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)]) = (1)/(sqrt(K_(c))`
`= (1)/(sqrt(4xx10^(-4)))`
`= (1)/(2xx10^(-2)) = (100)/(2) = 50`