of the molecules in the reaction. The catalyst isn't affected, and can go on to catalyze another reaction.

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  • $\begingroup$ And workers sometimes strain their wrists (or something) that is why you sometimes need to add more catalyst. $\endgroup$
    – Ajasja
    Jul 7 '15 at 7:35
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The problem you could be experiencing could be more related to the phenomenon of "mass transport" than anything else.

Basically, for a reaction to occur the two reactants need to get close to each other. It could be that a catalyst works by forming a reactive complex with the reactant and carrying it through multiple reactions. Essentially, the catalyst is speeding up the reaction but is caught up in the reaction while it is occuring. If there isn't enough catalyst for every single reactant, some of them are occuring at the catalyzed speed and others are proceeding normally; adding a little more catalyst just ensures that a higher percentage of substrates are being assisted by the catalyst at the same time.

Imagine the following scenario: A room full of people trying to cut sheets of paper in half. Everyone is using scissors, and there is one movable, industrial paper cutter being passed around. The guy who has the paper cutter is going to move much more quickly while the paper cutter is at his station. Additionally, the paper cutter has to be with him the entire time he is cutting, or else he slows down. The paper cutter is the catalyst. If you add more paper cutters, you would keep speeding up. It isn't because the paper cutter disappears or stops working; it's because not everybody has a paper cutter.

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    $\begingroup$ Please use italics or bold for emphasis instead of capitals. It is much more friendly to read. $\endgroup$
    – bon
    Jul 9 '15 at 15:42
  • $\begingroup$ That's very interesting. Is this just a theory or has it also been practically proven with observations and experiments? Also: "It isn't because the paper cutter disappears or stops working; it's because not everybody has a paper cutter." If I recall correctly from my own experiments with acids, when the reaction came to a halt, more of the catalyst was needed. Implying that it had actually gone somewhere in the reaction(s). $\endgroup$
    – Sork
    Jul 11 '15 at 12:21
  • $\begingroup$ That is an excellent point; the catalyst might have lost some of its activity since its introduction such as in your experiments with acids. In some systems the catalyst could be deactivated or outright destroyed by a secondary process. Here is a link to a paper about mass transport in catalysis that gives a fairly decent introduction cdn.intechopen.com/pdfs-wm/23539.pdf Really I guess it is highly system dependent. If a rate degrades over time, the catalyst is probably degrading and if the rate increases with addition of catalyst then mass transfer was probably limiting. $\endgroup$ Jul 14 '15 at 11:49

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