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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?

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  • $\begingroup$ Related: chemistry.stackexchange.com/questions/14058/… $\endgroup$ Aug 9, 2020 at 10:29
  • $\begingroup$ It will be good if someone can actually give explanation that why his argument (based on relative rates) is wrong, but not just telling why the correct concept is correct, unless contradiction between the two is apparent. I always have the same question too, why won't more gas produced as the equality between forward and backward rate got broken. $\endgroup$
    – TheLearner
    Aug 9, 2020 at 12:54

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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 $1$.

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  • $\begingroup$ Maurice, sorry but I don't get how the last expression shows that the pressure of gases does not depend on the amount of ammonium chloride in solid state. $\endgroup$
    – TheLearner
    Aug 9, 2020 at 12:27
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    $\begingroup$ @ The99sLearner. The concentration $\ce{[NH_4Cl]}$ in the solid state does not depend on the amount of salt in the container. It is the same in $1$ gram or in $10$ grams pure $\ce{NH_4Cl}$ $\endgroup$
    – Maurice
    Aug 9, 2020 at 12:37
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    $\begingroup$ It is exactly the same for water. The concentration of $\ce{H_2O}$ in pure water is $55.55$ mol/L in any sample of water, big or small. $\endgroup$
    – Maurice
    Aug 9, 2020 at 12:43
  • $\begingroup$ @ Maurice. Why is the concentration of solid taken as unity? Why do we take its concentration proportional to its density? Why don't we define its concentration to be its no. of moles/volume of container? $\endgroup$
    – Param_1729
    Aug 9, 2020 at 13:16
  • $\begingroup$ @ Param. Of course you may define the concentration of the solids as you did. But this concentration never changes, whatever the experiment. It is equal to $28.5$ M. So why bother about a useless constant ? You better introduce it into $\ce{K_3}$ to make $\ce{K_3' }$ = $28.5$ $\ce{K_3}$, which is a useful constant. This K'3 can be immediately used for studying mixtures of HCl and NH3 gases. $\endgroup$
    – Maurice
    Aug 9, 2020 at 13:52

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