The esterification of acetic acid and ethanol is given by the reaction below: C2
ID: 1009706 • Letter: T
Question
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?
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?
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?
Explanation / Answer
C2H5OH(aq) + CH3COOH(aq)===>CH3COOC2H5(aq) + H2O(l)
I 1 2 0
C - 0.86 -0.86 0.86
E 0.14 1.14 0.86
Kc = [CH3COOC2H5]/[C2H5OH][CH3COOH]
= 0.86/0.14*1.14 = 5.388 >>>>> answer
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