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What to do with the Solids and Liquids in Equilibrium Constant Expressions

posted Dec 7, 2016, 12:01 AM by Grace Ong   [ updated Dec 7, 2016, 12:19 AM ]
The following post was first posted on Blogger on Saturday, 13 July 2013.



In a previous post, I discussed why it is not appropriate to assume the molar concentrations of solids and liquids to be one in equilibrium constant expressions.

What should we do with the solids and liquids in equilibrium constant expressions then?

Before I attempt to answer this question, let's review the Equilibrium Law:

For a general chemical reaction aA + bB ⇌ cC + dD, we can define an equilibrium constant, Kc, where

equilibrium constant expression for a general reaction

when the reaction is at equilibrium at a certain temperature.

Let's see how this law is applied to different equilibria.

Case 1: Homogeneous Equilibria

This is where all species are present in the same phase, and the application of the Equilibrium Law is straight-forward: the concentrations of all species present in the equilibrium are included in the Kc expression.

For example,
N2(g)+ 3H2(g) ⇌ 2NH3(g)
equilibrium constant expression for the reaction between nitrogen and hydrogen (Haber Process)


CH3CO2H(l) + CH3CH2OH(l) ⇌ CH3CO2CH2CH3(l) + H2O(l)
equilibrium constant for the esterification reaction between ethanoic acid and ethanol

Case 2: Heterogeneous Equilibria

This is where the species are present in different phases. These include reactions that involve solids and gases, or solids and liquids. For these cases, the concentrations of solids are omitted from their Kc expressions.

For example,
I     C(s)+ H2O(g) ⇌ CO(g) + H2(g)
equilibrium constant expression for steam-carbon reaction - without solid concentration


II    Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH(aq)
equilibrium constant (solubility product) expression for solubility equilibrium of magnesium hydroxide
 
Now, isn't this the same as assuming the concentrations of solids to be 1 mol dm−3?

It may seem so, but the answer is no. If we apply the Equilibrium Law directly to Reactions I and II, their Kc expressions should include the concentrations of the solid reactants:
I     C(s)+ H2O(g) ⇌ CO(g) + H2(g)
Kc expression for steam-water reaction - with solid concentration


II    Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH(aq)
Kc expression for the solubility equilbrium of magnesium hydroxide - with solid concentration

However these solid concentrations are meaningless in Kc expressions. As long as there is some solid present in the equilibrium, whether it is present in a small or large quantity does not affect equilibrium. We can therefore re-define a new Kc expression without the solid concentrations:

For Reaction I: C(s) + H2O(g) ⇌ CO(g) + H2(g)
equilibrium constant expression for carbon and steam reaction - without concentration of solid
where [C(s)] is a constant that can be calculated from the density of carbon used in the reaction

Similarly, for Reaction II: Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH(aq)

equilibrium constant expression for solubility equilibrium of magnesium hydroxide - without solid concentration
where [Mg(OH)2(s)] is a constant that can be calculated from the density of magnesium hydroxide

Note:
  1. The molar concentration of a solid may be calculated from its density as follows:

    how to determine concentration of a solid

  2. The Kc/ for Reaction II is known as the solubility product of magnesium hydroxide, and is represented by the symbol, Ksp.

Case 3: Aqueous Equilibria

This is where all species are dissolved in water. By convention, only concentrations of the species involved in the equilibrium are included in the Kc expression; the concentration of the solvent, i.e. water, is excluded, even if the latter takes part in the equilibrium as a reactant or a product.

For example, ethanoic acid reacts with water as follows:

III    CH3CO2H(aq) + H2O(l) ⇌ CH3CO2(aq) + H3O+(aq)

As long as the solution is dilute, the amount of water present in the solution is hardly affected by the reaction with ethanoic acid.

Because the concentration of water remains essentially the same before and after the equilibrium is reached, we can re-define the Kc expression for Reaction III as follows:
simplifying the equilibrium constant expression for the dissociation of ethanoic acid in water

Note:
  1. The concentration of pure water was calculated using the density of water (1.00 g cm−3) as follows:

    converting density of water to concentration of water

  2. The Kc/ for Reaction III is known as the acid dissociation constant for ethanoic acid, and is represented by the symbol Ka.
  3. We can similarly define the base dissociation constant, Kb, for the reaction between a weak base and water:

    B(aq) + H2O(l) ⇌ BH+ + OH(aq)
    base dissociation constant expression

  4. While the concentration of water is generally omitted from Kc expressions for aqueous equilibria involving water (either as a reactant or a product), you should not apply this omission across the board without first considering what the equilibrium constant is used for. I shall illustrate some exceptions with examples in my next post.
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