0
$\begingroup$

Teacher told us that the concentration term for pure liquids and solids is always $1$. But online I found this reaction:

$$\ce{CH3COOH(l) + CH3CH2OH(l) <=> CH3COOCH2CH3(l) + H2O(l)}$$

According to this page on LibreTexts (search for "esterification" to quickly find it),

$$K_c = \frac{[\ce{CH3COOCH2CH3}][\ce{H2O}]}{[\ce{CH3COOH}][\ce{CH3CH2OH}]}$$

Shouldn't $K_c = 1$ as all reactants and products are liquids? The above expression even contains $[\ce{H2O}]$.

$\endgroup$
4
$\begingroup$

The key point is pure liquids. In context of the esterification equilibrium, it considers the reactant concentration in the common solvent - which may or may not be one of the reactants.

Note that if the solvent is at the same time a reactant or a product, which concentration (activity) can be considered constant, then there can be defined the equilibrium constant for this solvent, with the solvent excluded from the equilibrium.

$$K_{aqua} = \frac{[\ce{CH3COOCH2CH3}]}{[\ce{CH3COOH}][\ce{CH3CH2OH}]}$$

$$K_{ethanol} = \frac{[\ce{CH3COOCH2CH3}][\ce{H2O}]}{[\ce{CH3COOH}]}$$

$$K_{ethylacetate} = \frac{[\ce{H2O}]}{[\ce{CH3COOH}][\ce{CH3CH2OH}]}$$

$$K_{acetic acid} = \frac{[\ce{CH3COOCH2CH3}][\ce{H2O}]}{[\ce{CH3CH2OH}]}$$

$\endgroup$
  • 1
    $\begingroup$ Doesn't (l) signify pure liquid as opposed to (aq) which is used for solutions? $\endgroup$ – user76377 Mar 30 at 16:48
  • 2
    $\begingroup$ generally yes, but not in the reaction equilibrium contexts, as all compounds are miscible, with exclusion of ethylacetate, in any ratio. By other words, if you take these 4 compounds in "(l)" pure state and shake them, they are not pure any more, but you get 1 or 2 liquid phases. Then the equilibrium gets established in each of the phases. $\endgroup$ – Poutnik Mar 30 at 16:51
  • 1
    $\begingroup$ I have encountered several examples where a reactant or product is excluded from the equilibrium constant expression because it was a solid or liquid (denoted by s and l respectively). Is there a rule to decide when to include liquids and when to not? $\endgroup$ – user76377 Mar 30 at 16:55
  • 1
    $\begingroup$ The rule is, that the compound exists in separated phase in pure form, so its state is constant,not depending of ongoing reactions. $\endgroup$ – Poutnik Mar 30 at 17:12
  • $\begingroup$ It is typical for the solubility constants. $\endgroup$ – Poutnik Mar 30 at 17:14
1
$\begingroup$

Your teacher is correct. But sometimes it is difficult to figure out when the "standard state" condition applies.

For the reaction

$$\ce{CH3COOH(l) + CH3CH2OH(l) <=> CH3COOCH2CH3(l) + H2O(l)}$$

You have missed the context. The starting reactants are acetic acid and ethanol. Some water and some ethyl acetate are produced. The assumption here is that the solution stays as one phase. But there is so little water that the solution is not an "aqueous solution."

If the reaction was happening in an aqueous solution then the mole fraction of water should predominate. The reaction could be denoted as such:

$$\ce{CH3COOH + CH3CH2OH <=>[H2O] CH3COOCH2CH3 + H2O}$$

or

$$\ce{CH3COOH(aq) + CH3CH2OH(aq) <=> CH3COOCH2CH3(aq) + H2O(aq)}$$

With water predominating, the equilibrium would be forced far towards the left.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.