The equilibrium of esterification


It was experimentally found that the reaction at reflux between methanoic acid and ethanol in the presence of sulfuric acid results in an equilibrium with : $HCOOH$ $+$ $CH_3CH_2OH$ $\rightleftarrows$ $ HCOOCH_2CH_3$ $ +$ $H_2O$ $K$ $=$ $\frac{[ HCOOCH_2CH_3][H_2O]}{[HCOOH][CH_3CH_2OH]}$ $=$ $\frac{n_{HCOOCH_2CH_3}n_{H_2O}}{n_{HCOOH}n_{CH_3CH_2OH}}$ $\approx$ $4$

Equilibrium shift

An equilibrium is shifted to the right side by -adding a reagent -removing a product An equilibrium is shifted to the left side by -adding a product -removing a reagent

In our case we can improve the yield of ester (shift to the right) by -removing the ethyl methanoate from the reaction mixture by distillation. (it is more volatile than the other three substances) -removing water by fixing it with the sulfuric acid. -adding methanoic acid or ethanol to the reaction medium. promote hydrolysis (shift to the left) by -removing methanoic acid from the reaction mixture by neutralization with a base (hence, working in a basic medium) -adding water to the reaction mixture (thus diluting the medium)

Reaction rate

The esterification is a slow reaction. An increase in temperature (eg reflux) accelerates the reaction, but does not necessarily improve the yield!