What does pure liquids and pure solids mean in chemical equilibrium, why active mass of pure liquids is also zero? [closed]

Question

Background

The professor who taught us chemical equilibrium defined active mass as:

$$\mathrm{[activity~coefficient (\lambda)×[conc]~or~ [Partial~Pressure]}$$

and mentioned that, for the scope of our course, $\lambda=1$.

Questions

1. What do "pure solids" and "pure liquids" mean in this context?

More specifically, what qualifies as "impure" in this context? I’ve gone through several textbooks, but none explicitly discuss this part.

2. Why do pure solids and pure liquids have an active mass equal to 1?

Some books address this question with thermochemical or mathematical proofs, but I find them hard to follow at my current level of understanding. I did come across an explanation for why solids are excluded from the equilibrium constant that gave me a good intuitive understanding:

More solid means a faster forward reaction. At the same time, it means a faster reverse reaction (more surface area for iodine to deposit on). For that reason, the equilibrium constant does not change. In the example above with the sugar, finely granulated sugar dissolves faster than coarsely granulated sugar, but the solubility (and the equilibrium constant) remains the same.

This explanation helped me understand solids better, but it does not provide a similar intuition for liquids. Could someone provide a comparable intuition for liquids?

3. Equilibrium constant expressions

For the reaction:

$$\ce{2 H2O⇋H3O+ + OH-}$$

Is the equilibrium constant $(K_{C})$ correctly written as: $$K_{C}=\frac{ \ce{[OH-] [H3O+]}}{ \ce{[H2O]^2}}=\frac{ \ce{[OH-][H3O+]}}{1}= \ce{[OH-][H3O+]} ?$$

And, for the hypothetical reaction $$\ce{2A_(l) + 4C_(s) ⇋ B_(g) + 2D_(s) + 3E_(aq)}$$ $$K_{C}=\frac{ \ce{[B]^{1}[D]^{2}[E]^{3}}}{ \ce{[A]^{2}[C]^{4}}}= \ce{[B][E]^{3}} ?$$

4. Why is it said that $\ce{[H2O]}$ is approximately constant in dilute solutions but not in concentrated solutions?

I've often seen the statement that in dilute solutions, $\ce{[H2O]}$ is treated as constant, but I’m unsure why this is the case. Since water remains a liquid in both dilute and concentrated solutions, shouldn’t its active mass be considered constant in both cases? Why is the assumption about water's concentration only made for dilute solutions?

Answer 1

Solids/liquids

It is better to say “a pure compound in solid or liquid phase”. In our context, an impure compound is such a compound that noticeably differs in some properties from the pure compound due to the presence of other dissolved compounds. If the impurities are just dispersed without being dissolved, the phase of the major compound still contains the pure compound, as impurities form their own phase.

Active mass of a pure compound in a condensed phased (solid or liquid)

The active mass is usually expressed in the context of thermodynamics as the thermodynamic activity $a$, defined as:

$$\mu = \mu^\circ + RT \ln a$$

Where $\mu$ is the molar Gibbs energy of the compound, called the chemical potential.

$$\mu = \left(\frac{\partial G}{\partial n_i}\right)_{T,p,n_j, j \ne i}$$

$\mu^\circ$ is then the standard molar Gibbs energy = standard chemical potential, considered at the compound’s standard state. That is for condensed compounds in their pure form, usually in their most stable or best defined form.

This implies that pure liquids are considered as having $a=1$ by convention, as then $\mu = \mu^\circ$.

Generally, for ideal solutions, the activity is equal to the molar fraction, and

$$\mu = \mu^\circ + RT \ln{a} = \mu^\circ + RT \ln{x}$$

OTOH, the standard chemical potential for compounds dissolved in water, like ions, is frequently formally defined in such a way, that

$$\lim_{c \to 0}{a} = c$$

For liquids, a larger phase surface obviously leads to faster absolute rates of chemical or physical processes, with equilibrium constants remaining the same. This is not fully true in all cases, as smaller droplets of volatile liquids have slightly bigger vapor tension, compared to a flat liquid surface.

Equilibrium constant expressions

True thermodynamic equilibrium constants are expressed in terms of activity $a$ or fugacity $f$. Generally, activity/fugacity is the molar fraction/partial pressure of a compound, it would have needed, if it had been behaving ideally and still had the same chemical potential.

So for the water auto-dissociation:

$$K_\text{w} = \frac{a\left(\ce{H3O+(aq)}\right)a\left(\ce{OH-(aq)}\right)}{[a(\ce{H2O(l)})^2]}$$

As $a(\ce{H2O(l)})=1$ for pure water:

$$K_\text{w} = a\left(\ce{H3O+(aq)}\right)a\left(\ce{OH-(aq)}\right)$$

As for very dilute ions, conventionally $a\left(\ce{H3O+(aq)}\right) = [\ce{H3O+(aq)}]$:

$$K_\text{w} = [\ce{H3O+(aq)}][\ce{OH-(aq)}]$$

And yes, for the hypothetical reaction you mentioned, $K_c = [\ce{B}][\ce{E}]^3$.

Water activity is about constant for diluted solution, but not for concentrated ones

I would provide a simple real-life mechanical analogy:

• You have a compressible coiled metal spring (water).
• It has its standard length without applied force (pure water activity - active mass).
• You compress it very little, so its length contraction is negligible and you can still consider its original length (diluted solution, water activity change can be considered negligible and pure water activity is taken).
• You compress it a lot, so its length contraction is significant and cannot be neglected (concentrated solution, water activity changed significantly and the value for pure water cannot be used as an approximation).

Answer 2

I would like to answer the question $4$, about the concentration of water in diluted aqueous solutions. The concentration of water will be calculated in two different concentrations of $\ce{HCl}$ in order to show that the differences in water concentrations are negligible.

In the next calculations, the molar mass of water is taken as exactly $18$ g/mol, and $1$ liter water weighs exactly $1000$ g.

Let's first consider $\ce{HCl}$ $0.01$ M. One liter of this solution contains $\pu{0.01 36.45 = 0.3645}$ g $\ce{HCl}$. One liter of this solution weighs $1000$ g (from the Handbook of Chemistry) and contains $1000 - 0.3645 = 999.635 $g $\ce{H2O}$. In mole, the concentration of water is $\pu{\frac{999.635}{18}}$ = $55.53$ M.

Let's now consider $\ce{HCl}$ $0$%, or pure water. The concentration of water in pure water is $\ce{\frac{1000}{18}} = 55.55$ M.

As a consequence, the water concentration changes from $55.55$ M in pure water to $55.53$ M in $\ce{HCl}$ $0.01$ M. For all $\ce{HCl}$ concentrations smaller than $0.01$ M, the concentration of water does not differ significantly from $55.54$ M or $55.55$ M. The differences between these values are negligible.