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))`

Answer

(i)0.024 , (ii)6.4 , (iii)1681

Explanation

`K_p=(p_(NH_3))^2/((p_(N_2))xx(p_(H_2)))^3, K_p`=41
(i)`2NH_3(g) hArr N_2(g) + 3H_2(g)`
`K_p=1/K_p=1/41` =0.024
(ii)`1/2N_2(g) + 3/2 H_2(g) hArr NH_3(g)`
`K.._p=(p_(NH_3))/((p_(N_2))^(1//2)(p_(H_2))^(3//2))=(41)^(1//2)` =6.4
(iii)`2N_2(g) + 6H_2(g) hArr 4NH_3(g)`
`K..._p=(p_(NH_3))^4/((p_(N_2))^2 (p_(H_2))^6)=(41)^2` =1681