K + B2O3 = K2O + B

K potassium + B_2O_3 boron oxide ⟶ K_2O potassium oxide + B boron

Balance the chemical equation algebraically: K + B_2O_3 ⟶ K_2O + B Add stoichiometric coefficients, c_i, to the reactants and products: c_1 K + c_2 B_2O_3 ⟶ c_3 K_2O + c_4 B Set the number of atoms in the reactants equal to the number of atoms in the products for K, B and O: K: | c_1 = 2 c_3 B: | 2 c_2 = c_4 O: | 3 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 = 6 c_2 = 1 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 K + B_2O_3 ⟶ 3 K_2O + 2 B

+ ⟶ +

potassium + boron oxide ⟶ potassium oxide + boron

Construct the equilibrium constant, K, expression for: K + B_2O_3 ⟶ K_2O + B 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 K + B_2O_3 ⟶ 3 K_2O + 2 B 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 K | 6 | -6 B_2O_3 | 1 | -1 K_2O | 3 | 3 B | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression K | 6 | -6 | ([K])^(-6) B_2O_3 | 1 | -1 | ([B2O3])^(-1) K_2O | 3 | 3 | ([K2O])^3 B | 2 | 2 | ([B])^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 = ([K])^(-6) ([B2O3])^(-1) ([K2O])^3 ([B])^2 = (([K2O])^3 ([B])^2)/(([K])^6 [B2O3])

Construct the rate of reaction expression for: K + B_2O_3 ⟶ K_2O + B 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 K + B_2O_3 ⟶ 3 K_2O + 2 B 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 K | 6 | -6 B_2O_3 | 1 | -1 K_2O | 3 | 3 B | 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 K | 6 | -6 | -1/6 (Δ[K])/(Δt) B_2O_3 | 1 | -1 | -(Δ[B2O3])/(Δt) K_2O | 3 | 3 | 1/3 (Δ[K2O])/(Δt) B | 2 | 2 | 1/2 (Δ[B])/(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[K])/(Δt) = -(Δ[B2O3])/(Δt) = 1/3 (Δ[K2O])/(Δt) = 1/2 (Δ[B])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium | boron oxide | potassium oxide | boron formula | K | B_2O_3 | K_2O | B name | potassium | boron oxide | potassium oxide | boron IUPAC name | potassium | | dipotassium oxygen(2-) | boron

| potassium | boron oxide | potassium oxide | boron molar mass | 39.0983 g/mol | 69.62 g/mol | 94.196 g/mol | 10.81 g/mol phase | solid (at STP) | solid (at STP) | | solid (at STP) melting point | 64 °C | 450 °C | | 2075 °C boiling point | 760 °C | 1860 °C | | 4000 °C density | 0.86 g/cm^3 | 2.46 g/cm^3 | | 2.34 g/cm^3 solubility in water | reacts | | | insoluble dynamic viscosity | | 85 Pa s (at 700 °C) | |