The decomposition of `N_(2)O_(4)` to `NO_(2)` is carried out at `280^(@)C` in chloroform. When equilibrium is reached, `0.2` mol of `N_(2)O_(4)` and `2xx10^(-3)` mol of `NO_(2)` are present in a `2L` solution. The equilibrium constant for the reaction
`N_(2)O_(4) hArr 2NO_(2)` is

Options

A. `1xx 10^(-2)`
B. `2xx10^(-3)`
C. `1xx10^(-5)`
D. `2xx10^(-5)`

Answer

c

Explanation

`N_(2)O_(4) hArr 2NO_(2)`,
`[N_(2)O_(4)]=(0.2 "mol")/(2L)=10^(-1) mol^(-1)L`
`[NO_(2)]=(2xx10^(-3)mol)/(2L)=10^(-3)mol^(-1)L`
`K_(p)=(P_(NO_(2))^(2))/(P_(N_(2)O_(4)))=((0.4)^(2))/((1.2)^(2))xx(1.2)/(0.8)=(1)/(6)`