For `I_(2)(g) hArr 2I(g), K_(p)=1.79xx10^(-10)`. The partial pressure of `I_(2)=1.0` atm and `I=0.5xx10^(-6)` atm after `50 min`. Comment on the status of equilibrium process.

Answer

A, C, D

Explanation

`{:(,I_(2)(g),hArr,2I(g),K_(p)=1.79xx10^(-10)),(t=t,1,,0.5xx10^(-6),),(("Partial P"),,,,):}`
`Q_(P)=P_(I)^(2)/P_(I_(2))=((0.5xx10^(-6))^(2))/(1)=2.5xx10^(-11)`
As `Q_(P) lt K_(p)`, reaction is moving in forword direction.