The value of `K_(p)` is `1xx10^(-3) atm^(-1)` at `25^(@)C` for the reaction: `2NO+Cl_(2)hArr2NOCl`. A flask contains `NO` at `0.02 atm` and at `25^(@)C`. Calculate the mole of `Cl_(2)` that must be added if `1%` of the `NO` is to be converted to `NOCl` at equilibrium. The volume of the flask is such that `0.2 mol e` of gas produce `1 atm` pressure at `25^(@)C`. (Ignore probable association of `NO` to `N_(2)O_(2)`.)

Explanation

`{:(,2NO,+,Cl_(2),hArr,2NOCl),("Pressure at t=0",0.02,,,,),("Pressure at eqm.",(0.02xx99)/(100),,P,,(0.02xx1)/(100)):}`
`:' K_(p)=([P_(NOCl)]^(2))/([P_(NO)]^(2)[P_(Cl_(2))])=((0.02xx0.01)^(2))/((0.99xx0.02)^(2)xxP)=1xx10^(-3)`
`:. P=0.102 atm`
For `Cl_(2)` in vessel, using `PV=nRT`
`0.102xxV=nxxRxxT` ....(`1`)
For a gas vessel,
`1xxV=0.2xxrxxT`....(`2`)
By eqs. (`1`) and (`2`),
`:. ` Mole of `Cl_(2)(n)` at equilibrium `=0.0204`
`2` atom of `NO` reacts with `1` atom of `Cl_(2)` (If `V`, `T` are constant)
`(0.02xx1 "atom")/(100)` of `NO` reacts with `(0.02xx1)/(2xx100)`
`=0.0001 "atom" ` of `Cl_(2)`
Thus, `0.0001xxV=n_(Cl_(2))xxRT`.....(`3`)
By eqs. (`2`) and (`3`),
`:. n_(Cl_(2))` which have reacted `=2xx10^(-5)`
`:. ` Total `Cl_(2)` added `=0.0204+2xx10^(-5)=0.02042`