In the esterification C2H5OH(l)+CH3COOH (l)leftharpoons CH3COOC2H5(l)+H2O(l) an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction 0.333 . The equilibrium constant is.

Question

In the esterification C2H5OH(l)+CH3COOH (l)leftharpoons CH3COOC2H5(l)+H2O(l) an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction 0.333 . The equilibrium constant is. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

As alcohol and acid are equimolar, so the initial concentration would also be the same. Suppose the initial concentration is 'C'. (x/(C - x) + (C - x) + x + x)=0.33=(1/3) 3x=2C x=(2 C/3) Kc=((2 C/3) × (2 C/3)/(C - (2 C)3) × (C - (2 C)3)=4

Q. In the esterification C2​H5​OH(l)+CH3​COOH(l)⇌CH3​COOC2​H5​(l)+H2​O(l) an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction 0.333 . The equilibrium constant is.

As alcohol and acid are equimolar, so the initial concentration would also be the same. Suppose the initial concentration is 'C'. (C−x)+(C−x)+x+xx​=0.33=31​ 3x=2C x=32C​ Kc​=(C−32C​)×(C−32C​)32C​×32C​​=4

Q. In the esterification $C_{2} H_{5} O H (l) + C H_{3} COO H (l) ⇌ C H_{3} COO C_{2} H_{5} (l) + H_{2} O (l)$ an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction $0.333$ . The equilibrium constant is.