Dissociation equilibrium of PCl5

Dissociation equilibrium of PCl5 :

Dissociation equilibrium of PCl5 in the gaseous state is written as,
PCl5(g) ↔ PCl3(g) + Cl2(g)

Assume ‘a’ moles of PCl5 vapor is present in ‘V’ litres initially. When x moles of PCl5 dissociate to PCl3 and Cl2 gases at equilibrium constant ‘V’ litres, then molar concentrations of PCl5, PCl3 and Cl2 gases at equilibrium will be:

(a – x)/ V, a/V, x/V correspondingly.

141_dissociation of pcl5.jpg

 Kc = x2/ (a –x) V

x is also termed as the degree of dissociation that symbolizes the fraction of total moles of reactant dissociated.
x = (Number of moles dissociated)/(Total number of moles present initially)

If at first 1 mole of PCl5 is existed then,

Kc = x2/(1–x)V = x2P/(1-x) RT

In terms of partial pressures of PCl5, PCl3 and PCl2 then,

Kp = (P PCl3. PCl2)/ P PCl5 atm
and Kp = x2P/(1-x2) atm

When the degree of dissociation is little as compared to unity, then (1–x) is around equivalent to 1.0.

∴ Kc = x2/V or
x2 = Kc x V
x α √V  But, V α 1/P
x α √1/P

Here x is small; degree of dissociation differs inversely as the square root of pressure or varies directly as the square root of the volume of system.


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