For the reaction $$\ce{ C3Cl3N3O3 + 3 H2O -> C3H3N3O3 + 3 HClO3}$$ write the expression for $K_\mathrm{c}$ for this equilibrium.

The mark scheme says it would be $$K_c = \frac{[\ce{C3H3N3O3}][\ce{HClO3}]^3}{[\ce{C3Cl3N3O3}][\ce{H2O}]^3}$$

I thought $K_c$ shouldn't include water, is the mark scheme wrong? Or is this an exception that water is steam instead of liquid water?

How is one supposed to know that?

Here is another question I have just come across. (Different question, but still related.) Please take a look at the image below. Now, the mark scheme has seriously taken the concentration of $\ce{H2O}$ into account in (iii) and (iv).

screenshot of question


1 Answer 1


1. Reaction of Trichloroisocyanuric acid with water:

Water is omitted from the equilibrium expression only if it is a solvent in that reaction because it is a pure liquid. We can't increase the concentration of a pure liquid or pure solid and hence they are omitted from the expression.

However water in gaseous state cannot be ignored. It would be included in the equilibrium expression.

The above reaction is actually mixing of Trichloroisocyanuric acid in water to work as a disinfectant. So in the above reaction H2O is basically pure liquid (not gas).

Though in the expression water concentration is mentioned but in the later steps it would be substituted as 1. Actually some authors tend to write the formula of equilibrium constant including all the reactants & products concentrations (including pure solids and liquids) first and then in the next step they substitute 1 in the place of concentration of pure solids and liquids and other given numerical concentration values in place of other reactants and products.

So that expression is just a formula and in the next step when you would be substituting the values, 1 will be substituted inplace of concentration of water.

2. Coming on to the reaction you have sent in the picture i.e. reaction of aldehyde and alcohol:

Here water is not going to be omitted.

Okay water is always not neglected from the equilibrium expression. When I said water concentration is assumed to be 1, I had actually assumed water to be a solvent. Remember if water is a solvent in your reaction, then you can neglect the water concentration term but if water is not a solvent, then water term needs to be included. In the above question, water is being created in the process .i.e it is a product. If water had been a reactant as well as product (i.e. a solvent), then you could have neglected it.

Water in the above reaction is not present/taken from initial situation. It was eventually formed during the reaction and hence it is a product ( not a solvent). Hence you need to include it in the Kc expression.

Do you know what actually happens, when water is a solvent, its concentration term appears on reactant as well as on the product side. Hence while writing Kc expression it is cancelled out from the numerator and from the denominator. But if water is a product (i.e. not a solvent), then reactant part wont contain any concentration of water term and hence it can't be cancelled out in Kc expression .i.e. it will appear in the numerator of the Kc expression.

Hope it helps.

  • $\begingroup$ Hi! Thanks a lot for your help. I should have asked this in my question earlier, but I've just seen it. Could you take a look at the additional bold statement and the image below in my question, please? Now, this one has actually used the value of the concentration of H2O at equilibrium in the calculation. I think H2O here is still water not gas. (It's a different question from what's I asked before) Thank again! $\endgroup$
    – anonymous
    Commented Mar 18, 2017 at 10:32
  • 1
    $\begingroup$ Hey Thanks for the appreciation! Read the answer I have added. Read the second point. You will come to understand. Hope it helps $\endgroup$
    – Yb609
    Commented Mar 18, 2017 at 14:24
  • $\begingroup$ What is the definition of pure liquid? $\endgroup$
    – ado sar
    Commented Jan 8, 2019 at 13:02

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