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

`{:(,N_(2)O_(4),hArr,2NO_(2)),("Initial",1,,0),("Final",1-x,,2x),(,1-x=0.2 "mol",,2x=2xx10^(-3) "mol"):}`
`[N_(2)O_(4)]=0.2/2=0.1 M [NO_(2)]=(2xx10^(-3))/(2)=10^(-3) M`
`K=([NO_(2)]^(2))/([N_(2)O_(4)])=((10^(-3))^(2))/0.1=10^(-5)`