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KOH + Be(OH)2 = H2O + K2BeO2

Input interpretation

KOH potassium hydroxide + Be(OH)_2 beryllium hydroxide ⟶ H_2O water + K2BeO2
KOH potassium hydroxide + Be(OH)_2 beryllium hydroxide ⟶ H_2O water + K2BeO2

Balanced equation

Balance the chemical equation algebraically: KOH + Be(OH)_2 ⟶ H_2O + K2BeO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 Be(OH)_2 ⟶ 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_2 = 2 c_3 K: | c_1 = 2 c_4 O: | c_1 + 2 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 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + Be(OH)_2 ⟶ 2 H_2O + K2BeO2
Balance the chemical equation algebraically: KOH + Be(OH)_2 ⟶ H_2O + K2BeO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 Be(OH)_2 ⟶ 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_2 = 2 c_3 K: | c_1 = 2 c_4 O: | c_1 + 2 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 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + Be(OH)_2 ⟶ 2 H_2O + K2BeO2

Structures

+ ⟶ + K2BeO2
+ ⟶ + K2BeO2

Names

potassium hydroxide + beryllium hydroxide ⟶ water + K2BeO2
potassium hydroxide + beryllium hydroxide ⟶ water + K2BeO2

Equilibrium constant

Construct the equilibrium constant, K, expression for: KOH + Be(OH)_2 ⟶ 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 + Be(OH)_2 ⟶ 2 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 Be(OH)_2 | 1 | -1 H_2O | 2 | 2 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) Be(OH)_2 | 1 | -1 | ([Be(OH)2])^(-1) H_2O | 2 | 2 | ([H2O])^2 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) ([Be(OH)2])^(-1) ([H2O])^2 [K2BeO2] = (([H2O])^2 [K2BeO2])/(([KOH])^2 [Be(OH)2])
Construct the equilibrium constant, K, expression for: KOH + Be(OH)_2 ⟶ 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 + Be(OH)_2 ⟶ 2 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 Be(OH)_2 | 1 | -1 H_2O | 2 | 2 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) Be(OH)_2 | 1 | -1 | ([Be(OH)2])^(-1) H_2O | 2 | 2 | ([H2O])^2 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) ([Be(OH)2])^(-1) ([H2O])^2 [K2BeO2] = (([H2O])^2 [K2BeO2])/(([KOH])^2 [Be(OH)2])

Rate of reaction

Construct the rate of reaction expression for: KOH + Be(OH)_2 ⟶ 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 + Be(OH)_2 ⟶ 2 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 Be(OH)_2 | 1 | -1 H_2O | 2 | 2 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) Be(OH)_2 | 1 | -1 | -(Δ[Be(OH)2])/(Δt) H_2O | 2 | 2 | 1/2 (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[Be(OH)2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[K2BeO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KOH + Be(OH)_2 ⟶ 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 + Be(OH)_2 ⟶ 2 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 Be(OH)_2 | 1 | -1 H_2O | 2 | 2 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) Be(OH)_2 | 1 | -1 | -(Δ[Be(OH)2])/(Δt) H_2O | 2 | 2 | 1/2 (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[Be(OH)2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[K2BeO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium hydroxide | beryllium hydroxide | water | K2BeO2 formula | KOH | Be(OH)_2 | H_2O | K2BeO2 Hill formula | HKO | BeH_2O_2 | H_2O | BeK2O2 name | potassium hydroxide | beryllium hydroxide | water | IUPAC name | potassium hydroxide | beryllium(+2) cation dihydroxide | water |
| potassium hydroxide | beryllium hydroxide | water | K2BeO2 formula | KOH | Be(OH)_2 | H_2O | K2BeO2 Hill formula | HKO | BeH_2O_2 | H_2O | BeK2O2 name | potassium hydroxide | beryllium hydroxide | water | IUPAC name | potassium hydroxide | beryllium(+2) cation dihydroxide | water |

Substance properties

| potassium hydroxide | beryllium hydroxide | water | K2BeO2 molar mass | 56.105 g/mol | 43.026 g/mol | 18.015 g/mol | 119.21 g/mol phase | solid (at STP) | | liquid (at STP) | melting point | 406 °C | | 0 °C | boiling point | 1327 °C | | 99.9839 °C | density | 2.044 g/cm^3 | 1.92 g/cm^3 | 1 g/cm^3 | solubility in water | soluble | | | 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 |
| potassium hydroxide | beryllium hydroxide | water | K2BeO2 molar mass | 56.105 g/mol | 43.026 g/mol | 18.015 g/mol | 119.21 g/mol phase | solid (at STP) | | liquid (at STP) | melting point | 406 °C | | 0 °C | boiling point | 1327 °C | | 99.9839 °C | density | 2.044 g/cm^3 | 1.92 g/cm^3 | 1 g/cm^3 | solubility in water | soluble | | | 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 |

Units

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