H2O + BaO = Ba(OH)2

H_2O (water) + BaO (barium oxide) ⟶ Ba(OH)_2 (barium hydroxide)

Balance the chemical equation algebraically: H_2O + BaO ⟶ Ba(OH)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 BaO ⟶ c_3 Ba(OH)_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and Ba: H: | 2 c_1 = 2 c_3 O: | c_1 + c_2 = 2 c_3 Ba: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O + BaO ⟶ Ba(OH)_2

+ ⟶

water + barium oxide ⟶ barium hydroxide

| water | barium oxide | barium hydroxide molecular enthalpy | -285.8 kJ/mol | -548 kJ/mol | -944.7 kJ/mol total enthalpy | -285.8 kJ/mol | -548 kJ/mol | -944.7 kJ/mol | H_initial = -833.8 kJ/mol | | H_final = -944.7 kJ/mol ΔH_rxn^0 | -944.7 kJ/mol - -833.8 kJ/mol = -110.9 kJ/mol (exothermic) | |

Construct the equilibrium constant, K, expression for: H_2O + BaO ⟶ Ba(OH)_2 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: H_2O + BaO ⟶ Ba(OH)_2 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 H_2O | 1 | -1 BaO | 1 | -1 Ba(OH)_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 1 | -1 | ([H2O])^(-1) BaO | 1 | -1 | ([BaO])^(-1) Ba(OH)_2 | 1 | 1 | [Ba(OH)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 = ([H2O])^(-1) ([BaO])^(-1) [Ba(OH)2] = ([Ba(OH)2])/([H2O] [BaO])

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

| water | barium oxide | barium hydroxide formula | H_2O | BaO | Ba(OH)_2 Hill formula | H_2O | BaO | BaH_2O_2 name | water | barium oxide | barium hydroxide IUPAC name | water | oxobarium | barium(+2) cation dihydroxide

| water | barium oxide | barium hydroxide molar mass | 18.015 g/mol | 153.326 g/mol | 171.34 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) melting point | 0 °C | 1920 °C | 300 °C boiling point | 99.9839 °C | | density | 1 g/cm^3 | 5.72 g/cm^3 | 2.2 g/cm^3 surface tension | 0.0728 N/m | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | odor | odorless | |