Question
The equilibrium constant Kc equals 5.90 for the reaction
CH3COOH + C2H5OH <==> CH3COOC2H5 + H2O
Find the molar concentrations of acetic acid, ethanol, ethyl acetate, and water at equilibrium when .10 M acetic acid and .20 M ethanol are allowed to equilibrate.
I started like this:
I assumed 1 L of solution so I had
.10 mol CH3COOH
.20 mol C2H5OH
and in terms of moles at equilibrium...
CH3COOH= a - x
C2H5OH= b - x
CH3COOC2H5= c + x
H2O= d + x
where a, b, c, and d = moles initially, and x= moles lost
But I'm not really sure where to go from here. Any help would be much appreciated.
CH3COOH + C2H5OH <==> CH3COOC2H5 + H2O
Find the molar concentrations of acetic acid, ethanol, ethyl acetate, and water at equilibrium when .10 M acetic acid and .20 M ethanol are allowed to equilibrate.
I started like this:
I assumed 1 L of solution so I had
.10 mol CH3COOH
.20 mol C2H5OH
and in terms of moles at equilibrium...
CH3COOH= a - x
C2H5OH= b - x
CH3COOC2H5= c + x
H2O= d + x
where a, b, c, and d = moles initially, and x= moles lost
But I'm not really sure where to go from here. Any help would be much appreciated.
Answers
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