H3BO3 = H2O + H4B6O11

B(OH)_3 boric acid ⟶ H_2O water + H4B6O11

Balance the chemical equation algebraically: B(OH)_3 ⟶ H_2O + H4B6O11 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 B(OH)_3 ⟶ c_2 H_2O + c_3 H4B6O11 Set the number of atoms in the reactants equal to the number of atoms in the products for B, H and O: B: | c_1 = 6 c_3 H: | 3 c_1 = 2 c_2 + 4 c_3 O: | 3 c_1 = c_2 + 11 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 = 6 c_2 = 7 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 B(OH)_3 ⟶ 7 H_2O + H4B6O11

⟶ + H4B6O11

boric acid ⟶ water + H4B6O11

Construct the equilibrium constant, K, expression for: B(OH)_3 ⟶ H_2O + H4B6O11 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: 6 B(OH)_3 ⟶ 7 H_2O + H4B6O11 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 B(OH)_3 | 6 | -6 H_2O | 7 | 7 H4B6O11 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression B(OH)_3 | 6 | -6 | ([B(OH)3])^(-6) H_2O | 7 | 7 | ([H2O])^7 H4B6O11 | 1 | 1 | [H4B6O11] 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 = ([B(OH)3])^(-6) ([H2O])^7 [H4B6O11] = (([H2O])^7 [H4B6O11])/([B(OH)3])^6

Construct the rate of reaction expression for: B(OH)_3 ⟶ H_2O + H4B6O11 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: 6 B(OH)_3 ⟶ 7 H_2O + H4B6O11 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 B(OH)_3 | 6 | -6 H_2O | 7 | 7 H4B6O11 | 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 B(OH)_3 | 6 | -6 | -1/6 (Δ[B(OH)3])/(Δt) H_2O | 7 | 7 | 1/7 (Δ[H2O])/(Δt) H4B6O11 | 1 | 1 | (Δ[H4B6O11])/(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 = -1/6 (Δ[B(OH)3])/(Δt) = 1/7 (Δ[H2O])/(Δt) = (Δ[H4B6O11])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| boric acid | water | H4B6O11 formula | B(OH)_3 | H_2O | H4B6O11 Hill formula | BH_3O_3 | H_2O | H4B6O11 name | boric acid | water |

| boric acid | water | H4B6O11 molar mass | 61.83 g/mol | 18.015 g/mol | 244.9 g/mol phase | solid (at STP) | liquid (at STP) | melting point | 160 °C | 0 °C | boiling point | | 99.9839 °C | density | | 1 g/cm^3 | surface tension | | 0.0728 N/m | dynamic viscosity | | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | odorless |