How to calculate the equilibrium constant for an esterification? I have tried to solve it algebraically, using the data I'm provided with, but still, I'm not capable of proving that in this particular case, the equilibrium constant is independent from the volume, maybe because it isn't; however, I don't know how can I possibly know the concentration of each substance in the reaction without the volume, so as to find the equilibrium constant.


  When a flask of constant capacity holds 1.2 mole each of acetic acid and ethanol with a catalyst and the temperature is kept at 25&nbsp;°C, 0.80 mole of ethylacetate is produced at equilibrium. Calculate the equilibrium constant of the following reaction:
  $$\ce{CH3COOH + C2H5OH &lt;=&gt; CH3COOC2H5 + H2O}$$
  (Answer option: 0.25, 0.64, 2.0, 4.0, 6.0)
It's given that this whole process is taking place at constant volume, so you can concentrate instead on what the concentrations of the species are (no pun intended) at equilibrium. The information is there, you just need to work through it.

You have been given initial conditions:
$[{\rm CH}_{3}{\rm COOH}] = [{\rm C}_{2}{\rm H}_{5}{\rm OH}] = 1.2\;{\rm mol}; [{\rm CH}_{3}{\rm COOC}_{2}{\rm H}_{5}] = [{\rm H}_{2}{\rm O}] = 0\;{\rm mol}$

and final conditions, where you are told that $[{\rm CH}_{3}{\rm COOC}_{2}{\rm H}_{5}] = 0.8\;{\rm mol}$. This means that at equilibrium $[{\rm H}_{2}{\rm O}] = 0.8\;{\rm mol}$. 

Your reactant equilibrium concentrations will be $1.2\;{\rm mol} - x$, where $x$ is given to you (I'm leaving that part to you. Hint: It's in the  sentence right above this one). Plug in your equilibrium values and you'll get ${\rm K}_{\rm eq}$ (also up to you).