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NEET CHEMISTRYEquilibriumMedium

Question

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 'xx' and is small compared to 1. The expression relating the degree of dissociation (xx) with equilibrium constant KpK_p and total pressure pp is:

A

(2Kp/p)(2K_p/p)

B

(2Kp/p)1/3(2K_p/p)^{1/3}

C

(2Kp/p)1/2(2K_p/p)^{1/2}

D

(Kp/p)(K_p/p)

Step-by-Step Solution

Let the initial moles of AB2AB_2 be 1. The dissociation reaction is: 2AB2(g)2AB(g)+B2(g)2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g) Initial moles: 1, 0, 0 Moles at equilibrium: 1x1-x, xx, x/2x/2 Total number of moles at equilibrium = 1x+x+x/2=1+x/21 - x + x + x/2 = 1 + x/2 Partial pressures of the species at equilibrium: pAB2=1x1+x/2×pp_{AB_2} = \frac{1-x}{1+x/2} \times p pAB=x1+x/2×pp_{AB} = \frac{x}{1+x/2} \times p pB2=x/21+x/2×pp_{B_2} = \frac{x/2}{1+x/2} \times p The equilibrium constant KpK_p is given by: Kp=(pAB)2(pB2)(pAB2)2K_p = \frac{(p_{AB})^2 (p_{B_2})}{(p_{AB_2})^2} Kp=(x1+x/2p)2(x/21+x/2p)(1x1+x/2p)2=x3p2(1x)2(1+x/2)K_p = \frac{\left(\frac{x}{1+x/2}p\right)^2 \left(\frac{x/2}{1+x/2}p\right)}{\left(\frac{1-x}{1+x/2}p\right)^2} = \frac{x^3 \cdot p}{2(1-x)^2(1+x/2)} Since xx is small compared to 1, we can approximate 1x11-x \approx 1 and 1+x/211+x/2 \approx 1. Therefore, Kp=x3p2K_p = \frac{x^3 p}{2} x3=2Kppx^3 = \frac{2K_p}{p} x=(2Kpp)1/3x = \left(\frac{2K_p}{p}\right)^{1/3}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Equilibrium. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYEquilibriumdissociationequilibriumrepresentedrightleftharpoonsdegree

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