SO3 + BaO = BaSO4

SO_3 sulfur trioxide + BaO barium oxide ⟶ BaSO_4 barium sulfate

Balance the chemical equation algebraically: SO_3 + BaO ⟶ BaSO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 SO_3 + c_2 BaO ⟶ c_3 BaSO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for O, S and Ba: O: | 3 c_1 + c_2 = 4 c_3 S: | c_1 = c_3 Ba: | c_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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | SO_3 + BaO ⟶ BaSO_4

+ ⟶

sulfur trioxide + barium oxide ⟶ barium sulfate

| sulfur trioxide | barium oxide | barium sulfate molecular free energy | -373.8 kJ/mol | -520.3 kJ/mol | -1362 kJ/mol total free energy | -373.8 kJ/mol | -520.3 kJ/mol | -1362 kJ/mol | G_initial = -894.1 kJ/mol | | G_final = -1362 kJ/mol ΔG_rxn^0 | -1362 kJ/mol - -894.1 kJ/mol = -468.1 kJ/mol (exergonic) | |

Construct the equilibrium constant, K, expression for: SO_3 + BaO ⟶ BaSO_4 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: SO_3 + BaO ⟶ BaSO_4 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 SO_3 | 1 | -1 BaO | 1 | -1 BaSO_4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression SO_3 | 1 | -1 | ([SO3])^(-1) BaO | 1 | -1 | ([BaO])^(-1) BaSO_4 | 1 | 1 | [BaSO4] 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 = ([SO3])^(-1) ([BaO])^(-1) [BaSO4] = ([BaSO4])/([SO3] [BaO])

Construct the rate of reaction expression for: SO_3 + BaO ⟶ BaSO_4 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: SO_3 + BaO ⟶ BaSO_4 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 SO_3 | 1 | -1 BaO | 1 | -1 BaSO_4 | 1 | 1 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 SO_3 | 1 | -1 | -(Δ[SO3])/(Δt) BaO | 1 | -1 | -(Δ[BaO])/(Δt) BaSO_4 | 1 | 1 | (Δ[BaSO4])/(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 = -(Δ[SO3])/(Δt) = -(Δ[BaO])/(Δt) = (Δ[BaSO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sulfur trioxide | barium oxide | barium sulfate formula | SO_3 | BaO | BaSO_4 Hill formula | O_3S | BaO | BaO_4S name | sulfur trioxide | barium oxide | barium sulfate IUPAC name | sulfur trioxide | oxobarium | barium(+2) cation sulfate

| sulfur trioxide | barium oxide | barium sulfate molar mass | 80.06 g/mol | 153.326 g/mol | 233.38 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) melting point | 16.8 °C | 1920 °C | 1345 °C boiling point | 44.7 °C | | density | 1.97 g/cm^3 | 5.72 g/cm^3 | 4.5 g/cm^3 solubility in water | reacts | | insoluble dynamic viscosity | 0.00159 Pa s (at 30 °C) | |