Ask Question
20 January, 06:26

Mercury and oxygen react to form mercury (II) oxide, like this: 2 Hg (l) + 02 (g) - -HgO (s) At a certain temperature, a chemist finds that a 6.9 L reaction vessel containing a mixture of mercury, oxygen, and mercury (II) oxide at equilibrium has the following composition: compound amount Hg 16.9 g O 10.9 g HgO 23.8 g Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

+3
Answers (1)
  1. 20 January, 07:20
    0
    Kc = 20

    Explanation:

    We have the equilibrium:

    2 Hg (l) + O₂ (g) ⇄ HgO (s)

    Kc = 1 / [O₂]

    The key here is to remember that pure solids and liquids do not enter into the calculation for the equilibrium expression, and Hg is a pure liquid and HgO is a solid

    So what we need to do to solve this question is to calculate the concentration of oxygen at equilibrium. We are given its mass, and the volume so we are equipped to calculate the concentration of oxygen as follows:

    [O₂] = # moles O₂ / V

    # moles O₂ = mass / molar mass = 10.9 g / 32g/mol = 0.34 mol

    [O₂] = 0.34 mol / 6.9 L = 0.049 M

    ⇒ Kc = 1 / 0.049 = 20 (rounded to 2 significant figures)
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Mercury and oxygen react to form mercury (II) oxide, like this: 2 Hg (l) + 02 (g) - -HgO (s) At a certain temperature, a chemist finds that ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers