Does the rate of hydrolysis of an ester depend on the concentration of both the ester and the acid, or only the ester?

I was told by my professor that hydrolysis of ester follows a pseudo-first order reaction: as the amount of water present for the reaction is very large, the change in concentration of acid/base will result in very negligible changes in concentration of $\ce{H+}$/$\ce{OH-}$ ion. So, the rate is only proportional to the concentration of ester taken.

Can someone explain if the above is in presence of basic medium or acidic medium?

  • $\begingroup$ If the reaction is acid catalyzed, the rate will depend on the hydrogen ion concentration, probably first order in that concentration. However, the overall reaction will be pseudo first order because neither hydrogen ions (catalyst) nor water (solvent) change in concentration over the course of the reaction. $\endgroup$
    – Karsten
    Apr 21, 2019 at 20:14

1 Answer 1


We need to consider 4 quantities for the equilibrium reaction:

$$\ce{R1COOR2 + H2O <=> R1COOH + R2OH}$$

  1. The equilibrium constant
  2. The reaction quotient
  3. The forward reaction rate
  4. The reverse reaction rate

The forward reaction above is of the pseudo first order, as the professor says.

But its rate constant depends on acidic catalytic factors and the net reaction rate is affected by the reverse reaction rate

The concentration of products- the organic acid and the alcohol - affects the net reaction rate by increased rate of the reverse reaction and by shifting of the reaction quotient.

The content of a mineral acid catalyzes the reaction, what means speeding up both reactions, therefore keeping the same equilibrium constant.(required as consequence of the energy conservation law).

The reaction is still the pseudo first order, but the rate constant depends on concentration of the catalysing acid.

The alternative approach is considering the mineral acid as self-regenerating reagent. But it must be then evaluated as reaction chain.

The content of a mineral base decreases reverse reaction rate and shifts the reaction quotient by reacting with the organic acid. It also directly reacts with the ester.

$$\ce{R1COOR2 + OH- -> R1COO- + R2OH}$$

This reaction is of the 2nd order.

  • $\begingroup$ so, the reaction will be 2nd order in both cases? $\endgroup$ Apr 22, 2019 at 5:09
  • $\begingroup$ See the update. $\endgroup$
    – Poutnik
    Apr 22, 2019 at 5:20

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