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प्रश्न
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
\[\ce{CH3COOH (l) + C2H5OH (l) ⇌ CH3COOC2H5 (l) + H2O (l)}\]
At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
उत्तर
Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess.
The given reaction is:
| CH3COOH(l) | + | C2H5OH(l) | ↔ | CH3COOC2H5(l) | + | H2O(l) | |
| Initial conc. | `1/"V"`M | `0.18/"V"`M | 0 | 0 | |||
| At equilibrium | `(1 - 0.171)/"V"` | `(0.18 - 0.171)/"V"` | `0.171/"V"`M | `0.171/"V"`M | |||
| `= 0.829/"V"`M | `= 0.009/"V"`M |
Therefore, equilibrium constant for the given reaction is:
`"K"_"C" = (["CH"_3"COOC"_2"H"_5]["H"_2"O"])/(["CH"_3"COOH"]["C"_2"H"_5"OH"])`
`= (0.171/"V" xx 0.171/"V")/(0.829/"V" xx 0.009/"V") = 3.919`
= 3.92 (approximately)
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संबंधित प्रश्न
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
\[\ce{CH3COOH (l) + C2H5OH (l) ⇌ CH3COOC2H5 (l) + H2O (l)}\]
Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction).
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
\[\ce{CH3COOH (l) + C2H5OH (l) ⇌ CH3COOC2H5 (l) + H2O (l)}\]
Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
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