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` in 1L flask, then `NO_(2) ` present at equilibrium is `:`

Options

A. 1.5 mol
B. 2.0 mol
C. 0.5 mol
D. 1.0 mol

Answer

B

Explanation

`{:(,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.")`