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Question

Derive the expressions for Kc and Kp for the decomposition of PCl5.

Solution
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The dissociation equilibrium of PCl5 is represented as
PCl5(g)PCl5(g)+Cl2(g)
Let 'a' moles of PCl5 be present in 'V' litres initially. If x moles of PCl5 dissociate to PCl3 and Cl2 then molar concentrations of PCl5, PCl3 and Cl2 at equilibrium will be
axV,xV and xV respectively.
Kc=[PCl3][Cl2]e[PCl5]2
Kc=xV×xVaxV=x2V2×V(ax)
x2(ax)V
x is known as degree of dissociation
x=No.ofmolesdossociatedTotalno.ofmolespresentintitially
if initially 1 mole of PCl5 is present then
Kc=x2(1x)V=x2P(1x)RT(V=RTP)
In terms of partial pressures of PCl5, PCl3 and Cl2
Kp=PPCl5×PCl2PPCl5atmKp=x3p(1x)2atm
If x is small compare to unity, then (1x) is approximately equal to 1.0
Kc=x2V or x2=Kc×V
=xαV
=xα1/P(Vα1P)
Where x is small, degree of dissociation varies inversely as the square root of P (or) varies directly as the square root of volume of the system.

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Q5

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