I am confused as to how an order of a reactant (n or m) in the rate law can be negative. This means that increasing the concentration of a reactant would actually decrease its rate of disappearance. How does this make sense? Doesn't a higher concentration always allow for more collisions? Also, when the order is 0, this means that concentration has no effect on the rate of disappearance. How does this make sense chemically/at the atomic level?
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$\begingroup$ related chemistry.stackexchange.com/questions/263/… $\endgroup$– MithoronCommented Mar 4, 2015 at 0:02
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1$\begingroup$ Increasing the concentration of an inhibitor can lower the rate of reaction. $\endgroup$– dupdupbCommented Jun 25, 2020 at 17:39
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In principle, there are no zeroth-order reaction. Experimentally, however, zeroth-order rate law occurs because of the reaction condition. For example, in catalytic reactions such as hydrogenation of an alkene using platinum as a catalyst, the hydrogenation occurs on the surface of platinum. There is a finite surface area of platinum, so at a some point, it doesn't matter at all how much more hydrogen is put in. This is because no reaction can occur if the hydrogen cannot bind to the platinum.
An example of inverse first order reaction can be found here.
Basically, you have a two-step reaction: $$\ce{A + B <=> C}$$ $$\ce{C <=> D\ (slow)}$$ Increasing concentration of $\ce{C}$ would decrease the rate of reaction.
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$\begingroup$ Can we say that before the saturation of the surface of catalyst, the rate of reaction is directly proportional to the concentration of reactant and when it reaches above the saturation level then the rates becomes independent of the concentration of reactants? So to achieve zero order kinetics, concentration of reactants should always be more than the saturation level of catalyst? $\endgroup$– ManuCommented Jun 26, 2020 at 7:33