Use of concentrated sulfuric acid in Fischer esterification
In fact, sulfuric acid is not a must. Many other H-acids, as well as Lewis acids and their combinations would be sufficient. 

The trick here is not to get concentrated acid, but to reduce amount of water in the system. Consider the simpliest transformation, which is typically performed with a large excess of methanol and trace amounts of acid:

$$\ce{R-C(O)OH + CH3OH_{(excess)} &lt;=&gt;&gt;[HCl] R-C(O)OCH3 + H2O}$$

This is an equilibrium reaction. The catalyst allows us significantly faster achieve the equilibrium state, but it does not shift it. Catalyst is equally promoting both direct and reverse reactions. A large excess of alcohol, in accordance with the law of mass action, shifts the equilibrium toward the formation of the ether and thus increases the degree of conversion of the acid into ether.

But if water is not eliminated from the system, or is initially present, hydrolysis occurs:

$$\ce{R-C(O)OCH3 + H2O_{(excess)} &lt;=&gt;&gt;[HCl] R-C(O)OH + CH3OH }$$

Hydrolysis of ether will happen even faster, if a strong base is present, as in this case reaction equilibrium shifts further to the right due to the salt formation.

The bottom line is, in order for the reaction to go to the end, you either add an excess of acid (or alcohol), or remove water as it is produced. Starting with concentrated or water-free acid you are not only simplifying your life, but you are also getting better yield.