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Home  >>  JEEMAIN and AIPMT  >>  Chemistry
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The dissociation equilibrium of a gas $AB_2$ can be represented as :$2AB_2(g) \leftrightharpoons 2 \: AB (g) +B_2(g)$ The degree of dissociation is 'x' and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant $K_p$ and total pressure P is :

$\begin {array} {1 1} (1)\;(K_p/P) & \quad (2)\;(2K_p/P) \\ (3)\;(K_p/P)^{\large\frac{1}{3}} & \quad (4)\;(K_p/P)^{\large\frac{1}{2}} \end {array}$

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