For the following gas phase equilibrium reaction at constant temperature, NH ₃( ~g ) 1 / 2 ~N ₂( ~g )+3 / 2 H ₂( ~g ) if the total pressure is 3 ~atm and the pressure equilibrium constant ( K _ p ) is 9 atm, then the degree of dissociation is given as (x 10⁻² )^ -1 / 2 . The value of x is _ _ _ _ . (nearest integer)

Step-by-step solution

For the reaction: NH₃(g) 1 2 N₂(g) + 3 2 H₂(g) At t = 0 : 1 mole, –, – At equilibrium: 1- , 2 , 3 2 Total moles = 1 + K_P = ( 2 )^ 1/2 ( 3 2 )^ 3/2 (1- ) ( P_T 1+ )^1 Given P_T = 3 atm and K_P = 9 atm: 9 = ( 2 )^ 1/2 ( 3 2 )^ 3/2 (1- ) (3)^ 1/2 1+ Simplifying: 9 = 9 ( 2 )^2 1- ^2 1 - ^2 = ^2 4 5 ^2 4 = 1 ^2 = 0.8 = (0.8)^ 1/2 = ( 1 0.8 )^ -1/2 = (125 10⁻²)^ -1/2 So x = 125 .