Derive the values of K_(p) and K_(C) for dissociation of PCl_(5) .

1.	Derive the values of K_(p) and K_(C) for dissociation of PCl_(5) .
Answer» 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 ^(Deltan_(g))` `Deltan_(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)` `K_(p) = (x^(2) xx PV)/((a - x) V ( a + x))` `K_(p) = (x^(2) P)/((a - x) (a + x))`

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