Is the saponification reaction endothermic or exothermic

Esterification and saponification

Esterification

Esterification (also known as ester formation) is a chemical reaction that was discovered by Emil Fischer. It is an equilibrium and condensation reaction in which an alcohol reacts with an organic or inorganic acid to form an ester. Esters are made up of uncharged molecules, so they don't conduct electricity.


Ester hydrolysis = saponification

A freshly prepared mixture of pure ester and water also shows no electrical conductivity. However, if the mixture is heated slowly, conductivity occurs, which increases slowly and remains constant after a certain time. So one can state the following:

  • When esters and water react with one another, ions are formed and electrical conductivity occurs.
  • The ion formation increases during the reaction and reaches a constant concentration after a certain time.

An ester with water is split into acid and alcohol. The reaction is the reverse of esterification and is called saponification. The products of the reaction are the alcohol and the acid salt (carboxylate ion) that made up the ester.


$ \ mathrm {\ underbrace {CH_3 \ - \ \! \! \! \! \! \! \! \! \! \! \! \! \! \! {\ overset {\ qquad \ \ quad {\ Large O}} {\ overset {\ quad \ \ diagup \! \! \ diagup \! \! } C}} \! \! \! \! \! \! \! \! \! - O \ - \ C_2H_5} _ {ethyl acetate} + H_2O \ \ {\ overset {\ large saponification} {\ underset {\ large esterification} {\ large \ rightleftharpoons}}} \ quad \ underbrace {CH_3COOH} _ {acetic acid} + \ underbrace {C_2H_5OH} _ {Ethanol}} $




$ \ mathrm {\ underbrace {CH_3COOH} _ {acetic acid \ \ color {blue} {no conductivity}} \; \; {\ overset {\ large H_2O} {\ large \ rightleftharpoons}} \; \; \ underbrace {CH_3COO ^ -_ {(aq)}} _ {acetate ion \ \ color {brown} {electrical conductivity}} + H ^ + _ {(aq)}} $

The process of ester cleavage is hydrolysis. The acidic ester hydrolysis corresponds to the reverse reaction of the acidic esterification and therefore very quickly leads to a state of equilibrium.

Originally it was understood by Saponification only the basic ester hydrolysis of animal fats or vegetable oils, with alkalis, primarily caustic soda or potassium hydroxide, in the soap boiler. This creates the trivalent alcohol glycerine and the respective alkali salts of the fatty acids found in the fats. The latter will beSoap called. The saponification with caustic soda provides curd soap, the potassium hydroxide soft soap.


The chemical equilibrium is a dynamic equilibrium

The acid formed during the saponification is responsible for the occurrence of electrical conductivity. Since the acid can partially recombine with alcohol to form ester and water, it is not possible that an ester can be completely split into acid and alcohol with water. Correspondingly, in an alcohol-acid mixture, not both reactants react completely to form esters (Chapter 106, 1). In the case of ester cleavage, as in ester formation, there is always a mixture of ester, water, alcohol and acid. If, on the one hand, the amounts of ester and water and, on the other hand, of acid and alcohol in this mixture have reached a certain ratio, it appears that no more reaction takes place. This state is called chemical equilibrium, since the formation and splitting of the ester are in balance.

Chemical equilibrium is a dynamic equilibrium because it is a state in which molecules are constantly reacting with one another. An equilibrium reaction is spelled with a double arrow:


$ \ mathrm {A + B \ \ leftrightharpoons \ AB} $


In the state of equilibrium, the mole fraction of the acid in the mixture no longer changes, and thus the electrical conductivity reaches a constant value.



The shift in equilibrium

Chemists are able to influence and control a chemical equilibrium reaction. Numerous processes have been developed for this in technical chemistry. If you want to have a high yield of ester, you have to shift the equilibrium to the side of the ester. According to the equation below, the equilibrium should be shifted to the right. In experiments 2 and 5 in Chapter 106, the conc. Sulfuric acid has a double effect:

  1. Hydrogen ions accelerate the achievement of the ester equilibrium.
  2. Since the conc. Sulfuric acid continuously removes water from the equilibrium, the reaction shifts to the right.

The shift of the equilibrium in favor of the ester cleavage can be achieved by alkaline solutions (see experiment 2). The OH -Ions of the lye combine with the hydrogen ions of the acid, so that the acid that is formed is continuously consumed. A state of equilibrium is therefore not reached.


$ \ mathrm {CH_3 \ - \ \! \! \! \! \! \! \! \! \! \! \! \! \! {\ overset {\ qquad \ \ \ qquad {\ Large O}} {\ overset {\ qquad \ \ diagup \! \! \ diagup \! \! } C}} \! \! \! \! \! \! \! \! \! - O \ - \ C_2H_5 + H_2O \ {\ large \ rightleftharpoons} \ CH_3COOH + C_2H_5OH} $

$ \ mathrm {CH_3COOH \ qquad \; \; + \; OH ^ - \ longrightarrow \ \; CH_3COO ^ - + H_2O} $


$ \ mathrm {CH_3 \ - \ \! \! \! \! \! \! \! \! \! \! \! \! \! {\ overset {\ qquad \ \ \ qquad {\ Large O}} {\ overset {\ qquad \ \ diagup \! \! \ diagup \! \! } C}} \! \! \! \! \! \! \! \! \! - O \ - \ C_2H_5 + OH ^ -_ {(aq)} \ \ xrightarrow {H_2O} \ C_2H_5OH + CH_3COO ^ -_ {(aq)}} $


This one from the OH -Ions caused state is irreversible, i.e. it can no longer be reversed. The equilibrium is finally shifted to one side.