Chemistry Assignment Help with Degree of Dissociation

5.8 Degree of Dissociation

Degree of dissociation is the fraction of a mole of the reactant that underwent dissociation. It is represented by 'a'.

molar concentration

5.9 Relation between Vapour Density and Degree of Dissociation

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The relation between vapour density and the degree of dissociation can be established only for a gaseous equilibrium who’s KP exists. For example,

law of mass action2

Total concentration at equilibrium = c – ca + nca = c [1 – a + na] = c [1 + a (n – 1)]

Assuming that all the gaseous components at equilibrium behave ideally, we can apply ideal gas equation.

PV = nR = reversible reactions

equilibrium constant

\ V.D = gas equation …(i) [since molar mass = 2 ´ V.D]

As pressure of the system is given by, equilibrium, so putting the value of P in equation (i) gives

V.D = chemistry help

where r is the density of the gas or gaseous mixture expressed in g/litre.

If the equilibrium reaction is established in a closed vessel, then vapor density will be inversely proportional to the number of moles of the gaseous species as the density of the gaseous mixture (r) is a constant.

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Let the initial vapour density and vapour density at equilibrium be ‘D’ and ‘d’ respectively, then for the given equilibrium

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or chemistry equations

or equilibrium reactions

\ pressure-equilibrium2

where ‘n’ represents the number of moles of gaseous product given by 1 mole of the gaseous reactant.
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