# Why a catalyst is not consumed in a reaction when all steps are reversible?

It is said that a catalyst speeds up the rate of a reaction but is not consumed (assuming no side reactions take place). Suppose we have the following reaction:

$$\ce{A + B <=> C}$$

catalysed by $$\ce{X}$$ in the following steps: $$\ce{A + X <=> A-X} \tag{1}$$ $$\ce{A-X + B <=> C + X} \tag{2}$$

Each step is reversible so every step has an equilibrium constant $$K$$. So an amount of $$\ce{X}$$ should be present as the intermediate $$\ce{A-X}$$ even after the catalysed reaction has reached equilibrium. That means that the amount of catalyst that we have added to catalyse the reaction is not the same after the reaction has reached equilibrium. So why we say that a catalyst is not consumed through the reaction?

• In the way you have written it X is not a catalyst but forms an intermediate species A-X that then reacts with B to make C. If X were a catalyst X would be released intact as C is formed. – porphyrin Feb 16 at 17:21
• @porphyrin Thanks for pointing it out, I have edited it. – Anton Feb 16 at 18:02
• X has 2 active forms: X and A-X, so it is not consumed. – Poutnik Feb 16 at 20:46
• What if you remove C? – Alchimista Feb 17 at 11:45
• @Poutnik it might be. But counting this way no reaction consume something. A-X could be written as Y. – Alchimista Feb 17 at 14:51

By saying Catalysts are not consumed by reactions. is meant there is no stoichimetric ratio to reactants, consuming catalysts and forming from them catalytically inactive compound.

By other words, some of reaction steps regenerates the original form of a catalyst, consumed by a prior step, so the net consumption is negligible. The transition between active states of a catalyst is not considered a catalyst consumption, as it does not affect its catalytic activity.

Catalysis means the catalyst is reversibly switching between its 2 – chemically different – active states, producing an alternative and faster reaction path. Each form catalyzes one of the mutually opposite direactions of the net reaction. It is like a kind of a ferry to move compounds between reactant and product shores, bypassing a faraway bridge. The ferry is not consumed if ends at the opposite shore.

Note that things get little complicated, if some of products or intermediates catalyse a part or whole net reaction. Not being particularly aware of the intermediate case, but for the product, the famous case is oxidation of oxalate by permanganate in acidic environment:

$$\ce{5(COO)2^2-(aq) + 2 MnO4-(aq) + 16 H3O+(aq) -> \\ ->10 CO2(g) + 2 Mn^2+(aq) + 24 H2O}$$

$$\ce{Mn^2+(aq)}$$ as the product acts as a catalyst, forming catalytic system $$\ce{Mn^3+(aq)/Mn^2+(aq)}$$ with a similar redox potential as $$\ce{MnO4-(aq)/Mn^2+(aq)}$$.

$$\ce{4 Mn^2+(aq) + MnO4-(aq) + 8 H3O+(aq) <=> 5 Mn^3+(aq) + 12 H2O}$$

$$\ce{(COO)2^2-(aq) + 2 Mn^3+(aq) -> 2 CO2(g) + 2 Mn^2+(aq) }$$

The effective result is an autocatalytic reaction. There is sometimes applied a small amount of Mn^{+II} salt to speed up the reaction, or it is waited until initial addition of $$\ce{KMnO4}$$ creates enough $$\ce{Mn^2+}$$.

• But in your example $\ce{MnO4-}$ is ultimately consumed. Is that a good example of catalysis? Is the point that $\ce{Mn^{2+}}$ is the catalyst and $\ce{MnO4-}$ in excess (reagent)? – Buck Thorn Feb 17 at 17:39
• @Poutnik So could we say that for a catalyst initial concentration equals $\displaystyle\sum_{i}^{} c_i$ where $c_i$ corresponds to every chemical species in the catalysed reaction that catalyst is part of? – Anton Feb 17 at 19:24
• @buckthorn Sure, I have thought it is obvious and was explicitly described. Mn2+ is catalyst. MnO4- is not. – Poutnik Feb 17 at 19:50
• @anton It can be said like that. I do not really understand why the catalyst role brings such troubles. – Poutnik Feb 17 at 19:52