Fe2O3 + N2O5 = Fe(NO3)3

Fe_2O_3 iron(III) oxide + N_2O_5 dinitrogen pentoxide ⟶ Fe(NO_3)_3 ferric nitrate

Balance the chemical equation algebraically: Fe_2O_3 + N_2O_5 ⟶ Fe(NO_3)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe_2O_3 + c_2 N_2O_5 ⟶ c_3 Fe(NO_3)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, O and N: Fe: | 2 c_1 = c_3 O: | 3 c_1 + 5 c_2 = 9 c_3 N: | 2 c_2 = 3 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 = 3 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Fe_2O_3 + 3 N_2O_5 ⟶ 2 Fe(NO_3)_3

+ ⟶

iron(III) oxide + dinitrogen pentoxide ⟶ ferric nitrate

Construct the equilibrium constant, K, expression for: Fe_2O_3 + N_2O_5 ⟶ Fe(NO_3)_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: Fe_2O_3 + 3 N_2O_5 ⟶ 2 Fe(NO_3)_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 Fe_2O_3 | 1 | -1 N_2O_5 | 3 | -3 Fe(NO_3)_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Fe_2O_3 | 1 | -1 | ([Fe2O3])^(-1) N_2O_5 | 3 | -3 | ([N2O5])^(-3) Fe(NO_3)_3 | 2 | 2 | ([Fe(NO3)3])^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 = ([Fe2O3])^(-1) ([N2O5])^(-3) ([Fe(NO3)3])^2 = ([Fe(NO3)3])^2/([Fe2O3] ([N2O5])^3)

Construct the rate of reaction expression for: Fe_2O_3 + N_2O_5 ⟶ Fe(NO_3)_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: Fe_2O_3 + 3 N_2O_5 ⟶ 2 Fe(NO_3)_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 Fe_2O_3 | 1 | -1 N_2O_5 | 3 | -3 Fe(NO_3)_3 | 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 Fe_2O_3 | 1 | -1 | -(Δ[Fe2O3])/(Δt) N_2O_5 | 3 | -3 | -1/3 (Δ[N2O5])/(Δt) Fe(NO_3)_3 | 2 | 2 | 1/2 (Δ[Fe(NO3)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 = -(Δ[Fe2O3])/(Δt) = -1/3 (Δ[N2O5])/(Δt) = 1/2 (Δ[Fe(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iron(III) oxide | dinitrogen pentoxide | ferric nitrate formula | Fe_2O_3 | N_2O_5 | Fe(NO_3)_3 Hill formula | Fe_2O_3 | N_2O_5 | FeN_3O_9 name | iron(III) oxide | dinitrogen pentoxide | ferric nitrate IUPAC name | | nitro nitrate | iron(+3) cation trinitrate

| iron(III) oxide | dinitrogen pentoxide | ferric nitrate molar mass | 159.69 g/mol | 108.01 g/mol | 241.86 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) melting point | 1565 °C | 30 °C | 35 °C boiling point | | 47 °C | density | 5.26 g/cm^3 | 2.05 g/cm^3 | 1.7 g/cm^3 solubility in water | insoluble | | very soluble odor | odorless | |