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If α is the fraction of HI dissociated at equilibrium in the reaction, 2 HI(g) ⇌ H2(g) + I2(g), starting with 2 moles of HI, the total number of moles of reactants and products at equilibrium are:

1. 2 + 2α
2. 2
3. 1 + α
4. 2 − α

Correct answer: 2 + 2α

Solution

Initially, there are 2 moles of HI. At equilibrium, α fraction of HI dissociates, producing α moles each of H2 and I2. The remaining HI is 2(1−α). Total moles = 2(1−α) + α + α = 2 + 2α.

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