The equilibrium constant for the reaction, `N_(2(g)) + O_(2(g)) hArr 2NO_((g))` at temperature T is `25 xx 10^(-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. 20
B. 50
C. `4 xx 10^(-4)`
D. `2.5 xx 10^(-2)`

Answer

B

Explanation

`N_(2) + O_(2) hArr 2NO`
`K_(c_(1)) = ([NO]^(2))/([N_(2)][O_(2)]) = 4 xx 10^(-4) " " NO hArr (1)/(2)N_(2) + (1)/(2)O_(2)`
`K_(c_(2)) = ([N_(2)]^(1//2)[O_(2)]^(1//2))/([NO]) " "K_(c_(2)) = sqrt((1)/(K_(c_(1)))) = sqrt((1)/(4 xx 10^(-4))) = 50`