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Q.

The dissociation equilibrium of a gas AB2 can be represented as2AB2(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

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a

2KP/P1/3

b

2KP/P1/2

c

KP/P

d

2KP/P

answer is A.

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Detailed Solution

2AB2(g)   ⇌  2AB(g)  +  B2(g)   1                     0                  01 - x                   x                  x/2∴          KP = x2.x21-x2 . P1+x2 = x3. P21-x  ≈ 1  and 1 + x2 ≈ 1,  since x < < 1Or         P = 2KPP3
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