NaOH + FCl3 = NaCl + F(OH)3

NaOH sodium hydroxide + FCl3 ⟶ NaCl sodium chloride + F(OH)3

Balance the chemical equation algebraically: NaOH + FCl3 ⟶ NaCl + F(OH)3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 FCl3 ⟶ c_3 NaCl + c_4 F(OH)3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O, F and Cl: H: | c_1 = 3 c_4 Na: | c_1 = c_3 O: | c_1 = 3 c_4 F: | c_2 = c_4 Cl: | 3 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 = 3 c_2 = 1 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 NaOH + FCl3 ⟶ 3 NaCl + F(OH)3

+ FCl3 ⟶ + F(OH)3

sodium hydroxide + FCl3 ⟶ sodium chloride + F(OH)3

Construct the equilibrium constant, K, expression for: NaOH + FCl3 ⟶ NaCl + F(OH)3 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: 3 NaOH + FCl3 ⟶ 3 NaCl + F(OH)3 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 NaOH | 3 | -3 FCl3 | 1 | -1 NaCl | 3 | 3 F(OH)3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 3 | -3 | ([NaOH])^(-3) FCl3 | 1 | -1 | ([FCl3])^(-1) NaCl | 3 | 3 | ([NaCl])^3 F(OH)3 | 1 | 1 | [F(OH)3] 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 = ([NaOH])^(-3) ([FCl3])^(-1) ([NaCl])^3 [F(OH)3] = (([NaCl])^3 [F(OH)3])/(([NaOH])^3 [FCl3])

Construct the rate of reaction expression for: NaOH + FCl3 ⟶ NaCl + F(OH)3 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: 3 NaOH + FCl3 ⟶ 3 NaCl + F(OH)3 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 NaOH | 3 | -3 FCl3 | 1 | -1 NaCl | 3 | 3 F(OH)3 | 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 NaOH | 3 | -3 | -1/3 (Δ[NaOH])/(Δt) FCl3 | 1 | -1 | -(Δ[FCl3])/(Δt) NaCl | 3 | 3 | 1/3 (Δ[NaCl])/(Δt) F(OH)3 | 1 | 1 | (Δ[F(OH)3])/(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 = -1/3 (Δ[NaOH])/(Δt) = -(Δ[FCl3])/(Δt) = 1/3 (Δ[NaCl])/(Δt) = (Δ[F(OH)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium hydroxide | FCl3 | sodium chloride | F(OH)3 formula | NaOH | FCl3 | NaCl | F(OH)3 Hill formula | HNaO | Cl3F | ClNa | H3FO3 name | sodium hydroxide | | sodium chloride |

| sodium hydroxide | FCl3 | sodium chloride | F(OH)3 molar mass | 39.997 g/mol | 125.3 g/mol | 58.44 g/mol | 70.019 g/mol phase | solid (at STP) | | solid (at STP) | melting point | 323 °C | | 801 °C | boiling point | 1390 °C | | 1413 °C | density | 2.13 g/cm^3 | | 2.16 g/cm^3 | solubility in water | soluble | | soluble | surface tension | 0.07435 N/m | | | dynamic viscosity | 0.004 Pa s (at 350 °C) | | | odor | | | odorless |