HF + Ba(OH)2 = H2O + BaF2

HF hydrogen fluoride + Ba(OH)_2 barium hydroxide ⟶ H_2O water + BaF_2 barium fluoride

Balance the chemical equation algebraically: HF + Ba(OH)_2 ⟶ H_2O + BaF_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HF + c_2 Ba(OH)_2 ⟶ c_3 H_2O + c_4 BaF_2 Set the number of atoms in the reactants equal to the number of atoms in the products for F, H, Ba and O: F: | c_1 = 2 c_4 H: | c_1 + 2 c_2 = 2 c_3 Ba: | c_2 = c_4 O: | 2 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 = 2 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HF + Ba(OH)_2 ⟶ 2 H_2O + BaF_2

+ ⟶ +

hydrogen fluoride + barium hydroxide ⟶ water + barium fluoride

| hydrogen fluoride | barium hydroxide | water | barium fluoride molecular enthalpy | -273.3 kJ/mol | -944.7 kJ/mol | -285.8 kJ/mol | -1207 kJ/mol total enthalpy | -546.6 kJ/mol | -944.7 kJ/mol | -571.7 kJ/mol | -1207 kJ/mol | H_initial = -1491 kJ/mol | | H_final = -1779 kJ/mol | ΔH_rxn^0 | -1779 kJ/mol - -1491 kJ/mol = -287.5 kJ/mol (exothermic) | | |

Construct the equilibrium constant, K, expression for: HF + Ba(OH)_2 ⟶ H_2O + BaF_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: 2 HF + Ba(OH)_2 ⟶ 2 H_2O + BaF_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 HF | 2 | -2 Ba(OH)_2 | 1 | -1 H_2O | 2 | 2 BaF_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HF | 2 | -2 | ([HF])^(-2) Ba(OH)_2 | 1 | -1 | ([Ba(OH)2])^(-1) H_2O | 2 | 2 | ([H2O])^2 BaF_2 | 1 | 1 | [BaF2] 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 = ([HF])^(-2) ([Ba(OH)2])^(-1) ([H2O])^2 [BaF2] = (([H2O])^2 [BaF2])/(([HF])^2 [Ba(OH)2])

Construct the rate of reaction expression for: HF + Ba(OH)_2 ⟶ H_2O + BaF_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: 2 HF + Ba(OH)_2 ⟶ 2 H_2O + BaF_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 HF | 2 | -2 Ba(OH)_2 | 1 | -1 H_2O | 2 | 2 BaF_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 HF | 2 | -2 | -1/2 (Δ[HF])/(Δt) Ba(OH)_2 | 1 | -1 | -(Δ[Ba(OH)2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) BaF_2 | 1 | 1 | (Δ[BaF2])/(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[HF])/(Δt) = -(Δ[Ba(OH)2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[BaF2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen fluoride | barium hydroxide | water | barium fluoride formula | HF | Ba(OH)_2 | H_2O | BaF_2 Hill formula | FH | BaH_2O_2 | H_2O | BaF_2 name | hydrogen fluoride | barium hydroxide | water | barium fluoride IUPAC name | hydrogen fluoride | barium(+2) cation dihydroxide | water | barium(+2) cation difluoride

| hydrogen fluoride | barium hydroxide | water | barium fluoride molar mass | 20.006 g/mol | 171.34 g/mol | 18.015 g/mol | 175.324 g/mol phase | gas (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) melting point | -83.36 °C | 300 °C | 0 °C | 1270 °C boiling point | 19.5 °C | | 99.9839 °C | 2260 °C density | 8.18×10^-4 g/cm^3 (at 25 °C) | 2.2 g/cm^3 | 1 g/cm^3 | 4.89 g/cm^3 solubility in water | miscible | | | surface tension | | | 0.0728 N/m | dynamic viscosity | 1.2571×10^-5 Pa s (at 20 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | | | odorless |