Verified Solution • 2025–26 Session moderate

QAIPMT-2008-Ch-2230: NEET 2008 Chemistry - Class 11 AIPMT-2008-Ch-2230 Solution: The dissociation equilibrium of a gas \(AB_2\) can be…

The dissociation equilibrium of a gas AB2AB_2 can be represented as: 2AB2(g)⇌2AB(g)+B2(g)2AB_2(g)\rightleftharpoons 2AB(g)+B_2(g). The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant KpK_p and total pressure P is

Multiple Choice Options

  • A (2Kp/P)1/2(2K_p/P)^{1/2}
  • B (Kp/P)(K_p/P)
  • C (2Kp/P)(2K_p/P)
  • D (2Kp/P)1/3(2K_p/P)^{1/3}

Step-by-Step Verified Solution

2AB2(g) | ⇌ | 2AB(g) | + | B2(g) Initial mole | 2 | 0 | 0 At equilibrium | 2(1 - x) | 2x | x Amount of moles at equilibrium = 2(1 � x) + 2x + x = 2 + x K p = [ p A B ] 2 [ p B 2 ] [ p A B 2 ] 2 = [ 2 x 2 + x _ P ] 2 [ x 2 + x _ P ] [ 2 ( 1 _ x ) 2 + x _ P ] 2 = [ 4 x 3 2 + x _ P ] 4 ( 1 _ x ) 2 _ K p = 4 x 3 _ P 2 _ 1 4 ( _ 1 � x � 1 and 2 + x � 2) _ x = ( 8 K p 4 P ) 1 3 _ x = ( 2 K p P ) 1 3

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