Question

Two moles of `PCl_(5)` were heated to `327^(@)C` in a closed two-litre vessel, and when equilibrium was achieved, `PCl_(5)` was found to be `40%` dissociated into `PCl_(3)` and `Cl_(2)`. Calculate the equilibrium constant `K_(p)` and `K_(c)` for this reaction.

Answer 1

Calculation of `K_(c)` for the reaction.
Number of moles of `PCl_(5)` initially present =2 mol
Perentage of `PCl_(5)` dissociated =40%
Number of moles of `PCl_(5)` dissociated `=(2xx40)/(100)=0.8 mol`
number of moles of `PCl_(5)` left =2-0.8 =1.2 mol
According to the available data , 0.8 mole of `PCl_(5)` will dissociate to form 0.8 mole of `PCl_(3)` and 0.8 mol of `Cl_(2).`
Therefore , molar conc.`//` litre of the reactants and products before the reaction and at the equilibrium point is :
`PCl_(5)(g) hArr PCl_(3)(g) +Cl_(2)(g)`
`"Initial moles"// "litre" " " 2/2 " "0" "0`
`"Moles"// "litre at the equilibrium point "" "2-0.8=(1.2)/(2)=0.6 " "(0.8)/(2)=0.4" "(0.8)/(2) =0.4`
Applying Law of chemical equilibrium ,
Equilibrium constant `(K_(c)) =((0.4 mol L^(-1)) xx (0.4 mol L^(-1)))/((0.6 mol L)) = 0.267 mol L^(-1)`
(II) Calcualtion of `K_(p)` for the reaction
`" We know that"" "K_(p) =K_(c)(RT)Delta^(ng) , K_(c) 0.267 mol L^(-1) , Delta^(ng) =2 -1=1,`
`R =0.082 L atm K^(-1) mol^(-1) , T =327^(@)C = (327 +273) = 600 K`
By substituting the values in the relation,
`K_(p) =(0.267 mol L^(-1)) xx (0.082 L atn K^(-1) xx 600 K)`
`=13.14 atm`