equilibrium constant question ?

2016-08-05 4:09 am

The esterification of acetic acid and ethanol is given by the reaction below:

C2H5OH(aq) + CH3COOH(aq) ⇌ CH3COOC2H5(aq) + H2O(l)

When 1.00 mol of ethanol was mixed with 2.00 mol of acid in a 1.00 L flask, 0.86 mol of ester was formed at room temperature. What is the value of the equilibrium constant, Kc?

Please explain.

回答 (1)

胡雪8°

2016-08-05 7:06 am

✔ 最佳答案

Note that 1 L is the volume of the solution, but not the volume of the reaction mixture.

C₂H₅OH(aq) + CH₃COOH(aq) ⇌ CH₃COOC₂H₅(aq) + H₂O(l)

Initial amounts :
n(C₂H₅OH) = 1 mol
n(CH₃COOH) = 2 mol
n(CH₃COOC₂H₅) = n(H₂O) = 0 mol

Amounts at equilibrium :
n(C₂H₅OH) = (1 - 0.86) mol = 0.14 mol
n(CH₃COOH) = (2 - 0.86) mol = 1.14 mol
n(CH₃COOC₂H₅) = n(H₂O) = 0.86 mol

Let V L be the volume of the equilibrium mixture.

Kc = [CH₃COOC₂H₅] [H₂O] / [C₂H₅OH] [CH₃COOH]
= (0.86/V)² / {(0.14/V) × (1.14/V)}
= 0.86² / (0.14 × 1.14)
= 4.6

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