The equilibrium constant for the reaction
`N_(2)(g)+O_(2)(g) hArr 2NO(g)`
at temperature T is `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 is

Options

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

Answer

B

Explanation

`N_(2(g))+O_(2(g))hArr2NO_((g))`
`K_(C)=([NO]^(2))/([N_(2)][O_(2)])=4xx10^(-4)`
`NO_(2)hArr1/2N_(2(g))+1/2O_(2(g))`
`K_(c)=([N_(2)]^(1//2)[O_(2)]^(1//2))/([NO])=(1)/(sqrt(Kc))=(1)/(sqrt(4xx10^(-4)))`
`=(1)/(2xx10^(-2))=100/2=50`