Why active mass of solids and pure liquids taken as unity

Question

There is a similar question already asked here and answers provided say that activity of solid is taken as unity because their density dosn't changes... but I still have some doubts in it.
Suppose we have a solid for example $ce{NH4Cl}$ which decomposes into gases $ce{NH3}$ and $ce{HCl}$ then $K_{mathrm{eq}} = [ce{NH3}][ce{HCl}]$.
Now if we take in a container which initially has no $ce{HCl}$ and $ce{NH3}$ and put some amount of $ce{NH4Cl}_mathrm{(s)}$ in it then after some time it will achieve equilibrium. Let the amount of solid left be $x$ mol. In equilibrium the  rate of production of $ce{NH4Cl}$ is equal to the rate of its decomposition.
Question:
Now if we add another $x$ mol of $ce{NH4Cl}_mathrm{(s)}$ in the container then the rate of production of $ce{NH3}$ and $ce{HCl}$ should be double and hence would exceed the rate of production of $ce{NH4Cl}_mathrm{(s)}$ which will alter the equilibrium but the equilibrium constant equation predicts this would not be the case. Where is the mistake?

Maurice · Answer

Maybe the following reasoning may help Param 1729. The thermal decomposition of $ce{NH_4Cl}$ may be pedagogically seen as occurring in two steps. First, solid $ce{NH_4Cl}$ is simply vaporized producing the hypothetical gaseous species $ce{NH_4Cl(g)}$

$$ce{NH_4Cl(s) <=> NH_4Cl(g)}label{rxn:nh4cl}tag{1}$$

And this transition is described by an equilibrium constant $ce{K_1}$ which is similar to Henry's law

$$ce{K_1 = frac{p(NH_4Cl)(g)}{[NH_4Cl(s)]}}$$

Immediately after this vaporisation phenomena, a gas phase decomposition happens with the following equation :

$$ce{NH_4Cl(g) <=> NH_3(g) + HCl(g)}label{run:nh3+hcl}tag{2}$$

with an equilibrium constant $ce{K_2}$ defined by :

$$ce{K_2 = frac{p(NH_3)·p(HCl)}{p{(NH_4Cl)(g)}}}$$

So what is really observed is the sum $(1) + (2)= (3)$ $$ce{NH_4Cl(s) <=> NH_3(g) + HCl(g)} label{run:dnh4cl}tag{3}$$ with a constant $ce{K_3}$ equal to

$$ce{K_3 = frac{p(NH_3)·p(HCl)}{[NH_4Cl](s)} = K_1·K_2}$$

This last expression shows that the pressure of the gases $ce{NH_3}$ and $ce{HCl}$ does not depend on the amount of $ce{NH_4Cl}$ in the solid state. And, as $ce{[NH_4Cl](s)}$ is a constant, it may be included in $ce{K_3}$ giving a new constant $ce{K_3'}$

$$ce{K_3' = p(NH_3)·p(HCl) = K_1·K_2·[NH_4Cl](s)}$$ This is "equivalent" to stating that the concentration of $ce{NH_4Cl(s)}$ is equal to $1$.

Why active mass of solids and pure liquids taken as unity

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