In the reaction COCl2(g) CO(g) + CI2(g) at 550°, when the initial pressure of CO & Cl2 are 250 and 280 mm of Hg respectively. The equilibrium pressure is found to be 380 mm of Hg. Calculate the degree of dissociation of COC12 at 1 atm. What will be the extent of dissociation, when N2 at a pressure of 0.4 atm is present and the total pressure is 1 atm.
CoCl2 (g) CO(g) + Cl2 (g)
I.Pr – 250 280
Eq.pr x 250–x 280–x
x + 250 – x + 280 – x = 380
x = 150
Kp = 0.114
Kp =
Kp =
0.114 =
In presence of N2 (constant pressure process)
Kp =
0.114 =
α =
α = 0.4
α–increases from 0.32 to 0.4.