2

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?

Buttonwood
  • 29,590
  • 2
  • 45
  • 108
Param_1729
  • 61
  • 4
  • Related: https://chemistry.stackexchange.com/questions/14058/why-are-solids-and-liquids-not-included-in-the-equilibrium-constant-what-about – Yusuf Hasan Aug 09 '20 at 10:29
  • 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. – TheLearner Aug 09 '20 at 12:54

1 Answers1

2

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 $1$.

Maurice
  • 28,241
  • 3
  • 29
  • 61
  • 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. – TheLearner Aug 09 '20 at 12:27
  • 1
    @ 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 $\ce{NH_4Cl}$ – Maurice Aug 09 '20 at 12:37
  • 1
    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. – Maurice Aug 09 '20 at 12:43
  • @ 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? – Param_1729 Aug 09 '20 at 13:16
  • @ 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. – Maurice Aug 09 '20 at 13:52
  • @ Maurice. If we define concentration of solid as no. of moles/volume of container then it will change as the reaction proceeds it will not be constant. I want to know that why rate of forward reaction (from solid to gases) is not dependent on amount of solid. It seems like if we increase the amount of solid then the rate of forward reaction should increase and hence the amount of solid should alter the equilibrium but it doesn't. – Param_1729 Aug 09 '20 at 17:32
  • @ Param. The concentration of the solid is not the number of mole/volume of the container. No ! It is the number of moles /volume of the solid. It is exactly like Henry's law : the pressure in the gas phase is proportional to the concentration in the liquid phase (not in the whole container) – Maurice Aug 09 '20 at 18:30
  • @ Maurice. Is dissociation of solid into gases a surface phenomenon? Does the reverse i.e. conversion of gases into solid also occurs on the surface? If it is a surface phenomenon i.e. the rate of both the forward and backward reaction dependent on the surface area of solid then I can see why concentration of solids is defined the way it is. If the rate of reactions is not dependent on the surface area but on the amount of solid then my question still persist. – Param_1729 Aug 10 '20 at 04:10
  • @ Param. The rate of the reaction is dependent on the surface of the solid. But the equilibrium constant is not. The final state may be obtained sooner. Why don't we continue this private discussion in chat ? – Maurice Aug 10 '20 at 13:57
  • Let us continue this discussion in chat. – Param_1729 Aug 10 '20 at 18:33