Reaction between ethyl acetate and water attain equilibrium in an open vessel but decomposition of caco3 does not.explain.
The difference in behavior between the reaction between ethyl acetate (CH3COOC2H5) and water, and the decomposition of CaCO3 (calcium carbonate), can be explained based on their respective chemical properties and reaction conditions.
In the case of the reaction between ethyl acetate and water, it involves the formation of alcohol and acetic acid:
CH3COOC2H5 (ethyl acetate) + H2O (water) → CH3COOH (acetic acid) + C2H5OH (alcohol)
This reaction is reversible, meaning that the formed products (acetic acid and alcohol) can react with each other to reform ethyl acetate and water. In an open vessel, where there is constant contact with air, the water in the reaction mixture evaporates due to its high vapor pressure, driving the equilibrium of the reaction towards the formation of more products. As a result, the reaction reaches an equilibrium state where the forward and backward reactions occur at the same rate, leading to a constant concentration of products and reactants.
On the other hand, the decomposition of CaCO3 does not attain equilibrium under normal conditions. Calcium carbonate decomposes when heated to form calcium oxide and carbon dioxide:
CaCO3 (calcium carbonate) → CaO (calcium oxide) + CO2 (carbon dioxide)
Unlike the reaction between ethyl acetate and water, the decomposition of CaCO3 is an irreversible reaction. Once CaCO3 is decomposed, it cannot easily reform back into its original form in the absence of external factors or a reverse reaction. Therefore, the decomposition of CaCO3 does not reach equilibrium because it proceeds in one direction only until all the CaCO3 is decomposed.
In summary, the difference in behavior can be attributed to the reversibility of the reaction between ethyl acetate and water, allowing it to reach equilibrium in an open vessel, while the decomposition of CaCO3 is an irreversible reaction and does not attain equilibrium.