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So I wanted to know why catalysts didn't affect concentration; since if they increased rate of reaction, presumably the concentration of the products would increase.

I did read online, however, that catalysts increase the rate of the forwards and backwards reactions equally; and then the concentration not changing makes sense, but there's another question now.

I thought the whole point of a catalyst was to increase the rate of reaction so you can get more of something in less time. If it increases only the rate of reaction and not concentration, the net output of the products would be the same as if you didn't use a catalyst — and in that case, what's the point of even using a catalyst?

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    $\begingroup$ You got it absolutely right. The point is not to get more of something in less time (which the catalysts can't do anyway). The point is to get as much of something in less time. Also, welcome to Chem.SE. $\endgroup$ – Ivan Neretin Apr 1 '17 at 6:03
  • $\begingroup$ Thanks very much for the welcome. :) I wasn't sure what you meant by that though; isn't getting as much of something as you can in less time the same as getting more of something in less time? I mean I intended it as getting as much of something as you can as fast as possible, but maybe my definitions are all messed up...? $\endgroup$ – Jasmine Apr 1 '17 at 9:56
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In a simplified representation, a catalyst lowers the overall activation energy while going from initial state A to the of the products B along the reaction coordinate. The interference of the catalyst on the reaction alters the reaction mechanism, renders alternative pathways accessible, which then may be more favourable than for the reaction in absence of the catalyst. Catalysts are deployed regardless if the overall reaction is exothermic, or endothermic. And, because of the relationship described in the Eyring equation, changes in the free activation enthalpy influence rate constants.

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For the instance of a heterogeneous catalysis, the picture is simplified as assuming energy chances for adsorption of the starting material(s) to and subsequent desorption of the product(s) from the catalyst were non-significant in the overall picture.

You are right that it may ease both the forward, as well as the backward reaction. However, in absence of catalyst, overcoming the activation energy at all were just too expensive, even if the reaction $\ce{A -> B}$ drawn here happens to be an exothermic one.

As examples, think about the hydrogenation of benzene; in absence of Nickel there were practically no reaction at all, even at elevated temperatures. Numerous biochemical reactions were unable to proceed in absence of a catalyst at ambient temperature, too.

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    $\begingroup$ I'm nit picking a bit, but the catalyst does not lower the activation energy as such but opens up a new pathway to products and this has a a lower activation energy. The thermodynamics are always the same as the product and reactant have the same energies as when un-catalysed. $\endgroup$ – porphyrin Apr 1 '17 at 9:37
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    $\begingroup$ Thanks for replying! I do understand that catalysts provide an alternate pathway with lower activation energy, which leads to a faster rate of reaction; but if more product isnt being produced through that, what is the point? I mean what would be the point of a faster reaction with the same net output as a slower reaction? I don't know, maybe I'm just confusing myself... 😭 $\endgroup$ – Jasmine Apr 1 '17 at 10:00
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    $\begingroup$ @Porphyrin I welcome your comment and consider it as useful nit picking. The answer given was revised. $\endgroup$ – Buttonwood Apr 1 '17 at 10:03
  • $\begingroup$ @Jasmine Keeping reaction rates aside, thanks to a catalyst's interaction the conversion along $\ce{A -> B}$ occurs e.g. already at a lower temperature, so you need less investment in heating the reaction volume. Or it occurs at a lower pressure, which lowers the cost of the reaction vessel, too. $\endgroup$ – Buttonwood Apr 1 '17 at 10:11
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    $\begingroup$ @Jasmine What is the point of driving someplace, when you can just walk there? Sure, it would be somewhat slower, but the end result is the same, isn't it? $\endgroup$ – Ivan Neretin Apr 1 '17 at 11:23

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