In a 0.25 L tube dissociation of 4 mol of NO is taking place. If its degree of dissociation is 10%. The value of K​P​ for the reaction `2NO_((g)) to N_(2(g)) + O_(2(g)) ` is :-

Options

A. `(1)/((18)^(2))`
B. `(1)/((8)^(2))`
C. `(1)/(16)`
D. `(1)/(32)`

Answer

A

Explanation

`{:(,2NO" "hArr" "N_(2)" "+" "O_(2),alpha=10%),(t=0,4-0.4" " 0.2" " 0.2,),(,3.60" "0.2" "0.2,),(Deltan=0,,):}`
`Delta n =0`,
`:. K_(p) = K_(c), K_(c) = ((0.2//V)^(2))/((3.6//V)^(2)) = (4)/(36xx36)`