The dissociation equilibrium of a gas -AB-2- can be represented as- 2A- B - 2 -left- g right- rightleftharpoons 2ABleft- g right- - B - 2 -left- g right- The degree of dissociation is x and is small compared to 1- The expression relating the degree of dissociation x with equilibrium constant -K-p- and total pressure p is- left- 2- - K - p - - p - right- - - 1 - 3 - -left- - - K - p - - p - right- - left- 2- - K - p - - p - right- - - 1 - 2 - -left- 2- - K - p - - p - right-

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Toppr Admin · Accepted Answer

Let P be the initial pressure of AB2The equilibrium pressures of AB2- AB and B2 are P-1-x2212-x- -xA0-xP and 0-5xP respectively-

The dissociation equilibrium of a gas AB2 can be represented as: 2AB2(g)⇌2AB(g)+B2(g)The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation x with equilibrium constant Kp and total pressure p is:(2Kp/p)1/3(Kp/p)(2Kp/p)1/2(2Kp/p)

Let P be the initial pressure of AB2The equilibrium pressures of AB2, AB and B2 are P(1−x), xP and 0.5xP respectively.

Let $P$ be the initial pressure of $A B_{2}$
The equilibrium pressures of $A B_{2},$ $A B$ and $B_{2}$ are $P (1 - x),$ $x P$ and $0.5 x P$ respectively.