Equilibrium

用戶7821

2010-03-02 10:31 pm

1. Calculate the equilibrium constant,Kc,for the reaction at 298K.

CH3COOH + C2H5OH-----> CH3COOC2H5 +H2O

12.01g OF THE ETHANOIC ACID ARE TRATED WITH 4.61g OF ETHANOL IN
THE PRESENCE OF A CATALYST. WHEN THE REACTION REACHES
EQUILIBRIUM AT 298K, 5.04g OF THE ETHANOIC ACID ARE FOUND TO
HAVE REACTED.

THANKS SO MUCH

回答 (1)

Kalvin

2010-03-02 10:53 pm

✔ 最佳答案

no of mole of ethanoic acid = 12.01/12x2+1x4+16x2=0.2mol
no of mole of ethanol = 4.61/12x2+1x6+16=0.1mol

no of mole of ethanoic acid reacted = 5.04/60=0.084mol
= no of mole of ethyl ethanoate and H2O
no of mole of ethanoic acid remained=0.2-0.084=0.116mol
no of mole of ethanol remained=0.1-0.084=0.016mol

Kc = [CH3COOC2H5][H2O] / [CH3COOH][C2H5OH]
H2O is included because all of the species are in liquid phase and H2O is not a solvent
as their volume are the same, we can juz use no of mole instead of molarity for calculation
therefore Kc = (0.084)^2 / (0.116)(0.016)
=3.80 (no unit)

參考: me

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