N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)), The K_p of this reaction is 35 at 500 K temp. Calculate K_p of following reaction at this temp. (i)4NH_(3(g)) hArr 2N_(2(g)) + 6H_(2(g)) (ii)1/2N_(2(g)) +3/2H_(2(g)) hArr NH_(3(g))

N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)), The K_p of this reaction is 35 a

1.	N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)), The K_p of this reaction is 35 at 500 K temp. Calculate K_p of following reaction at this temp. (i)4NH_(3(g)) hArr 2N_(2(g)) + 6H_(2(g)) (ii)1/2N_(2(g)) +3/2H_(2(g)) hArr NH_(3(g))
Answer» Solution :Given equation x 2 `2(N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)))(K_p=35)^2` `therefore 2N_(2(g)) + 6H_(2(g)) hArr 4NH_(3(g)) , K._p =(35)^2` Backward equation of this equation = `[4NH_(3(g)) hArr 2N_(2(g)) + 6H_(2(g)) ]` = equation (i) `therefore K_p` (i) of eq. (i) `=1/(K._p)^2=1/(35)^2=8.16xx10^(-4)` .... (Given equation x `1/2`) `=1/2(N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)))` `=(1/2N_(2(g)) + 3/2H_(2(g)) hArr NH_(3(g)))` = eq.(II) `therefore K_p` of eq.(ii) = `"(K of given equation)"^(1/2)` `=(35)^(1/2)`=5.916