First slide

Chemical Equilibrium

Question

If $α$ is the fraction of HI dissociation at equilibrium in the reaction,
$2 HI ⇌ H_{2} + I_{2}$
then starting with 2 mole of HI, the total number of moles of reactants and products at equilibrium is:

Moderate

Solution

$Total mole = 2 (1 - α) + α + α = 2$

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