O2 + Sb = Sb2O3

O_2 oxygen + Sb gray antimony ⟶ Sb_2O_3 antimony trioxide

Balance the chemical equation algebraically: O_2 + Sb ⟶ Sb_2O_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 Sb ⟶ c_3 Sb_2O_3 Set the number of atoms in the reactants equal to the number of atoms in the products for O and Sb: O: | 2 c_1 = 3 c_3 Sb: | c_2 = 2 c_3 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3/2 c_2 = 2 c_3 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 4 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 O_2 + 4 Sb ⟶ 2 Sb_2O_3

+ ⟶

oxygen + gray antimony ⟶ antimony trioxide

Construct the equilibrium constant, K, expression for: O_2 + Sb ⟶ Sb_2O_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: 3 O_2 + 4 Sb ⟶ 2 Sb_2O_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i O_2 | 3 | -3 Sb | 4 | -4 Sb_2O_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 3 | -3 | ([O2])^(-3) Sb | 4 | -4 | ([Sb])^(-4) Sb_2O_3 | 2 | 2 | ([Sb2O3])^2 The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([O2])^(-3) ([Sb])^(-4) ([Sb2O3])^2 = ([Sb2O3])^2/(([O2])^3 ([Sb])^4)

Construct the rate of reaction expression for: O_2 + Sb ⟶ Sb_2O_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: 3 O_2 + 4 Sb ⟶ 2 Sb_2O_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i O_2 | 3 | -3 Sb | 4 | -4 Sb_2O_3 | 2 | 2 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term O_2 | 3 | -3 | -1/3 (Δ[O2])/(Δt) Sb | 4 | -4 | -1/4 (Δ[Sb])/(Δt) Sb_2O_3 | 2 | 2 | 1/2 (Δ[Sb2O3])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -1/3 (Δ[O2])/(Δt) = -1/4 (Δ[Sb])/(Δt) = 1/2 (Δ[Sb2O3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | gray antimony | antimony trioxide formula | O_2 | Sb | Sb_2O_3 Hill formula | O_2 | Sb | O_3Sb_2 name | oxygen | gray antimony | antimony trioxide IUPAC name | molecular oxygen | antimony | oxo-oxostibanyloxystibane

| oxygen | gray antimony | antimony trioxide molar mass | 31.998 g/mol | 121.76 g/mol | 291.517 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) melting point | -218 °C | 630 °C | 655 °C boiling point | -183 °C | 1587 °C | 1550 °C density | 0.001429 g/cm^3 (at 0 °C) | 6.69 g/cm^3 | 5.2 g/cm^3 solubility in water | | | insoluble surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | | odor | odorless | |