If equilibrium constant for the reaction N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) at 298 K is 2.54 , the value of equilibrium constant for the reaction (1)/(2) N_(2) + (3)/(2) H_(2) hArr NH_(3) will be ……..

If equilibrium constant for the reaction N_(2) (g) + 3 H_(2) (g) hArr

1.	If equilibrium constant for the reaction N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) at 298 K is 2.54 , the value of equilibrium constant for the reaction (1)/(2) N_(2) + (3)/(2) H_(2) hArr NH_(3) will be ……..
Answer» Solution :`K_(P)` for `N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) = ((P_(NH_(3)))^(2))/(P_(N_(2)) xx (P_(H_(2)))^(3)) = 2.54` `K.._(P)` for `(1)/(2) N_(2) (g) + (3)/(2) H_(2) (g) hArr NH_(3) = ((P_(H_(3))))/((P_(N_(2)))^((1)/(2)) xx (P_(H_(2)))^((3)/(2)))` SINCE the REACTION is halved `K.._(p) = SQRT(K._(p)) = sqrt((2.54)) = 1.59`