Determine `K_(p)` for following reaction at same Temperature if A flask contains an equilibrium mixture of `I_(2)(g)` and I atomic (g) as
`I_(2)(g)hARr2l(g)`
The equilibrium pressure was 2.4 atm at constant volume and temperature. At constnat volume and Temperature if `I_(2)(g)` at a partial pressure of 3 atm is added to the new equlibrium pressure was 5.66 atm.

Answer

`0.4`

Explanation

`I_(2)(g)hArr2l(g),K_(p)=(p^(2))/(2.4-p)`
`2.4-p" "p`……i
After `I_(2)(g),P_(I_(2))=2.4-P+3=5.4-P`
As reactant concentration increases equilibrium favoured in forward direction.
`:.P_(I_(2))=5.4-P-P_(1)` and `P_(I)=P+2P_(1)`
`P_("total")=5.4-p-p_(1)+p+2p_(1)=5.4`
`+p_(1)=5.66`
`:.p_(1)=0.26` atm
`K_(p)=(p_(1))/(p_(2))=((p+0.52)^(2))/((5.14-p))`...........ii
`I=II=K_(p)`
`(P^(2))/(2.4-p)=((p+0.52)^(2))/(5.14-p)=(p^(2)+0.27+1.04p)/(5.14-p)`
`5.14p^(2)-p^(3)=2.4p^(2)+0.648+2`
`.496p-p^(3)-0.27p-1.04`
`3.78p^(2)-2.226p-0.648=0`
`p=(2.226+-sqrt((2.226)^(2)+4xx378xx0.648))/(2xx3.78)`
`=0.8` atm
`K=(P^(2))/(2.4-p)=((0.8)^(2))/1.6=0.4` atm