The equilibrium constant for the reacction `N_(2)(g)+O_(2)(g)hArr2NO(g)` at temperature (T) `4xx10^(-4)` The value of `K_(c)` for the reaction `NO(g)hArr(1)/(2) N_(2)(g)+(1)/(2)O_(2)(g) at the same temperature ?
`(1)/(2)F_(2)(g)hArrF(g)`

Options

A. `2.5xx10^(2)`
B. `0.02`
C. `4xx10^(-4)`
D. `50`

Answer

D

Explanation

`N_(2)(g)+O_(2)(g)hArr2NO(g): K_(C)=([NO]^(2))/([N_(2)][O_(2))=4xx10^(-4)`
`NO(g)hArr(1)/(2)N_(2)(g)+(1)/(2)O_(2)(g), K_(C)=([N_(2)]^(1//2)[O_(2)]^(1//2))/([NO])=(1)/(sqrtK_(C))=(1)/(sqrt4xx10^(-4))=(1)/(2xx10^(-2))=(100)/(2)=50`