KOH + BeO = H2O + K2BeO2

KOH potassium hydroxide + BeO beryllium oxide ⟶ H_2O water + K2BeO2

Balance the chemical equation algebraically: KOH + BeO ⟶ H_2O + K2BeO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 BeO ⟶ c_3 H_2O + c_4 K2BeO2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O and Be: H: | c_1 = 2 c_3 K: | c_1 = 2 c_4 O: | c_1 + c_2 = c_3 + 2 c_4 Be: | 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 = 2 c_2 = 1 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + BeO ⟶ H_2O + K2BeO2

+ ⟶ + K2BeO2

potassium hydroxide + beryllium oxide ⟶ water + K2BeO2

Construct the equilibrium constant, K, expression for: KOH + BeO ⟶ H_2O + K2BeO2 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: 2 KOH + BeO ⟶ H_2O + K2BeO2 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 | 2 | -2 BeO | 1 | -1 H_2O | 1 | 1 K2BeO2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 2 | -2 | ([KOH])^(-2) BeO | 1 | -1 | ([BeO])^(-1) H_2O | 1 | 1 | [H2O] K2BeO2 | 1 | 1 | [K2BeO2] 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])^(-2) ([BeO])^(-1) [H2O] [K2BeO2] = ([H2O] [K2BeO2])/(([KOH])^2 [BeO])

Construct the rate of reaction expression for: KOH + BeO ⟶ H_2O + K2BeO2 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: 2 KOH + BeO ⟶ H_2O + K2BeO2 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 | 2 | -2 BeO | 1 | -1 H_2O | 1 | 1 K2BeO2 | 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 KOH | 2 | -2 | -1/2 (Δ[KOH])/(Δt) BeO | 1 | -1 | -(Δ[BeO])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) K2BeO2 | 1 | 1 | (Δ[K2BeO2])/(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 = -1/2 (Δ[KOH])/(Δt) = -(Δ[BeO])/(Δt) = (Δ[H2O])/(Δt) = (Δ[K2BeO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium hydroxide | beryllium oxide | water | K2BeO2 formula | KOH | BeO | H_2O | K2BeO2 Hill formula | HKO | BeO | H_2O | BeK2O2 name | potassium hydroxide | beryllium oxide | water | IUPAC name | potassium hydroxide | oxoberyllium | water |

| potassium hydroxide | beryllium oxide | water | K2BeO2 molar mass | 56.105 g/mol | 25.011 g/mol | 18.015 g/mol | 119.21 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | melting point | 406 °C | 2410 °C | 0 °C | boiling point | 1327 °C | 4300 °C | 99.9839 °C | density | 2.044 g/cm^3 | 3.01 g/cm^3 | 1 g/cm^3 | solubility in water | soluble | insoluble | | surface tension | | | 0.0728 N/m | dynamic viscosity | 0.001 Pa s (at 550 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | | | odorless |