O2 + W = WO3

O_2 oxygen + W tungsten ⟶ WO_3 tungsten trioxide

Balance the chemical equation algebraically: O_2 + W ⟶ WO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 W ⟶ c_3 WO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for O and W: O: | 2 c_1 = 3 c_3 W: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3/2 c_2 = 1 c_3 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 2 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 O_2 + 2 W ⟶ 2 WO_3

+ ⟶

oxygen + tungsten ⟶ tungsten trioxide

| oxygen | tungsten | tungsten trioxide molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -842.9 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -1686 kJ/mol | H_initial = 0 kJ/mol | | H_final = -1686 kJ/mol ΔH_rxn^0 | -1686 kJ/mol - 0 kJ/mol = -1686 kJ/mol (exothermic) | |

Construct the equilibrium constant, K, expression for: O_2 + W ⟶ WO_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 + 2 W ⟶ 2 WO_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 W | 2 | -2 WO_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) W | 2 | -2 | ([W])^(-2) WO_3 | 2 | 2 | ([WO3])^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) ([W])^(-2) ([WO3])^2 = ([WO3])^2/(([O2])^3 ([W])^2)

Construct the rate of reaction expression for: O_2 + W ⟶ WO_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 + 2 W ⟶ 2 WO_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 W | 2 | -2 WO_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) W | 2 | -2 | -1/2 (Δ[W])/(Δt) WO_3 | 2 | 2 | 1/2 (Δ[WO3])/(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) = -1/2 (Δ[W])/(Δt) = 1/2 (Δ[WO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | tungsten | tungsten trioxide formula | O_2 | W | WO_3 Hill formula | O_2 | W | O_3W name | oxygen | tungsten | tungsten trioxide IUPAC name | molecular oxygen | tungsten |

| oxygen | tungsten | tungsten trioxide molar mass | 31.998 g/mol | 183.84 g/mol | 231.84 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) melting point | -218 °C | 3410 °C | 1473 °C boiling point | -183 °C | 5660 °C | density | 0.001429 g/cm^3 (at 0 °C) | 19.3 g/cm^3 | 7.16 g/cm^3 solubility in water | | insoluble | insoluble surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | | odor | odorless | |