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)
A. A
For the reaction: NH_3(g) 1 2 N_2(g) + 3 2 H_2(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.
Related: Chemistry — Chemical Equilibrium · All PYQ Banks