Răspuns :
CH3COOH + C2H5OH ⇄ CH3COOC2H5 + H2O
initial : 2mol 1mol 0,5 moli 8 moli
au reactionat x x
la echilibru : (2-x) (1-x) (0,5+x) (8+x)
m amestec = 200+ 110 + 44 = 354g
(0,5+x)·88·100/354= 17,41 0,5+x = 1,427 x = 0,9278
Kc = [ester][apa]/{[acid][alcool]} = 1,427·8,9278/(1,0722·0,0722) = 164,57
initial : 2mol 1mol 0,5 moli 8 moli
au reactionat x x
la echilibru : (2-x) (1-x) (0,5+x) (8+x)
m amestec = 200+ 110 + 44 = 354g
(0,5+x)·88·100/354= 17,41 0,5+x = 1,427 x = 0,9278
Kc = [ester][apa]/{[acid][alcool]} = 1,427·8,9278/(1,0722·0,0722) = 164,57