Al + B2O3 = B + AlO

Al aluminum + B_2O_3 boron oxide ⟶ B boron + AlO

Balance the chemical equation algebraically: Al + B_2O_3 ⟶ B + AlO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al + c_2 B_2O_3 ⟶ c_3 B + c_4 AlO Set the number of atoms in the reactants equal to the number of atoms in the products for Al, B and O: Al: | c_1 = c_4 B: | 2 c_2 = c_3 O: | 3 c_2 = c_4 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 c_2 = 1 c_3 = 2 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 Al + B_2O_3 ⟶ 2 B + 3 AlO

+ ⟶ + AlO

aluminum + boron oxide ⟶ boron + AlO

Construct the equilibrium constant, K, expression for: Al + B_2O_3 ⟶ B + AlO 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 Al + B_2O_3 ⟶ 2 B + 3 AlO 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 Al | 3 | -3 B_2O_3 | 1 | -1 B | 2 | 2 AlO | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Al | 3 | -3 | ([Al])^(-3) B_2O_3 | 1 | -1 | ([B2O3])^(-1) B | 2 | 2 | ([B])^2 AlO | 3 | 3 | ([AlO])^3 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 = ([Al])^(-3) ([B2O3])^(-1) ([B])^2 ([AlO])^3 = (([B])^2 ([AlO])^3)/(([Al])^3 [B2O3])

Construct the rate of reaction expression for: Al + B_2O_3 ⟶ B + AlO 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 Al + B_2O_3 ⟶ 2 B + 3 AlO 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 Al | 3 | -3 B_2O_3 | 1 | -1 B | 2 | 2 AlO | 3 | 3 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 Al | 3 | -3 | -1/3 (Δ[Al])/(Δt) B_2O_3 | 1 | -1 | -(Δ[B2O3])/(Δt) B | 2 | 2 | 1/2 (Δ[B])/(Δt) AlO | 3 | 3 | 1/3 (Δ[AlO])/(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[Al])/(Δt) = -(Δ[B2O3])/(Δt) = 1/2 (Δ[B])/(Δt) = 1/3 (Δ[AlO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| aluminum | boron oxide | boron | AlO formula | Al | B_2O_3 | B | AlO name | aluminum | boron oxide | boron |

| aluminum | boron oxide | boron | AlO molar mass | 26.9815385 g/mol | 69.62 g/mol | 10.81 g/mol | 42.981 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | melting point | 660.4 °C | 450 °C | 2075 °C | boiling point | 2460 °C | 1860 °C | 4000 °C | density | 2.7 g/cm^3 | 2.46 g/cm^3 | 2.34 g/cm^3 | solubility in water | insoluble | | insoluble | surface tension | 0.817 N/m | | | dynamic viscosity | 1.5×10^-4 Pa s (at 760 °C) | 85 Pa s (at 700 °C) | | odor | odorless | | |