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

If 0.2 mole of a gas produces 1 atm pressure at `25^@C` then n moles (say) of `Cl_2` under similar conditions shall produce 5n atm pressure.
Initial pressure : `"0.02 atm 5n atm 0"`
`2NO+Cl_(2) Leftrightarrow 2NOCl`
Eqb. Pressure `"(0.02-0.0002) (5n-0.0001) 0.0002"`
`K_(p)=(p_(NOCl)^2)/(p_(NO).p_(Cl_2)^2)=1 xx 10^(-3)`
`or =((0.0002))^2/((0.0198)^2 (5n-0.0001))=1 xx 10^(-3)`
`n=0.0204"mole of Cl"_2`