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For reaction : PCl5(g) ---> PCl3(g) + Cl2(g) . The degree of dissociation of PCl5 is alpha at equilibrium pressure ‘P’.What is the equilibrium constant(Kp) for the reaction?

rahul mandviya , 7 Years ago
Grade 11
anser 2 Answers
Arun
PCl5 dissociates as:
PCl5 ⇌ PCl3 Cl2
If p is the degree of dissociation at certain temperature under atmospheric pressure, then
Initial concentration:
PCl5 = 1
PCl3 = 0
Cl2 = 0
At Equilibrium:
PCl5 = 1 – p
PCl3 = p
Cl2 = p
Total number of moles at equilibrium = 1 – p + p + p = 1 + p
Partial pressures of PCl5, PCl3 and Cl2 will be:
p (PCl3) = p p / 1 + p
p (Cl2) = p p / 1 + p
p (PCl5) = (1 – p) p / 1 + p
Kp = p (PCl3) X p (Cl2) / p (PCl5)
Kp = [(p p / 1 + p) X (p p / 1 + p)]/[ (1 – p) p / (1 + p)]
= p2 p / (1 – p) 2
Last Activity: 7 Years ago
ankit singh
        PCl5(g)   PCl3(g)  Cl2(g)
Initial number of moles         1          0          0
Number of moles reacted         x       -          -
Number of moles at equilibrium 1-x           x        x
Equilibrium concentration       1-x/V x/V    x/V
According to law of mass action,
 
 Kc = [PCl3][Cl2] / [PCl5]
Substituting  the  values  of  equilibrium  concentrations  in  the  above equation we get,                Kc = (x/V)(x/V)  / (1-x)/V
 = x2V/V2(1-x)
Last Activity: 5 Years ago
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