CBSE (Science) Class 11CBSE
Share

# 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. - CBSE (Science) Class 11 - Chemistry

ConceptApplications of Equilibrium Constants Predicting the Extent of a Reaction

#### Question

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)

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.

#### Solution

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:

CH_3COOH_(l) + C_2H_5OH_(l) ↔ CH_3COOC_2H_(5(l)) + H_2O_(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_3COOC_2H_5][H_2O])/([CH_3COOH][C_2H_5OH])

= (0.171/V xx 0.171/V)/(0.829/V xx 0.009/V) = 3.919

= 3.92 (approximately)

Is there an error in this question or solution?

#### APPEARS IN

Solution 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. Concept: Applications of Equilibrium Constants - Predicting the Extent of a Reaction.
S