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Questions  >>  JEEMAIN and NEET  >>  NEET PAST PAPERS  >>  2008
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The dissociation equilibrium of a gas $AB_2$ can be represented as : <br> $2AB_2 (g)\rightleftharpoons 2AB(g) + B_2 (g)$ <br> The degree of dissociation is 'x' and is small s compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant $K_p$ and total pressure P is -


( A ) $(\frac{2K_p}{P})^\frac{1}{2}$
( B ) $(\frac{K_p}{P})$
( C ) $(\frac{2K_p}{P})$
( D ) $(\frac{2K_p}{P})^\frac{1}{3}$

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