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A certain amount of H2S was added to a 2.0 L flask and allowed to come to equilibrium. At equilibrium, 0.072 mol of H2 was found. How many moles of H2S were originally added to the flask?

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Answer:

0.098 moles H₂S

Explanation:

The reaction that takes place is

  • 2H₂(g) + S₂(g) ⇄ 2H₂S(g)  keq = 7.5

We can express the equilibrium constant as:

  • keq = [H₂S]² / [S₂] [H₂]² = 7.5

With the volume we can calculate the equilibrium concentration of H₂:

  • [H₂] = 0.072 mol / 2.0 L = 0.036 M

The stoichiometric ratio tells us that the concentration of S₂ is half of the concentration of H₂:

  • [S₂] = [H₂] / 2 = 0.036 M / 2 = 0.018 M

Now we can calculate [H₂S]:

  • 7.5 = [H₂S]² / (0.018*0.036²)
  • [H₂S] = 0.013 M

So 0.013 M is the concentration of H₂S at equilibrium.

  • This would amount to (0.013 M * 2.0 L) 0.026 moles of H₂S
  • The moles of H₂ at equilibrium are equal to the moles of H₂S that reacted.

Initial moles of H₂S - Moles of H₂S that reacted into H₂ = Moles of H₂S at equilibrium

Initial moles of H₂S - 0.072 mol = 0.026 mol

Initial moles of H₂S = 0.098 moles H₂S