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Degree of Dissociation - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • 11 Questions around this concept.

Solve by difficulty

If the equilibrium constant for 

N_2(g) + O_2(g)\rightleftharpoons 2NO(g) is K,  the equilibrium constant for 1/2N_2(g) + 1/2O_2(g)\rightleftharpoons NO(g)  will be

Concepts Covered - 2

Degree of Dissociation

Degree of dissociation: It is the extent to which an electrolyte gets dissociated in a solvent. It is shown by ?.

\mathrm{\alpha=\frac{number \: of\: molecules \: dissociated}{total \: number\: of\: molecules }}

Degree of dissociation(?) depends on the following factors:

  • Nature of solute and solvent: For strong electrolytes, ? is more than that for weak electrolytes.
  • ? ∝ Dielectric constant of solvent
    That is, greater the dielectric constant of a solvent more will be ionization of electrolyte in it.
  • The degree of dissociation of weak electrolyte ∝ Dilution that is ? is maximum at infinite dilution.
  • ? ∝ 1/Concentration
  • ? ∝ Temperature
Observed Density and Molar Mass

In equilibrium, observed molar mass or average molar mass of the reactant is the total mass of the mixture divided by the total number of moles.

             \mathrm{A\: \rightleftharpoons \: nB}
Initially:  1            0
Equil:    1 - ?       n?

\\\mathrm{M_{observed}\: =\: \frac{Total\: mass\: of\: mix}{Total\: number\: of\: moles\: of\: mix}}\\\\\mathrm{M_{obs}\: =\: \frac{M_{real}}{1-\alpha +n\alpha }\: =\: \frac{M_{real}}{1+\alpha (n-1)}}

In the equilibrium system, the observed molar mass of the reactant is always different than the actual mass. Thus, when reaction is reversible, then observed mass vary. In a chemical reaction, some amount of this reactant gets convert into product, thus observed mass is different than actual mass.

For example:

\mathrm{N_{2}O_{_{4}}(g)\: \rightleftharpoons \: \: 2NO_{2}(g)}
In this reaction, original molar mass of N2O4 = 92g/mol. But thee observed molar mass at equilibrium is 80g/mol. The observed molar mass is less than original molar mass as during the reaction some amount of N2O4 is converted into NO2.

Vapour Density

Similarly, observed density of the substance is different than the actual density.

Thus, we know: 

Vapour density = Molar mass/2

\\\mathrm{Thus,\: 2\: x\: (V.D)_{obs}\: =\: \frac{2\, x\, (V.D)_{real}}{1+\alpha (n-1)}}\\\\\mathrm{\Rightarrow\: \: d\: =\: \frac{D}{1+\alpha (n-1)}}

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Degree of Dissociation
Observed Density and Molar Mass

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Degree of Dissociation

Physical Chemistry

Page No. : 769

Line : 1

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