For the following gas phase equilibrium reaction at constant temperature, NH3(g) 1/2 N2(g) + 3/2 H2(g) if the total pressure is 3 atm and the pressure equilibrium constant (Kp) is 9 atm, then the degree of dissociation is given as (x 10-2 )-1/2. The value of x is . (nearest integer)
Answer
125
Solution
Let α = degree of dissociation (start with 1 mol NH3):
At equilibrium: NH3 = 1 - , N2 = 2, H2 = 3 2; total = 1 +
Kp = pN21/2 pH23/2pNH3 = ( 2 )1/2 ( 3 2 )3/21 - P1 +
Kp = 3 3 24 P1 - 2
Put P = 3 atm, Kp = 9 atm:
9 = 9 24(1 - 2)
2 = 4 - 4 2
2 = 45 = 0.8
Given = (x 10-2 )-1/2
2 = 100x = 0.8
x = 125
Answer: 125