Question

A system at equilibrium contains I₂(g) at a pressure of 0.19 atm and I(g) at a pressure of 0.22 atm . The system is then compressed to half its volume.

Part A: Find the pressure of I2 when the system returns to equilibrium

Part B: Find the pressure of I when the system returns to equilibrium

Answer 1

The reaction is I2(g) ----> 2I(g)

Total pressure = 0.19+0.22= 0.41 atm

since the temperature and moles remained constant

P1V1= P2V2

V1= V ( initial) and P1= 0.41atm since V2 is halved =V/2

P2= 0.41*2=0.82 atm

K =equilibrium constant = [PI]²/ [PI2] =0.22*0.22/0.19=0.254

PI2 and PI refer to Partialp ressures of I2 and I respectively

for the second case K= [PI]²/ [PI2] = 0.254 = (x*P)²/ (1-x)*P= 0.254

x= mole fraction of I and 1-x =mole fraction of I2

P* x²/(1-x) =0.254

x²/(1-x)= 0.254/0.82

x²/(1-x)= 0.31

The equation can be solved using solver which gives x=0.423

So partial pressure of I =0.82*0.423=0.35 atm and partial pressure of I2= 0.83-0.35=0.48 atm