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Question

Derive the expressions Kc and Kp for decomposition of PCl5.

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Solution

The dissociation equilibrium of PCl5 in gaseous state is written as PCl5(g)PCl3(g)+Cl2(g)
Let 'a' moles of PCl5 vapour be present in V, liters initially. If x moles of PCl5 dissociate to PCl3 and Cl2 gases at equilibrium then, the molar concentrations of PCl5, PCl3 and l2 gases at equilibrium will be axV,xV and xV respectively.
Kc=[PCl3][Cl2][PCl5]
=xV×xVaxV
=x2V2×Vax
Kc=x2(ax)V
x= degree of dissociation
x=NumberofmolesdissociatedTotalnumberofmolespresentinitially
Kc=x2(1x)V
[ initially the no. of PCl5 present=1 mole ]
=x2P(1x)RT [PV=RT]
In terms of partial pressure of PCl5,PCl3 and Cl2, then
Kp=pPCl3pCl2pPCl5atm
Kp=x2P1x2
atm.

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