KOH + B = H2 + K3BO3

KOH potassium hydroxide + B boron ⟶ H_2 hydrogen + K3BO3

Balance the chemical equation algebraically: KOH + B ⟶ H_2 + K3BO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 B ⟶ c_3 H_2 + c_4 K3BO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O and B: H: | c_1 = 2 c_3 K: | c_1 = 3 c_4 O: | c_1 = 3 c_4 B: | 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 = 3/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 6 c_2 = 2 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 KOH + 2 B ⟶ 3 H_2 + 2 K3BO3

+ ⟶ + K3BO3

potassium hydroxide + boron ⟶ hydrogen + K3BO3

Construct the equilibrium constant, K, expression for: KOH + B ⟶ H_2 + K3BO3 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 KOH + 2 B ⟶ 3 H_2 + 2 K3BO3 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 KOH | 6 | -6 B | 2 | -2 H_2 | 3 | 3 K3BO3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 6 | -6 | ([KOH])^(-6) B | 2 | -2 | ([B])^(-2) H_2 | 3 | 3 | ([H2])^3 K3BO3 | 2 | 2 | ([K3BO3])^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 = ([KOH])^(-6) ([B])^(-2) ([H2])^3 ([K3BO3])^2 = (([H2])^3 ([K3BO3])^2)/(([KOH])^6 ([B])^2)

Construct the rate of reaction expression for: KOH + B ⟶ H_2 + K3BO3 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 KOH + 2 B ⟶ 3 H_2 + 2 K3BO3 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 KOH | 6 | -6 B | 2 | -2 H_2 | 3 | 3 K3BO3 | 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 KOH | 6 | -6 | -1/6 (Δ[KOH])/(Δt) B | 2 | -2 | -1/2 (Δ[B])/(Δt) H_2 | 3 | 3 | 1/3 (Δ[H2])/(Δt) K3BO3 | 2 | 2 | 1/2 (Δ[K3BO3])/(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[KOH])/(Δt) = -1/2 (Δ[B])/(Δt) = 1/3 (Δ[H2])/(Δt) = 1/2 (Δ[K3BO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium hydroxide | boron | hydrogen | K3BO3 formula | KOH | B | H_2 | K3BO3 Hill formula | HKO | B | H_2 | BK3O3 name | potassium hydroxide | boron | hydrogen | IUPAC name | potassium hydroxide | boron | molecular hydrogen |

| potassium hydroxide | boron | hydrogen | K3BO3 molar mass | 56.105 g/mol | 10.81 g/mol | 2.016 g/mol | 176.1 g/mol phase | solid (at STP) | solid (at STP) | gas (at STP) | melting point | 406 °C | 2075 °C | -259.2 °C | boiling point | 1327 °C | 4000 °C | -252.8 °C | density | 2.044 g/cm^3 | 2.34 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | solubility in water | soluble | insoluble | | dynamic viscosity | 0.001 Pa s (at 550 °C) | | 8.9×10^-6 Pa s (at 25 °C) | odor | | | odorless |