7.18. 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)

(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)

(ii) 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.

(iii) 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?      

6 Views|Posted 8 months ago
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1 Answer
V
9 months ago

(i) The concentration ratio (Concentration quotient), Qc for the reaction is:

Qc = [CH3COOC2H5] [ H2O] / [CH3COOH] [C2H5OH]

 

(ii)

 

CH3COOH

C2H5OH

CH3COOC2H5

H2O

Initial molar concentration

1.0 mol

0.18 mol

0

0

Molar concentration at equilibrium

(1 – 0.171) = 0.829 mol

(0.18 – 0.171) = 0.009 mol

0.171 mol

0.171 mol

Applying

Kc = [CH3COOC2H5] [H2O] / [CH3COOH] [C2H5OH]

= (0.171 mol) x (0.171 mol) / (0.829 mol) (0.009 mol)

 = 3.92

 

(iii)

 

CH3COOH

C2H5OH

CH3COOC2H5

H2O

Initial molar concentration

1.0 mol

0.5 mol

0.214 mol

0.214 mol

Molar concentration at equilibrium

(1 – 0.214) = 0.786 mol

(0.5 – 0.214) = 0.286 mol

0.214 mol

0.214 mol

 

Qc = [CH3COOC2H5] [H2O] / [CH3COOH] [C2H5OH]

     = (

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Chemistry Ncert Solutions Class 11th 2023

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