For the reactions, `N_(2(g))+3H_(2(g))hArr2NH_(3(g))`. At `400 K`, `K_(p)=41 atm^(-2)`. Find the value of `K_(p)` for each of the following reactions at the same temperature:
(`i`) `2NH_(3(g))hArrN_(2(g))+3H_(2(g))`,
(`ii`) `(1)/(2)N_(2(g))+(3)/(2)H_(2(g))hArrNH_(3(g))`,
(`iii`) `2N_(2(g))+6H_(2(g))hArr4NH_(3(g))`

Explanation

(i) It is the reverse of the given reaction. Hence , `K_(p) = 1/41`
(ii) It is obtained by dividing the given equation by 2. Hence, `K_(p) = sqrt(41)`
(iii) It is obtained by multiplying the given equation by 2. Hence, `K_(p) = (41)^(2)`.