A system at equilibrium contains I2(g) at a pressure of 0.16 atm and I(g) at a pressure of 0.20 atm. The system is then compressed to half its volume. a)Find the pressure of I2 when the system returns to equilibrium. b)Find the pressure of I when the system returns to equilibrium.

Sol:-

I₂ (g)  <----> 2 I (g)
given

At equilibrium

p[I2] = 0.16 atm and p [I] = 0.20 atm

Now expression of K_p is :
Kp = p[I]² / p[I2]
= (0. 20)² / (0.16)
= 0.25
When volume is half the equilibrium will shift towards the side with less number of moles i.e. reactants side. When volume is half, pressure is doubled. therefore , we have ICE table is :
I₂ (g) <----> 2 I (g)

I 0.32 atm 0.40 atm
C + x - 2x

E ( 0.32 + x) atm ( 0.40 - 2x ) atm

again expression of k_p is :

k_p = ( 0.40 - 2x )² / 0.32 + x

0.25 = ( 0.40 - 2x )² / 0.32 + x
on solving

we have x = 0.0483

therefore

p[I2] = 0.32 + x = 0.32 + 0.0483 = 0.3683 atm

and

p[I] = 0.40 - 2x = 0.40 - 2(0.0483) = 0.40 - 0.0966 = 0.3034 atm
Hence

p[I2] = 0.3683 atm and

p[I] = 0.3034 atm