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)`.

Answer

A

Explanation

`{:(1,,0,"Initial mole (suppose)"),(I_(2),hArr,2I,),(1-x,,2x,"mole at equilibrium"):}`
x is the degree of dissociation.
Total moles at equilibrium `=1-x+2x=1+x`
Since pressure is proportional to the number of moles,
`("Experimantal value of pressure")/("Expected value of pressure")=(1+x)/(1)`
or `0.112/0.074=1+x rArr x=0.51`
`K_(p)=p_(l)^(2)/p_(l_(2))=(((2x)/(1+x)xxp)^(2))/(((1-x)/(1+x)xxp))`
or `K_(p)=(4x^(2))/(1-x^(2))xxp`
Putting `x=0.51` and `p=0.112 "atm"`, we get
`K_(p)=0.1575`.