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8.CHEMICAL EQUILIBRIUM eLearn.Punjab

Let ‘a’ moles of PCl5 present initially are decomposed by ‘x’ moles. So, at equilibrium stage, ‘a-
x’ moles of PCl5 are left behind while ‘x‘ moles of PCI3 and ‘x’ moles of Cl2 are produced. If the volume
of equilibrium mixture is ‘V’ dm3, then

PCl5(g)  PCl3 (g) + Cl2 (g)

‘a’moles ‘O’moles ‘O’moles t = 0sec

(a - x) moles ‘x‘ moles ‘x‘ moles t = teq
Dividing the number of moles by total volume of reactants and products at equilibrium.

 a-x  moles dm-3   x  moles dm-3 +  x  moles dm-3
 V   V   V 

Since Kc = [PCl3][Cl2 ]
PCl5

Putting the concentrations at equilibrium

X . X
V V
Kc = (a-x)

V

Simplifying the right hand side, we get

K c = x2
V(a-x)

The final expression is not independent of the factor of volume. So, the change of volume
at equilibrium stage disturbs the equilibrium position of the reaction. We will discuss this reaction
in Le-Chatelier’s principle with reference to effect of volume change and its effect on change of
equilibrium position.

iii. Decomposition of N2O4 (gaseous phase reaction)

Similarly, for decomposition of N2O4 (g). the expression of Kc involves the factor of volume.

N2O4 (g)  2NO2 (g)

K c = 4x 2
(a-x)V

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