In the reaction COCl_2(g)hArr CO(g)+Cl_2(g) at 550^@C, when the initial pressure of CO & Cl_2 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 COCl_2 at 1 atm.What will be the extent of dissociation, when N_2 at a pressure of 0.4 atm is present and the total pressure is 1 atm.

In the reaction COCl_2(g)hArr CO(g)+Cl_2(g) at 550^@C, when the initia

1.	In the reaction COCl_2(g)hArr CO(g)+Cl_2(g) at 550^@C, when the initial pressure of CO & Cl_2 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 COCl_2 at 1 atm.What will be the extent of dissociation, when N_2 at a pressure of 0.4 atm is present and the total pressure is 1 atm.
Answer» 0.32 and no change 0.32 and 0.4 0.4 and 0.3 In presence of `N_2` DISSOCIATION cannot take place Solution :`{:(,CoCl_2(G)hArr,CO(g)+Cl_2(g)),("Initial PRESSURE",-,250""280),("Equilibrium pressure",x,250-x " " 280-x):}` x+250-x+280-x=380 x=150 `K_P=0.114` `K_P=(P_(CO).P_(Cl_2))/(P_(COCl_2))implies K_P=(alpha^2 . P)/(1-alpha^2) implies 0.114 =(alpha^2 . 1)/(1-alpha^2)` hence `alpha=0.32 " " alpha=0.32` In presence of `N_2` (CONSTANT pressure process) `K_P=(alpha^2xx0.6)/(1-alpha^2)implies alpha=sqrt(0.114/0.714)implies alpha=0.4` `alpha` increases from 0.32 to 0.4