# Are pseudo-zeroth order reactions possible?

Are pseudo zeroth-order reactions possible?

I have been taught that pseudo first-order reactions take place and they can be considered pseudo zeroth-order reactions because one reactant is in excess of the other(s).

What are the necessary conditions to label a reaction pseudo-zeroth order?

• I've added this as a comment, below, but it warrants attention here: "Hence, even for elementary reactions, the molecularity of a reaction does not have to be the same as the order of the reaction. When the molecularity is not the same as the order of a reaction because the concentration of one or two species is kept constant either due to the concentration of the species is high or because the concentration of the species is buffered, the reaction order is also referred to as pseudo-order." Jul 21, 2021 at 14:21

I think to some extent, you can think about all zero-order reactions as "pseudo-zero-order" reactions. This is because it's not possible for a reaction to be true zero-order. How can it be that the rate of a reaction does not depend on any one of the reactants? So, anything that is zero-order means that there is some artifact in the system, or that something else is present.

In general, zero-order reactions happen when the reactant is not able to react at the theoretical maximum rate due to something preventing it. For example, for a catalytic process, the reaction happens at the surface of the catalyst, so if you have a large excess of reactant, all of the surface of the catalyst will be covered up with the reactant, and the rate will become constant. When the concentration of the reactant goes down so that all of the catalyst surface is not in use, the reaction will show a first/second or higher order rate equation.

An example of such reaction is the decomposition of nitrous oxide: $$\ce{2N2O\overset{\Delta, Ni}{->}2N2 + O2}$$ When nickel catalyst is used, the rate is limited by the total surface area of the catalyst. Once the catalyst is working at the full capacity, the rate won't increase, even if the concentration of the reactants is increased further.

Without the catalyst, the reaction is slower, but it shows the standard second-order rate equation that we would expect by looking at the equation.

For reactions that are driven only by light (not just initiated by light!), the rate would be limited by the intensity of the light. Once there are enough species that can absorb all of the light, the reaction rate won't go up even if more reactants are added. Then there are biochemical reactions which are diffusion controlled.

So, in summary, all zero-order reactions are zero-order because there is some reactant which is in excess than something else the reaction depends upon (which can be catalyst, light, diffusion etc.). You can say they are probably all pseudo zero-order.

• This is incorrect. There are plenty of reactions where the rate does not depend on one or more of the reactants, which is the defining part of "pseudo zeroth-order." Example: decomposition of $\ce{PH3}$ on hot tungsten at high pressure (taken from this). Jul 21, 2021 at 14:24
• @ToddMinehardt Yes, because the tungsten is the catalyst here, and as long as all the catalyst surface sites are engaged, the rate doesn't depend on the conc. of PH3. Nothing in my answer contradicts anything you say. Did you read my whole answer? Please do so before downvoting. Jul 21, 2021 at 14:28
• @ToddMinehardt I have added another word in the answer, hopefully that makes it clearer. Jul 21, 2021 at 14:29

A standard example in biochemistry are enzyme-catalyzed reactions with a single substrate (e.g. ATPase, where we ignore the second substrate, water, because it is the solvent). At low substrate concentration (low reactant concentration), the reaction is first order (actually, pseudo-first order because of the water). At high susbtrate concentration (high reactant concentration) the reaction is pseudo-zero order. The theoretical expression was derived by Maude Leonora Menten and Leonor Michaelis, and the results are shown as a graph below:

As to whether it is pseudo or not, here is a textbook quote that shares the OPs view:

Zero-order kinetics is always an artifact of the conditions under which the reaction is carried out. For this reason, reactions that follow zero-order kinetics are often referred to as pseudo-zero-order reactions. Clearly, a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero as depicted at the upper left.

Of course, if you disagree with that view, you can always call the source a pseudo-textbook...

In the case of the enzyme-catalyzed reactions, the reaction would proceed at almost constant speed until the substrate concentration is depleted to a concentration in the range of the parameter K$$_\mathrm{M}$$, at which time it would slow down appreciably and continually until the substrate is depleted.

