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Arrive at the expressions of K_(P) and K_(C) for the dissociation of PCl_(5). |
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Answer» Solution :Consider that 'a' moles of `PCl_(5)` is taken in a container of volume V. Let 'x' moles of `PCl_(5)` be dissociated into x moles of `PCl_(3)` and x moles of `Cl_(2)`. `PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)`   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)` The EQUILIBRIUM CONSTANT `K_(P)` can also be calculated as follows : We KNOW the relationship between the `K_(C)` and `K_(P)` `K_(P)=K_(c)(RT)^((Deltan_(g)))` Here the `""Deltan_(g)=n_(P)-n_(r)=2-1=1` Hence `K_(P)=K_(C)(RT)` 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))/((a-x)V)(PV)/((a+x))=(x^(2)P)/((a-x)(a+x))` |
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