Chemical Equation:
C2H5OH + CH3COOH + Conc. H2SO4 → Ester + H2O

Explanation:
When ethyl alcohol (C2H5OH) and acetic acid (CH3COOH) react with concentrated sulfuric acid (H2SO4), an ester is formed along with the production of water.

Reaction Mechanism:
1. The sulfuric acid (H2SO4) acts as a catalyst in this reaction.
2. The first step involves the protonation of the carbonyl group in acetic acid (CH3COOH) by the sulfuric acid (H2SO4), resulting in the formation of an acylium ion.
- CH3COOH + H2SO4 → CH3CO+ + H3SO4-
3. The acylium ion then reacts with the alcohol (C2H5OH) through a nucleophilic substitution reaction known as esterification.
- CH3CO+ + C2H5OH → CH3COOC2H5 + H+
4. The esterification reaction produces the ester ethyl acetate (CH3COOC2H5) and a hydronium ion (H+).
5. Finally, water (H2O) is formed when the hydronium ion (H+) reacts with the sulfuric acid (H2SO4-).
- H+ + H2SO4- → H2O + HSO4-

Overall Reaction:
C2H5OH + CH3COOH + H2SO4 → CH3COOC2H5 + H2O

Key Points:
- The reaction between ethyl alcohol and acetic acid is an esterification reaction.
- Concentrated sulfuric acid acts as a catalyst in this reaction.
- The first step involves the protonation of acetic acid by sulfuric acid to form an acylium ion.
- The acylium ion then reacts with ethyl alcohol through esterification to form ethyl acetate.
- The reaction also produces water as a byproduct.
- The overall reaction can be represented as C2H5OH + CH3COOH + H2SO4 → CH3COOC2H5 + H2O.

CH3COOC2H5 + H20
( ETHYL ETHANOATE
&
ETHYL ACETATE)