The pressure of iodine gas at `1273 K` is found to be `0.112` atm whereas the expected pressure is `0.074` atm. The increased pressure is due to dissociation `I_(2) hArr 2I`. Calculate `K_(p)`.

Explanation

`overset(1)(I_(2)) Leftrightarrow overset(0)(2I) "Initial moles (suppose)"`
`"1-x 2x Moles at equilibrium"`
x is the degree of dissociation.
Total moles at equilibrium =1-x+2x=1+x
Since pressure is proportional to number of moles,
`=("experimental value of pressure")/("expected value of pressure")=(1+x)/1`
`or (0.112)/(0.074)=1+x, x=0.51`
`K_(p)=(p_(1)^(2))/(p_(I_(2)))=(((2x)/(1+x) xx p)^2)/((1-x)/(1+x) xx p)`
`or K_(p)=(4x^(2))/(1-x^2) xx p`
Putting x=0.51 and p =0.112 atm, we get
`K_(p)=0.1575`