Na2O + HF = H2O + NaF

Na_2O sodium oxide + HF hydrogen fluoride ⟶ H_2O water + NaF sodium fluoride

Balance the chemical equation algebraically: Na_2O + HF ⟶ H_2O + NaF Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Na_2O + c_2 HF ⟶ c_3 H_2O + c_4 NaF Set the number of atoms in the reactants equal to the number of atoms in the products for Na, O, F and H: Na: | 2 c_1 = c_4 O: | c_1 = c_3 F: | c_2 = c_4 H: | c_2 = 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 = 2 c_3 = 1 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Na_2O + 2 HF ⟶ H_2O + 2 NaF

+ ⟶ +

sodium oxide + hydrogen fluoride ⟶ water + sodium fluoride

Construct the equilibrium constant, K, expression for: Na_2O + HF ⟶ H_2O + NaF 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: Na_2O + 2 HF ⟶ H_2O + 2 NaF 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 Na_2O | 1 | -1 HF | 2 | -2 H_2O | 1 | 1 NaF | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Na_2O | 1 | -1 | ([Na2O])^(-1) HF | 2 | -2 | ([HF])^(-2) H_2O | 1 | 1 | [H2O] NaF | 2 | 2 | ([NaF])^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 = ([Na2O])^(-1) ([HF])^(-2) [H2O] ([NaF])^2 = ([H2O] ([NaF])^2)/([Na2O] ([HF])^2)

Construct the rate of reaction expression for: Na_2O + HF ⟶ H_2O + NaF 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: Na_2O + 2 HF ⟶ H_2O + 2 NaF 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 Na_2O | 1 | -1 HF | 2 | -2 H_2O | 1 | 1 NaF | 2 | 2 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 Na_2O | 1 | -1 | -(Δ[Na2O])/(Δt) HF | 2 | -2 | -1/2 (Δ[HF])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NaF | 2 | 2 | 1/2 (Δ[NaF])/(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[Na2O])/(Δt) = -1/2 (Δ[HF])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[NaF])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium oxide | hydrogen fluoride | water | sodium fluoride formula | Na_2O | HF | H_2O | NaF Hill formula | Na_2O | FH | H_2O | FNa name | sodium oxide | hydrogen fluoride | water | sodium fluoride IUPAC name | disodium oxygen(-2) anion | hydrogen fluoride | water | sodium fluoride

| sodium oxide | hydrogen fluoride | water | sodium fluoride molar mass | 61.979 g/mol | 20.006 g/mol | 18.015 g/mol | 41.98817244 g/mol phase | | gas (at STP) | liquid (at STP) | solid (at STP) melting point | | -83.36 °C | 0 °C | 993 °C boiling point | | 19.5 °C | 99.9839 °C | 1700 °C density | 2.27 g/cm^3 | 8.18×10^-4 g/cm^3 (at 25 °C) | 1 g/cm^3 | 2.558 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) | 0.00105 Pa s (at 1160 °C) odor | | | odorless | odorless