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I am a high school student and I am very confused in Equilibrium expression, My confusion is that "Why we don't write concentration of solids and pure liquids in equilibrium expression?" Many people says that its because their concentration don't changes but how does it related to this? If concentration is same for them then what's the issue in it? I am not able to picturize how does it affect the equilibrium expression if their concentration remain the same?

please Help me regarding this using simple language that I can understand because I am at high school level. If anyone could help me to picturize this thing in my mind then it would be very helpful.

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  • $\begingroup$ Well, it doesn't, so it can be ignored, as simple as that. $\endgroup$
    – Mithoron
    Nov 10, 2020 at 15:20
  • $\begingroup$ Well, you can either define an equilibrium constant with the formal constant concentration of a pure condensed phase in mol/dm3, or you can implicitly involve them in the equilibrium constant value, pretending their concentration is 1 mol/dm3. Why way is more convenient ? The same for the pressure variant of the equilibrium constant. $\endgroup$
    – Poutnik
    Nov 10, 2020 at 15:22
  • $\begingroup$ why we have to pretend that their concentration is 1mol/dm^3,.it can be of any value.. $\endgroup$ Nov 10, 2020 at 15:28
  • $\begingroup$ We do not have to. Neither we have to go along stairs forwards. We can go backwards as well. Multiplication or division by 1 is very convenient. You can omit it totally. $\endgroup$
    – Poutnik
    Nov 10, 2020 at 15:35

1 Answer 1

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Suppose you want to study the equilibrium of the reaction of a metal like zinc $\ce{Zn}$ with water to produce $\ce{ZnO(s)}$ and $\ce{H2}$ :$$\ce{Zn(s) + H2O(g) <=> ZnO(s) + H2(g)}$$ At high temperature, the equilibrium constant may be written $$K_\mathrm{1}=\frac{[\ce{ZnO}]·p\ce{(H2)}}{[\ce{Zn}]·p\ce{(H2O)}}$$ In this expression three parameters are not changing : the constant $K_\mathrm{1}$, and the two concentrations of the solid stuffs $\ce{Zn}$ and $\ce{ZnO}$ in their own phases. The mass of these stuffs may change but not their concentrations, as they are in the solid phase. So it is no use keeping such a formula where three parameters are constant. It is better to group all constant parameters on the same side of the = sign, and define a new constant like $K_\mathrm{2}$ : $$K_\mathrm{2} = K_\mathrm{1}\frac{[\ce{Zn}]}{[\ce{ZnO}]}= \frac{p\ce{(H2)}}{p\ce{(H2O)}}$$ With this formula and a numerical value for $K_\mathrm{2}$ you may calculate immediately how $p\ce{(H2)}$ changes when $p\ce{(H2O)}$ is varied. No need to know the concentration of the solids. This may be useful if you want to optimize the yield in producing $\ce{H2}$ gas through this reaction.

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  • $\begingroup$ so this new constant include all the constants in it, this also creates a problem in my mind that the equilibrium constant tells us whether the concentration of product is high or low than the reactant at equilibrium, but if we are including the concentration of solid in this new constant then we would have get some higher value for Keq. which we MIGHT not get if the a reaction has some solid reactants in it, so it creates a new problem , $\endgroup$ Nov 11, 2020 at 8:28
  • $\begingroup$ I don't see your point. The ratio of the concentrations of Zn in metallic zinc and ZnO in the solid ZnO is a constant, which should not be important, and probably between $0.1$ and $10$. I don't speak of the amounts, in mass or in moles. I just speak about the concentrations. $\endgroup$
    – Maurice
    Nov 11, 2020 at 9:18
  • $\begingroup$ I am not only talking about this reaction, I am talking about those reactions in which those solids are involved which have a higher concentration. $\endgroup$ Nov 12, 2020 at 8:41
  • $\begingroup$ Whatever the nature of the solids appearing in a chemical reaction, their concentrations stay constant $\endgroup$
    – Maurice
    Nov 12, 2020 at 11:13
  • $\begingroup$ I know their concentration stay constant, but what I am saying is that, if we consider their concentration to be the part of Keq. itself and do not count them in the expression then how can we predict that at equilibrium the concentration of product is high or reactant is high? we should consider them in the expression. $\endgroup$ Nov 15, 2020 at 8:26

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