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Characteristics of Equilibrium Constant

Kc for PCl5...

Question

$K_{c}$ for $P C l_{5 (g)} ⇌ P C l_{3 (g)} + C l_{2 (g)}$ , is $0.04$ at $250^{o} C$ . How many moles of $P C l_{5}$ must be added to a $3$ litre flask to obtain a $C l_{2}$ concentration of $0.15 M$ ?

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Solution

$P C l_{5}$ dissociates according to
$P C l_{5} \to P C l_{3} + C l_{2}$
Initial conc. : a(let) 0 0
Final conc : a(1-x) ax ax
where x is the degree of dissociation
$[C l_{2}]$ = 0.1 = ax ….......(i)
$k_{c} = ([P C l_{3}] [C l_{2}]) / [P C l_{5}]$
= (ax. ax)/a(1-x) = 0.0414
= a $x^{2}$ /(1-x) = 0.0414
= a $x^{2}$ =0.0414 { 1-x=1 , since x
= 0.1 x = 0.0414 { From (i) }
= x=0.414
From (i) ,
a = 0.1/0.414 = 0.2415
So, moles of $P C l_{5}$ added =0.2415 moles.

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