Question
1 mole of acetic acid and 1 mole of ethyl alcohol were mixed at 298K. At equilibrium, 0 - 33 mole of acetic acid was found unreacted. Calculate the equilibrium constant for the reaction
Initial conc.
1 mole 1 mole 1 mole 0 mole
Equilibrium conc.
0.333 0.333 0 mole 0 mole
Solution
Since at equilibrium 0·33 mole of acetic acid is left, we say that 1 – 0·333 = 0·667 mole of acetic acid is used up.
Now according to the equation, it is clear that 0·667 mole of acetic acid combines with a 0·667 mole of ethyl alcohol to form a 0·667 mole of ethyl acetate and a 0·667 mole of water.
At equilibrium [CH3COOH] = 0·333 mole ; [C2H5OH] = 0·333 mole [CH3COOC2H5] = 0·667 mole;
[H2O] = 0·667 mole
Applying the law of chemical equilibrium, we have
