Derive the value of K_(P and K_(C) for dissociation of PCI_(5).

1.	Derive the value of K_(P and K_(C) for dissociation of PCI_(5).
Answer» <P> Solution :Consider that .a. moles of `PCl_(5)` is taken in container of volume .V. LET X moles of `PCl_(5)` be dissociated into x moles of `PCl_(3)` and x moles of `Cl_(2)`. Applying law of mass action `K_(C) = ([PCl_(3) ][Cl_(2)])/([PCl_(5)]) = (((x)/(V))((x)/(V)))/((a-x)/(V)) = (x^(2))/((a-x) V)` `K_(P)` Calculation `K_(P) = K_(C) .RT^(Delta n_(g)) , Delta n_(g)= 2-1=1` We know that PV= nRT `RT= (PV)/(n)` Where .n. is the TOTAL number of moles at equilibrium `n= a-x + x+x= a+x` `K_(P) = (x^(2))/((a-x) V) .(PV)/(n) rArr K_(P) = (x^(2) xx PV)/((a-x) V(a+x))` `K_(P) = (x^(2)P)/((a-x) (a+x))`

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