`K_(c)` for `N_(2)O_(4)(g) hArr 2NO_(2)(g)` is `0.00466` at `298 K`. If a `1-L` container initially contained `0.8` mol of `N_(2)O_(4)`, what would be the concentrations of `N_(2)O_(4)` and `NO_(2)` at equilibrium? Also calculate the equilibrium concentration of `N_(2)O_(4)` and `NO_(2)` if the volume is halved at the same temperature.

Options

A. `9 atm^(-1)`
B. 9 atm
C. `4.5 atm^(2)`
D. 10 atm

Answer

B

Explanation

`N_(2)O_(4) Leftrightarrow 2NO_(2) K_(P)=(P_(NO_(2))^(2))/(P_(N_(2)O_(4))`
`2" 3"`
total moles =5