Question

Mercury and oxygen react to form mercury(II) oxide, like this: 2Hg (l) + O2 (g) → 2HgO (s) At a certain temperature, a chemist finds that a 5.2L reaction vessel containing a mixture of mercury, oxygen, and mercury(II) oxide at equilibrium has the following composition: compound amount Hg 14.7g O2 13.4g HgO 17.8g calculate Kc for this reaction

Answer 1

• Kc = 12.4 M⁻¹

Step-by-step explanation:

1) Chemical equilibrium

• 2Hg (l) + O₂ (g) ⇄ 2HgO (s) (the double arrow indicates an equilibrium reaction)

2) Equilibrium constant, Kc:

• The equilibrium constant, Kc, is equal to product of the concentrations of each product,each raised to its stoichiometric coefficient / product of the concentrations of each reactant each raised to its stoichiometric coefficient.

• Since the concentrations of liquid and solid substances remain practically constant, their value is incorporated into the constant Kc, and the equation only uses the concentrations of the aqueous or gaseous substances.

Thus, the equation to use is:

• Kc = 1 / [O₂(g) ]

3) Determine the concentration of O₂ (g)

• M = number of moles / volume in liters

• Number of moles = mass in grams / molar mass

• Number of moles of O₂ (g) = 13.4 g / 32.00 g/mol = 0.419 mol

• M = 0.419 mol / 5.2 liter = 0.0806 mol / liter = 0.0806 M

4) Compute Kc

• Kc = 1 / 0.0 806 M = 12.4 M⁻¹ ← answer