Mg(OH)2 + HF = H2O + MgF2

Mg(OH)_2 magnesium hydroxide + HF hydrogen fluoride ⟶ H_2O water + MgF_2 magnesium fluoride

Balance the chemical equation algebraically: Mg(OH)_2 + HF ⟶ H_2O + MgF_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Mg(OH)_2 + c_2 HF ⟶ c_3 H_2O + c_4 MgF_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Mg, O and F: H: | 2 c_1 + c_2 = 2 c_3 Mg: | c_1 = c_4 O: | 2 c_1 = c_3 F: | c_2 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Mg(OH)_2 + 2 HF ⟶ 2 H_2O + MgF_2

+ ⟶ +

magnesium hydroxide + hydrogen fluoride ⟶ water + magnesium fluoride

| magnesium hydroxide | hydrogen fluoride | water | magnesium fluoride molecular enthalpy | -924.5 kJ/mol | -273.3 kJ/mol | -285.8 kJ/mol | -1124 kJ/mol total enthalpy | -924.5 kJ/mol | -546.6 kJ/mol | -571.7 kJ/mol | -1124 kJ/mol | H_initial = -1471 kJ/mol | | H_final = -1696 kJ/mol | ΔH_rxn^0 | -1696 kJ/mol - -1471 kJ/mol = -224.8 kJ/mol (exothermic) | | |

| magnesium hydroxide | hydrogen fluoride | water | magnesium fluoride molecular free energy | -833.5 kJ/mol | -275.4 kJ/mol | -237.1 kJ/mol | -1071 kJ/mol total free energy | -833.5 kJ/mol | -550.8 kJ/mol | -474.2 kJ/mol | -1071 kJ/mol | G_initial = -1384 kJ/mol | | G_final = -1545 kJ/mol | ΔG_rxn^0 | -1545 kJ/mol - -1384 kJ/mol = -161 kJ/mol (exergonic) | | |

| magnesium hydroxide | hydrogen fluoride | water | magnesium fluoride molecular entropy | 64 J/(mol K) | 173.8 J/(mol K) | 69.91 J/(mol K) | 57.2 J/(mol K) total entropy | 64 J/(mol K) | 347.6 J/(mol K) | 139.8 J/(mol K) | 57.2 J/(mol K) | S_initial = 411.6 J/(mol K) | | S_final = 197 J/(mol K) | ΔS_rxn^0 | 197 J/(mol K) - 411.6 J/(mol K) = -214.6 J/(mol K) (exoentropic) | | |

Construct the equilibrium constant, K, expression for: Mg(OH)_2 + HF ⟶ H_2O + MgF_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: Mg(OH)_2 + 2 HF ⟶ 2 H_2O + MgF_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 Mg(OH)_2 | 1 | -1 HF | 2 | -2 H_2O | 2 | 2 MgF_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Mg(OH)_2 | 1 | -1 | ([Mg(OH)2])^(-1) HF | 2 | -2 | ([HF])^(-2) H_2O | 2 | 2 | ([H2O])^2 MgF_2 | 1 | 1 | [MgF2] 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 = ([Mg(OH)2])^(-1) ([HF])^(-2) ([H2O])^2 [MgF2] = (([H2O])^2 [MgF2])/([Mg(OH)2] ([HF])^2)

Construct the rate of reaction expression for: Mg(OH)_2 + HF ⟶ H_2O + MgF_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: Mg(OH)_2 + 2 HF ⟶ 2 H_2O + MgF_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 Mg(OH)_2 | 1 | -1 HF | 2 | -2 H_2O | 2 | 2 MgF_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 Mg(OH)_2 | 1 | -1 | -(Δ[Mg(OH)2])/(Δt) HF | 2 | -2 | -1/2 (Δ[HF])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) MgF_2 | 1 | 1 | (Δ[MgF2])/(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[Mg(OH)2])/(Δt) = -1/2 (Δ[HF])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[MgF2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| magnesium hydroxide | hydrogen fluoride | water | magnesium fluoride formula | Mg(OH)_2 | HF | H_2O | MgF_2 Hill formula | H_2MgO_2 | FH | H_2O | F_2Mg name | magnesium hydroxide | hydrogen fluoride | water | magnesium fluoride IUPAC name | magnesium dihydroxide | hydrogen fluoride | water | magnesium difluoride

| magnesium hydroxide | hydrogen fluoride | water | magnesium fluoride molar mass | 58.319 g/mol | 20.006 g/mol | 18.015 g/mol | 62.302 g/mol phase | solid (at STP) | gas (at STP) | liquid (at STP) | solid (at STP) melting point | 350 °C | -83.36 °C | 0 °C | 1266 °C boiling point | | 19.5 °C | 99.9839 °C | 2260 °C density | 2.3446 g/cm^3 | 8.18×10^-4 g/cm^3 (at 25 °C) | 1 g/cm^3 | 3.15 g/cm^3 solubility in water | insoluble | miscible | | slightly soluble 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 |