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Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as: CH3COOH (l) + C2H5OH (l) ⇌CH3COOC2H5 (l) + H2O (l) - Chemistry

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प्रश्न

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

\[\ce{CH3COOH (l) + C2H5OH (l) ⇌ CH3COOC2H5 (l) + H2O (l)}\]

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?

संख्यात्मक

उत्तर

Let the volume of the reaction mixture be V.

  CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)
Initial conc. `1.0/"V"`M   `0.5/"V"`M   0   0
After some time `(1.0 - 0.214)/"V"`   `(0.5 - 0.214)/"V"`   `0.214/"V"`M   `0.214/"V"`M
  `= 0.786/"V"`M   `= 0.286/"V"`M        

Therefore, the reaction quotient is,

`"Q"_"C" = (["CH"_3"COOC"_2"H"_5]["H"_2"O"])/(["CH"_3"COOH"]["C"_2"H"_5"OH"])`

`=  (0.214/"V" xx 0.214/"V")/(0.786/"V" xx 0.286/"V")`

= 0.2037

= 0.204 (approximatly)

Since `"Q"_"c" < "K"_"c"` equilibrium has not been reached.

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Applications of Equilibrium Constants - Predicting the Extent of a Reaction
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अध्याय 7: Equilibrium - EXERCISES [पृष्ठ २३४]

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एनसीईआरटी Chemistry - Part 1 and 2 [English] Class 11
अध्याय 7 Equilibrium
EXERCISES | Q 7.18 - (iii) | पृष्ठ २३४
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