Question

From equation 4CH3COOH + 12CH3OH <--> 5C4H8O2 + 10H2O,

If 0.5mol of alcohol and 0.2 mol of acid are initially used, what is the equilibrium composition if the equilibrium coefficient(K) is 8. Use the mole balance and then solve it again with the compound balance.

Answer 1

Given: K = 8 I think the reaction is not balanced and the stoichiometric coefficients are wrong. The correct balanced chemical reaction is : Reaction is 4CH3COOH + 8 CH3OH -> 4C4H2O2 + 8 H20 Initial moles of CH3COOH Initial moles of CH3OH = = 0.5 moles 0.2 moles Let & be extent of reaction Thus, Moles of CH3COOH reacted = 4 Moles of CH3OH reacted = 8 Moles of C4H8O2 formed = 45 Moles of H20 formed = 8 Moles of B reacted = Moles of C formed = Moles of D formed = 5 C .H8O2 H20 0.2 0.5 By stoichiometry, we can create following table: CH3COOH CH3OH Initial No of moles Moles reacted/ formed | 43 till equilibrium Final No of moles 0.2-45 0.5-8 Total Moles at equilibrium + 4 +8} = = 0.2 -48 +0.5-8 0.7 Thus, Mole fractions at equilibrium are: 0.2 - 45 Уснэсоон 0.7 Уснзон 0.5 - 85 YC4H802 YH2O YH2O

Here, An = v = -4 -8 + 4 + 8 = 0 (vi is stoichiometric coefficients) For equilibrium reactions, K*p-An = YC4H802* YH2O Үснэсоон •Үснзон Thus, K* po = (024) 65-86) K = 4-86 (0.2-45) (0.5-85) 3242 8 = (0.2-4. (0.5-86) On solving above equation, * = 0.0406 (Extent of Reaction at equilibrium) Thus, Moles at equilibrium are: CH3COOH = 0.2 - 45 CH3COOH = 0.5 - 87 CH3COOH = CH3COOH = 87 = = 0.2 - 4 + 0.0406 0.5 - 8+ 0.0406 4 * 0.0406 8 * 0.0406 0.0376 moles 0.1752 moles 0.1624 moles 0.3248 moles Mole Fractions at equilibrium are: YCH3COOH = 0.2-4 = 0.2 - 4. 0.0406 0.0537 0.7 0.5-8. 0.0406 уснзон 0.2503 YCH3CH YCAHBO2 VC4H802 - = = 05.05 e = = 0.5-8-0.0405 0.0406 8• 0.0406 4.0.0406 0.7 0.232 YH2O 0.464 0.7