Question

Ab dissociates as 2ab=2a+b2 when initial pressure becomes 625mm when equilibrium is attained calculate kp for reaction assuming volume remains constant

Answer 1

Considering the equation :
2AB---->2A + B2
Initial 500. 0 0
final. 500-P. P P/2
total pressure=625=500-P+ P+ P/2
P =250 mm
Pab=250
Pa=250
Pb2=125
[tex]Kp=\frac{Pb2}\times Pa^{2}}{Pab}[\tex]
putting all the values in the equation
[tex]Kp=\frac{125}\times 250^{2}}{250}[\tex]
Kp=125mm

Answer 2

Hey dear,

◆ Answer-
Kp = 125

◆ Explaination-
# Given-
Initial pressure = 500 mm Hg
Total final pressure = 625 mm Hg

# Solution-
Conversion of pressure in reaction occurs as-
Reaction 2ab ---> 2a + b2
Initially 500 0 0
Finally 500-2p 2p p

Total pressure is given by-
625 = 500-2p + 2p + p
p = 125 mm Hg

Therefore,
p(ab) = 500-2×125 = 250 mm Hg
p(a) = 2p = 250 mm Hg
p(b2) = p = 125 mm Hg

Kp = p(a)^2.p(b2) / p(ab)^2
Kp = 250^2 × 125 / 250^2
Kp = 125

Therefore value of Kp = 125.

Hope this helps...