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

Answer

C

Explanation

`K = ([NO_(2)]^(2))/([NO_(2)O_(4)]) = ([2xx(10^(-3))/(2)]^(2))/([(.2)/(2)]) = (10^(-6))/(10^(-1)) = 10^(-5)`