For the following equilibrium N_(2)O_(4) hArr 2NO_(2) K_(c)=0.67. If we start with 3 moles of NO_(2) and 1 mole of N_(2)O in1L flask, then NO_(2)present at equilibrium is :

For the following equilibrium N_(2)O_(4) hArr 2NO_(2) K_(c)=0.67. If w

1.	For the following equilibrium N_(2)O_(4) hArr 2NO_(2) K_(c)=0.67. If we start with 3 moles of NO_(2) and 1 mole of N_(2)O in1L flask, then NO_(2)present at equilibrium is :
Answer» 1.5 mol 2.0 mol 0.5 mol 1.0 mol Solution :`{:(,N_(2)O_(4),hArr,2NO_(2),),("Initial",1" mol",,2" mol",),("At eqm.",m(1-x)"mol",,(3+2x)"mol",):}` `K_(c)=((3+2x)^(2))/(1-x)=0.67` `(3+2x)^(2)=0.67(1-x)` `4x^(2)+9+12x=0.67-0.67x` `4x^(2)+12.67x+8.33=0` `x=(-12.67+-sqrt((12.67)^(2)-16xx8.33))/(8)` `=(-12.67+-sqrt(160.53-133.28))/(8)` `=(-12.67+-sqrt(27.25))/(8)` `=(-12.67+-5.22)/(8)` Taking `+ve` SIGN `x=(-17.89)/(8)=-2.24` Taking `+ve` sign `x=-0.93` As `x CANCEL(=)-2.24"":.x=-0.93` `:. `Moles of `N_(2)O_(4)` at eqm `=1-x` `=1+0.93=2("Approx.")`