This is for the hydrolysis of p-nitrophenyl phosphate by alkaline phosphatase, whereby the enzyme is in a solution of distilled water. We also add glycine buffer to the solution. It exhibits first-order kinetics, even though there is a second reactant (being water) - how is this possible?


1 Answer 1


It is because water is the solvent in the reaction and therefore its concentration remains practically constant throughout the course of the reaction. Hence, it is absorbed in the rate constant:

rate=k[p-nitrophenyl phosphate]$^1$[H$_2$O]$^x$=k'[p-nitrophenyl phosphate]$^1$

Hence, we say the reaction is peudo-first order. Also, in general, you could have a reaction with two reactants and it could still be first order overal since it could be 0 order in reactant A and first order in reactant B.

  • $\begingroup$ Why do you write $[\ce{H^2O}]^n$ ? $\endgroup$
    – Karl
    Commented Feb 11, 2018 at 16:51
  • 1
    $\begingroup$ Because we dont really know the order in water due to it being the solvent. Stricktly speaking, you shouldnt have water in that rate equation at all. I just wanted to show the logic behind it and if i used [water] it would imply first order in water which is incorrect. The correct form of the equation is the one on the right. $\endgroup$
    – Outlander
    Commented Feb 11, 2018 at 17:07
  • $\begingroup$ OK, makes sense. I'd recommend to replace $n$ by $x$. And really x is very close to zero, and then the equation is in good approximation correct. Right? $\endgroup$
    – Karl
    Commented Feb 11, 2018 at 17:39
  • $\begingroup$ Actually, thinking about it again i would say that leaving n or x (where n or x is some number) there is ok. Say for example that we have a reaction in water, some kind of hydrolysis, in which we know the mechanism and the rate determinig step involves one molecule of substrate and one molecule of water. This makes the reaction second order overall. However because the concentration of water always remains constant being the solvent, the observed kinetics will be 1st order. Hence, pseudo 1st order. $\endgroup$
    – Outlander
    Commented Feb 11, 2018 at 23:17
  • $\begingroup$ $n$ implies an integer number, and that's surely not the case here, hence $x$. $\endgroup$
    – Karl
    Commented Feb 12, 2018 at 17:24

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