0.1 mole of `N_(2O_(4)(g)` was sealed in a tude under one atmospheric conditions at `25^(@)C` Calculate the number of moles of `NO_(2)(g)` preesent , if the equilibrium `N_(2)O_(4)(g)hArr2NO_(2)(g)(K_(P)=0.14)` is reached after some time :

Explanation

` {: (,N_(2)O_(4)(g),hArr,2NO_(2)(g),) ,("Intial amounts",0.1 "mole",,0,), ("At equilibrium " ,(0*1 -x),,2x," Total " 0*1 +x "moles"):}`
As P=1 atm, ` p_(N_(2)O_(4)) = (0*1 -x)/(0*1 +x), p_(NO_(2))= 2/(0*1 + x) atm `
` K_(p) - (p_(N_(O_(4)))^(2))/p_(N_(2)O_(4)) = (2x // (0*1 + x))^(2)/((0*1 -x) //( 0*1 +x))= (4x^(2))/((0*1- x) ( 0*1 +x) )= (4x^(2))/(0*01 - x^(2) )`
` :. (4x^(2))/(0*01 - x^(2) = 0*14 )or 4*14x^(2) = 0*0014 or x = 0* 018 `
` :. " No. of moles of " NO_(2) " at equilibrium " = 2x = 2 xx 0* 018 = 0* 036 "mole" `