ohe09nz81r的部落格

標題:

chem eqm

發問:

C3H7COOH+C2H5OH-->ester+H2O Kc=2.45 2.00 mol acid 2.00 mol ethanol 2.4 mol H2O find the equilibrium concentration of the ester

最佳解答:

• 二條問題 數學@1@
• f.1math轉移策略一問(急!!)
• 26個細楷英文字母像什麼-

• 二條問題 數學@1@
• f.1math轉移策略一問(急!!)
• 26個細楷英文字母像什麼-

 

此文章來自奇摩知識+如有不便請留言告知

LetV dm3 bethe volume of the reaction mixture at eqm. Let y mol be the number of moles of ester at eqm. C3H7COOH + C2H5OH ?ester + H2O At eqm : [C3H7COOH] = (2 - y)/V mol [C2H5OH] = (2 - y)/V mol [ester] = y/V mol [H2O] = (2.4 + y)/V mol Kc = [ester] [ H2O] / [C3H7COOH] [C2H5OH] (y/V) [(2.4 + y)/V] / [(2 - y)/V] [(2 - y)/V] = 2.45 y(2.4 + y) / (2 - y)2 = 2.45 (2.4y + y2) /(4 - 4y + y2) =2.45 2.4y + y2 = 9.8- 9.8y + 2.45y2 1.45y2 - 12.2y + 9.8 = 0 y = [12.2 ± √(12.22 - 4x1.45x9.8)] / (2x1.45) y = 7.51 (rejected) or y = 0.899 The number of moles of the ester at eqm = 0.899mol ≈ 0.9 mol The equilibrium concentration of the ester = 0.899/V mol dm?3 ≈ 0.9/Vmol dm?3 where V dm3 isthe volume of the equilibrium mixture.

其他解答: