# Law of Mass Action and Heterogeneous System

Law of Mass Action and Heterogeneous System:

(i) Dissociation of calcium carbonate $(CaCO_3)$:

If $CaCO_3$ is heated in a closed vessel, a portion of it usually decomposes forming CaO (solid) and $CO_2$ (gas) and ultimately an equilibrium is set up according to the equation; $\underset{\text{solid}}{CaCO_3} \leftrightharpoons \underset{\text{solid}}{CaO} + \underset{\text{gas}}{CO_2}$

The active mass of a solid is constant at a definite temperature and independent of the amount of solid present.

We may, therefore, regard it as a homogeneous equilibrium in the gas phase, in which solids are present with a constant partial pressure or active mass. $[\text{Solid}] = P_{\text{solid}} = 1$ convention (at all temperatures)

The active mass of all solids is equal to unity at all temperatures. Now applying the law of mass action to the above heterogeneous system: $K = \dfrac{P_{CaO}\times P_{CO_2}}{P_{CaCO_3}} \\[3mm] \text{or} \hspace{3mm}P_{CO_2} = \dfrac{K.P_{CaCO_3(s)}}{P_{CaO (S)}} = K$

The very interesting and unexpected result comes out that the pressure of $CO_2$ is constant at constant temperature independent of the amount of $CaCO_3$ present in the system.

At every temperature, the system will attain its corresponding dissociation pressure and then it is impossible by any means, even by adding to the system, lime, $CaCO_3 \text{or} CO_2$ from outside to change the dissociation pressure.

(ii) Dissociation of ammonium hydrosulphide $(NH_4.HS)$:

When solid ammonium hydrosulphide is heated, it dissociates forming the gaseous products, ammonia and hydrogen sulphide. $\underset{\text{solid}}{(NH_4)HS} \leftrightharpoons \underset{\text{gas}}{NH_3} + \underset{\text{Gas}}{H_2S}$

According to law of mass action, $\dfrac{P_{NH_3} \times P_{H_2S}}{P_{(NH_4HS)}} = K \\[3mm] \text{but} \hspace{3mm} P_{(NH_4HS) \hspace{2mm} \text{Solid}} \hspace{2mm} = 1\\[3mm] \therefore P_{NH_3} \times P_{H_2S} = K$

Hence in the gaseous mixture the product of the partial pressure of ammonia and hydrogen sulphide is constant at constant temperature. This relationship has been experimentally verified.

(iii) Dissociation of salt hydrates $(CuSO_4.5H_2O)$:

Copper sulphate (blue vitriol) if kept in a desiccator connected to a manometer dissociates as follows: $\underset{\text{Solid}}{CuSO_4.5H_2O} \leftrightharpoons \underset{\text{solid}} + \underset{\text{Gas}}{2H_2O}$

Manometer indicates the constant pressure as long as the temperature is constant. This dissociation is quite similar to the thermal dissociation of $CaCO_3$. $CaCO_3 \leftrightharpoons CaO + CO_2$

Similarly, here the pressure of water vapour in the system would be constant. $\dfrac{P^2_{H_2O} \times P_{CuSO_4. 3H_2O}}{P_{CuSO_4. 5H_2O}} = K \\[3mm] \text{or}, P_{H_2O} = \text{Constant}$

Calculation of degree of dissociation from vapour density:

Let us consider the dissociation of AB, which occurs as follows: $\underset{\underset{1- x} 1}{AB} \leftrightharpoons \underset{\underset{x} 0}{A} + \underset{\underset{x} 0}{B}\underset{\underset{\hspace{10mm} \cdots (\text{at equilibrium})} \cdots (\text{Initial Conc.})}{}$

Now, since vapour density is inversely proportional to volume, therefore, $D \propto \dfrac{1}{V}$

And $d \propto \dfrac{1}{(1 + x) V}$

where ‘D’ and ‘d’ are the densities of the reactants before and after equilibrium. Thus, $\dfrac{D}{d} = 1 + x \\ \text{or} \hspace{3mm} x = \dfrac{D- d}{d}$

If 1 mole of a substance dissociates into ‘n’ ions, then the above equation may be written as: $x = \dfrac{D- d}{(n- 1)d}$

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