Why active mass of solids and pure liquids taken as unity
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 $1$.