Advertisement Remove all ads

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. - Chemistry

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.

Advertisement Remove all ads

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)

Concept: Applications of Equilibrium Constants - Predicting the Extent of a Reaction
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 7 Equilibrium
Q 18.2 | Page 226
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×