Question

pls answer the question with explanation. question is in attachment​

Answer 1

Answer:

Explanation:

Let P be the initial pressure of AB2​

The equilibrium pressures of AB2​, AB and B2​ are P(1−x),  xP and 0.5xP respectively.

The equilibrium constant expression is

Kp​=PAB2​2​PAB2​ PB2​​​

Kp​=[P(1−x)]2[xP]2×0.5xP​

Kp​=P[(1−x)]20.5x3​

Since the degree of dissociation is small compared to 1,1−x can be approximated to 1.

Kp​=P×0.5x3 .... (1)

The total pressure is

P(1−x)+Px+0.5Px=P(1+0.5x)=p

P=1+0.5xp​.....(2)

Substitute (2) in (1),

Kp​=1+0.5xp​×0.5x3

Since the degree of dissociation is small compared to 1,1+0.5x can be approximated to 1.

Kp​=p×0.5x3

x=(2Kp​/p)1/3

Hence, the expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure p is  x=(2Kp​/p)1/3

HOPE IT HELPS YOU...........