NaOH + Fe2O3 = H2O + NaFeO2

NaOH sodium hydroxide + Fe_2O_3 iron(III) oxide ⟶ H_2O water + NaFeO2

Balance the chemical equation algebraically: NaOH + Fe_2O_3 ⟶ H_2O + NaFeO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 Fe_2O_3 ⟶ c_3 H_2O + c_4 NaFeO2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O and Fe: H: | c_1 = 2 c_3 Na: | c_1 = c_4 O: | c_1 + 3 c_2 = c_3 + 2 c_4 Fe: | 2 c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 1 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 NaOH + Fe_2O_3 ⟶ H_2O + 2 NaFeO2

+ ⟶ + NaFeO2

sodium hydroxide + iron(III) oxide ⟶ water + NaFeO2

Construct the equilibrium constant, K, expression for: NaOH + Fe_2O_3 ⟶ H_2O + NaFeO2 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 NaOH + Fe_2O_3 ⟶ H_2O + 2 NaFeO2 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 | 2 | -2 Fe_2O_3 | 1 | -1 H_2O | 1 | 1 NaFeO2 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 2 | -2 | ([NaOH])^(-2) Fe_2O_3 | 1 | -1 | ([Fe2O3])^(-1) H_2O | 1 | 1 | [H2O] NaFeO2 | 2 | 2 | ([NaFeO2])^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 = ([NaOH])^(-2) ([Fe2O3])^(-1) [H2O] ([NaFeO2])^2 = ([H2O] ([NaFeO2])^2)/(([NaOH])^2 [Fe2O3])

Construct the rate of reaction expression for: NaOH + Fe_2O_3 ⟶ H_2O + NaFeO2 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 NaOH + Fe_2O_3 ⟶ H_2O + 2 NaFeO2 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 | 2 | -2 Fe_2O_3 | 1 | -1 H_2O | 1 | 1 NaFeO2 | 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 NaOH | 2 | -2 | -1/2 (Δ[NaOH])/(Δt) Fe_2O_3 | 1 | -1 | -(Δ[Fe2O3])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NaFeO2 | 2 | 2 | 1/2 (Δ[NaFeO2])/(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[NaOH])/(Δt) = -(Δ[Fe2O3])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[NaFeO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium hydroxide | iron(III) oxide | water | NaFeO2 formula | NaOH | Fe_2O_3 | H_2O | NaFeO2 Hill formula | HNaO | Fe_2O_3 | H_2O | FeNaO2 name | sodium hydroxide | iron(III) oxide | water |

| sodium hydroxide | iron(III) oxide | water | NaFeO2 molar mass | 39.997 g/mol | 159.69 g/mol | 18.015 g/mol | 110.83 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | melting point | 323 °C | 1565 °C | 0 °C | boiling point | 1390 °C | | 99.9839 °C | density | 2.13 g/cm^3 | 5.26 g/cm^3 | 1 g/cm^3 | solubility in water | soluble | insoluble | | surface tension | 0.07435 N/m | | 0.0728 N/m | dynamic viscosity | 0.004 Pa s (at 350 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | | odorless | odorless |