Chemistry - Chemical Equilibrium Question with Solution | TestHub

ChemistryChemical EquilibriumMiscellaneous/MixedMedium2 minQB

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$28 g N_{2}$ and $6 g H_{2}$ were mixed. At equilibrium $17 g NH_{3}$ was formed. The weight of $N_{2}$ and $H_{2}$ at equilibrium are respectively.

Answer:

Solution:

N₂(g) + 3H₂(g) $⇌$ 2NH₃(g)

Initially moles: 1 3 0

At equilibrium: $1 - \frac{1}{2}$ $3 - \frac{3}{2}$ 1

i.e., N₂ is 14 g and H₂ is 3 g.

Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: QB

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