Zero-order kinetics can also appear in certain industrial settings. In some steel strip annealing processes where improved bendability of the product is required, steam is applied to decarburize the steel near the surfaces of the strip according to the reaction

$$\ce{C(s) + H2O(g) -> CO(g) +H2(g)}$$

The decarburization process requires carbon to diffuse out of the steel in order to reach the water molecules at the surface, and this process has a limited rate no matter how much steam is introduced outside the strip. Thus at a high concentration of steam in the atmosphere the decarburization process becomes zero-order in the steam concentration, dependent only on the diffusion of the carbon.

• In this case, wouldn't the diffusion rate of the carbon to the surface be directly proportional to its concentration, making the overall reaction (ideally) first order with respect to it? Jul 21, 2021 at 14:21
• Still zero-order with respect to steam. Jul 21, 2021 at 14:33
• Wouldn't the reaction rate still increase at very high steam concentrations as steam starts to diffuse into the steel? Jul 21, 2021 at 17:04
• Steam would not diffuse into the steel as water molecules, and if individual oxygen atoms were to enter there would be no way to form CO molecules in the solid matrix either. Also I am ignoring a side reaction where the steam oxidizes the iron, which the manufacturers avoid by having hydrogen in the atmosphere. Jul 21, 2021 at 17:54
• @OscarLanzi By that reasoning, every reaction is "zero-order with respect to" one of the reactants that's in excess, which is a fairly uninteresting fact Jul 23, 2021 at 11:47

The best example of a zero-order reaction is the combustion of a candle. If the candle weighs $$m_o$$ grams at the beginning, and $$m$$ at any time afterwards, and if $$a$$ grams of it are burned per minute, the reaction rate $$r$$ is constant from the very beginning to the end of the candle, and it is : $$r = dm/dt = a$$ in grams per minute. The integrated rate law is : $$m = m_o - at$$. This is a zeroth-order kinetics.

• Is this a real candle or a theoretical candle? Seems pretty contrived. Jul 21, 2021 at 13:59
• Contrived or not, a zeroth-order reaction is one where the rate is independent of the quantity of reactants/products. This is a good write-up. I've encountered these (and the defined pseudo-type, which I don't have time to write up now) in the context of geochemical kinetics. Jul 21, 2021 at 14:19
• @ToddMinehardt I am not saying zero-order reactions don't exist. But I have seen real wax candles burn and they don't burn at the same rate from the start to finish. At the beginning it will burn slowly, and after a while as more of the wick comes out, it will burn faster. It's okay if the candle example is given as a thought experiment, but actual candles definitely don't burn at the same rate from start to finish. Jul 21, 2021 at 14:23
• @SRMaiti A bit of calculus might help here. The reaction is 0th order with respect to mass of the candle but non linear with respect to temperature and wax shape. If i give you access to a candle of unknown length who's end you can't and see and its pushed upward such that the flame stays in the same place there is no measurement you can perform even when allowed to turn the candle off and on again that allows you measure the length of the candle without just burning it completely The non linearity with regard to candle warming yields no information about the total length of the candle. => 0th Jul 21, 2021 at 20:04
• @SRMaiti The rate of mass loss (~length loss) of a candle that has been turned on for a while is independent of the remaining fuel. The wick has an influence, as has a lighting a cold vs a hot candle however the wick effect becomes irrelevant really fast. If we compare two identically unlight candles of different lenght and light them we see that the burn rate is independent of the total mass, so the effect of the wick is not order 1 or higher either. The burn rate at the wick therefor does not depend on mass remaining but mass burnt and quickly levels of once the constant thickness is reached Jul 23, 2021 at 8:07

First, be aware that the order of a chemical reaction depends on its mechanism and a great deal of chemical reactions have more than one mechanism, depending on the conditions.

The traditional example for zeroth order reaction is a reaction fully dependent on a catalyst (e.g. enzymes in biochemistry).

When you have enough reactants and a scarce amount of the catalyst, you get a reaction rate that depends only on the concentration of the catalyst. It is zeroth order in regard to reactants.

Another example: topochemical reactions. The site(s) of the reaction may limit the reaction rate.

Related example: reactions limited by the mixing rate.