Why active mass of solids and pure liquids taken as unity 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?
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 $\ce{NH_4Cl(g)}$
$$\ce{NH_4Cl(s) &lt;=&gt; NH_4Cl(g)}\label{rxn:nh4cl}\tag{1}$$
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 :
$$\ce{NH_4Cl(g) &lt;=&gt; NH_3(g) + HCl(g)}\label{run:nh3+hcl}\tag{2}$$
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 $(1) + (2)= (3)$ $$\ce{NH_4Cl(s) &lt;=&gt;  NH_3(g) + HCl(g)} \label{run:dnh4cl}\tag{3}$$ 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 &quot;equivalent&quot; to stating that the concentration of $\ce{NH_4Cl(s)}$ is equal to $1$.