0.020 g of selenium vapour at equilibrium occupying a volume of 2.463 mL at 1 atm and 27 o C. The selenium is in a state of equilibrium according to reaction 3 Se 2 g ⇌ Se 6 g What is the degree of association of selenium ? (Atomic weight of Se = 79)

Step-by-step solution

3 Se 2 g ⇌ Se 6 g At equilibrium a 1 - α a α 3 Total moles at equilibrium = a 1 - 2 α 3 a 1 - 2 α 3 = n Total = PV RT = 1 atm × 2.463 × 1 0 - 3 0.0821 × 3 0 0 = 1 0 - 4 a = 0.020 g 79 × 2 g / mole (Initial moles) = 0.020 1 5 8 = 1.266 × 1 0 - 4 1.266 × 1 0 - 4 1 - 2 α 3 = 1 0 - 4 1 - 2 α 3 = 1 1.266 = 0.79 2 α 3 = 0.21 ⇒ α = 0.63 2 = 0.315