Naveen Kumar
Last Activity: 10 Years ago
First write the reaction:
C2H5OH......+......CH3COOH......................=.CH3COOC2H5.....+.....H2O
a mole...................a mole.......................................0........................0......(initially)
a-x......................a-x............................................x.................x..(At equilibrium)
Here equimolar of both the reactants are mixed. Let the number of mole of reactant is a, V is the volume of the solution and x is mole reacted till equilibrium is reached.
according to the question, x=2/3 of a=(2/3)*a
Hence, a-x=a-(2/3)a=a/3
[Here I have written the amount of water formed and it will also apear in the expression of equilibrium.]
So
![K=\frac{[H2O]*[CH3COOC2H5]}{[CH3COOH]*[C2H5OH]}](https://files.askiitians.com/cdn1/cms-content/common/latex.codecogs.comgif.latexk_frach2och3cooc2h5ch3coohc2h5oh.jpg)

Here put value of x=2a/3 and calculate the value of K