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

Suppose intial pressure of `Cl_(2)` added is p atm .
`{:("Then ",2NO,+,Cl_(2),hArr,2 NOCl),("Intial",0*02 "atm",,p " atm",,),("A eqm. " ,0*02 - (0*02)/100,,(p-(0*01)/100),,(0*02)/100),(,=2 xx 10^(-2)-2 xx10^(-4),,=p-10^(-4),,-2 xx 10^(-4)"atm"),(,=2 xx 10^(-4)(100-1),,,,),(,=198 xx10^(-4) "atm",,,,):}`
` K_(p) = (p_(NOCl)^(2))/(p_(NO)^(2)xx p_(Cl_(2)) )`
` 10^(-3) = ( 2xx 10^(-4))^(2)/ ((198 xx 10^(-4))^(2)xx(p-10^(-4))) `
or ` (p- 10^(-4)) = 4/(198)^(2) xx 1/ (10^(-3))= 0* 102 or p= 0* 102 + 0* 0001 = 0* 1021 "atm "`
Volume of the vessel can be calculated as follows :
` PV = nRT or V = (nRT)/P = (0* 2 xx 0* 082 xx 273 )/1 L = 4* 887 L `
To calculate the number of moles of `Cl_(2)` , again apply
` PV = nRT or n = (PV)/ (RT) = (0*1021 xx 4* 887 )/(0*082 xx 298) = 0* 204 "mol"`