標題:
Equilibrium
發問:
最佳解答:
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)
其他解答:
Equilibrium
發問:
此文章來自奇摩知識+如有不便請留言告知
1. Calculate the equilibrium constant,Kc,for the reaction at 298K.CH3COOH + C2H5OH-----> CH3COOC2H5 +H2O12.01g 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... 顯示更多 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最佳解答:
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)
其他解答:
文章標籤
全站熱搜
留言列表
