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In a heterogeneous reaction (where the states are varied) we do not include liquids and solids in the equilibrium equation because their concentrations do not change.

E.g. Chemguide.co.uk

Heterogenous

However, when it is a homogenous equation we DO include solids and liquids.

E.g. Chemguide.co.uk enter image description here

1) Why do we include liquids and solids in the Kc equation in homogeneous equilibrium equations? I get that if we didn't there would be nothing on the right hand side of Kc=.... , but why is it physically different to the case where there are varied phases?

2) What if there was an equation that involved only liquids and solids (if this is possible)? As this is a heterogeneous equation would we still not include solids and liquids? If so how would you write the Kc equation where there is nothing but solids and liquids?

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  • $\begingroup$ Because experimentally solids and liquids in the first case hardly affect the rate of the reaction. $\endgroup$
    – DHMO
    Commented Sep 10, 2016 at 5:04
  • $\begingroup$ Thanks for the response. So in the case where it is say all liquids (or all solids) the rate of reaction would be effected by the liquids and solids since there is nothing else right? Does that mean if we have a reaction which includes both liquids and solids (but not gases or aqueous solutions) then we would include both the liquids and solids in the equilibrium constant, even though it is a heterogeneous equilibrium equation? $\endgroup$
    – K-Feldspar
    Commented Sep 10, 2016 at 5:08
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    $\begingroup$ According to the website you cited, "you don't include any term for a solid in the [heterogeneous] equilibrium expression". $\endgroup$
    – DHMO
    Commented Sep 10, 2016 at 5:12
  • $\begingroup$ Thanks for pointing that out. Is there any reason why liquids would be different? $\endgroup$
    – K-Feldspar
    Commented Sep 10, 2016 at 5:14
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    $\begingroup$ Because liquids are fluid whose concentration still affects the rate of effective collisions. $\endgroup$
    – DHMO
    Commented Sep 10, 2016 at 5:16

1 Answer 1

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Chemguide is simplified for A-levels and therefore in this case is strictly speaking incorrect.

The equilibrium constant $K$ is defined as a product of activities. I described this in a previous answer here.

The crux of the matter is that the activity of a pure solid or pure liquid is equal to 1, which means that it can be omitted from the expression for $K$ without affecting the value.

In your first reaction

$$\ce{H2O(g) + C(s) <=> H2(g) + CO(g)}$$

the chunks of carbon in the reaction are necessarily pure because they don't mix with the gases.

In the second reaction (yes, I am lazy, please feel free to edit for me)

$$\ce{EtOAc(l) + H2O(l) <=> AcOH(l) + EtOH(l)}$$

none of the liquids are pure, hence their activities deviate from unity.

Just as a final example, in the dissociation of a weak acid

$$\ce{HA(aq) + H2O(l) <=> H3O+(aq) + A-(aq)}$$

water is omitted from the expression for $K_\mathrm{a}$ because water, as the solvent, is in large excess over $\ce{HA}$ and is therefore effectively "pure".

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  • $\begingroup$ I don't like the statement "The crux of the matter is that the activity of a pure solid or pure liquid is equal to 1, which means that it can be omitted from the expression for K without affecting the value." In the reaction shown for carbon and water the carbon doesn't necessarily need to be "pure." Rather I think it is better to say that the equilibrium is based on the gas phase and since the carbon is in a different phase (a solid) the equilibrium is independent of the amount of carbon present - be it 1 milligram or 1 metric ton. $\endgroup$
    – MaxW
    Commented Sep 11, 2016 at 4:02
  • $\begingroup$ @MaxW The equilibrium constant uses activities, not amounts. You are of course right that the amount of carbon doesn't affect the equilibrium but I felt that the direct link is that it doesn't affect the activity of the solid, hence doesn't affect equilibrium constant. $\endgroup$
    – orthocresol
    Commented Sep 11, 2016 at 4:10
  • $\begingroup$ The point that I'm trying to make is that new students to chemistry haven't probably heard the word "activity." If K-Feldspar had any notion of what activity was this question wouldn't have been asked. I think explaining in terms of phases wouldn't introduce a new concept. $\endgroup$
    – MaxW
    Commented Sep 11, 2016 at 4:17
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    $\begingroup$ Fair enough, feel free to add your own answer - I'd +1. $\endgroup$
    – orthocresol
    Commented Sep 11, 2016 at 4:21
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    $\begingroup$ @user8718165 I'm sorry, but I don't entirely get what you're asking. The equilibrium constant $K$ and the dissociation constant $K_\mathrm{a}$ are both defined in terms of activities, and the activity of the solvent is 1, as mentioned above. So in no circumstances do we actually "set" a concentration to be equal to 1. The only reason why concentrations come into play is because the activities of the solutes can be approximated by their (molar) concentration. So we can write [HA], [H+], [A-] as approximations for their activities (notwithstanding units). But you should not write [H2O]. $\endgroup$
    – orthocresol
    Commented Dec 8, 2019 at 15:28

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