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.
Kc = x2/ (a –x) Vx 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 atmand Kp = x2P/(1-x2) atmWhen 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 Vx α √V But, V α 1/Px α √1/PHere 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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