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If $α$ is the fraction of $H I$ dissociated at equilibrium in the reaction, $2 H I ⇌ H_{2} (g) + I_{2} (g)$ , starting with 2 moles of $H I$ ,the total number of moles of reactants and products at equilibrium are

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$1 + α$

$2 - α$

$2 + 2 α$

answer is C.

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Detailed Solution

$2 H I ⇌ H_{2} (g) + I_{2} (g)$

$2 - 2 α$ $α$ $α$ ….. Moles at equilibrium

Total no. of moles $= 2 - 2 α + α + α = 2$ .

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