HNO3 + Fe = H2O + N2 + Fe(NO3)3

HNO_3 (nitric acid) + Fe (iron) ⟶ H_2O (water) + N_2 (nitrogen) + Fe(NO_3)_3 (ferric nitrate)

Balance the chemical equation algebraically: HNO_3 + Fe ⟶ H_2O + N_2 + Fe(NO_3)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Fe ⟶ c_3 H_2O + c_4 N_2 + c_5 Fe(NO_3)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Fe: H: | c_1 = 2 c_3 N: | c_1 = 2 c_4 + 3 c_5 O: | 3 c_1 = c_3 + 9 c_5 Fe: | c_2 = c_5 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 12 c_2 = 10/3 c_3 = 6 c_4 = 1 c_5 = 10/3 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 36 c_2 = 10 c_3 = 18 c_4 = 3 c_5 = 10 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 36 HNO_3 + 10 Fe ⟶ 18 H_2O + 3 N_2 + 10 Fe(NO_3)_3

+ ⟶ + +

nitric acid + iron ⟶ water + nitrogen + ferric nitrate

Construct the equilibrium constant, K, expression for: HNO_3 + Fe ⟶ H_2O + N_2 + 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: 36 HNO_3 + 10 Fe ⟶ 18 H_2O + 3 N_2 + 10 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 HNO_3 | 36 | -36 Fe | 10 | -10 H_2O | 18 | 18 N_2 | 3 | 3 Fe(NO_3)_3 | 10 | 10 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 36 | -36 | ([HNO3])^(-36) Fe | 10 | -10 | ([Fe])^(-10) H_2O | 18 | 18 | ([H2O])^18 N_2 | 3 | 3 | ([N2])^3 Fe(NO_3)_3 | 10 | 10 | ([Fe(NO3)3])^10 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 = ([HNO3])^(-36) ([Fe])^(-10) ([H2O])^18 ([N2])^3 ([Fe(NO3)3])^10 = (([H2O])^18 ([N2])^3 ([Fe(NO3)3])^10)/(([HNO3])^36 ([Fe])^10)

Construct the rate of reaction expression for: HNO_3 + Fe ⟶ H_2O + N_2 + 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: 36 HNO_3 + 10 Fe ⟶ 18 H_2O + 3 N_2 + 10 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 HNO_3 | 36 | -36 Fe | 10 | -10 H_2O | 18 | 18 N_2 | 3 | 3 Fe(NO_3)_3 | 10 | 10 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 HNO_3 | 36 | -36 | -1/36 (Δ[HNO3])/(Δt) Fe | 10 | -10 | -1/10 (Δ[Fe])/(Δt) H_2O | 18 | 18 | 1/18 (Δ[H2O])/(Δt) N_2 | 3 | 3 | 1/3 (Δ[N2])/(Δt) Fe(NO_3)_3 | 10 | 10 | 1/10 (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 = -1/36 (Δ[HNO3])/(Δt) = -1/10 (Δ[Fe])/(Δt) = 1/18 (Δ[H2O])/(Δt) = 1/3 (Δ[N2])/(Δt) = 1/10 (Δ[Fe(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | iron | water | nitrogen | ferric nitrate formula | HNO_3 | Fe | H_2O | N_2 | Fe(NO_3)_3 Hill formula | HNO_3 | Fe | H_2O | N_2 | FeN_3O_9 name | nitric acid | iron | water | nitrogen | ferric nitrate IUPAC name | nitric acid | iron | water | molecular nitrogen | iron(+3) cation trinitrate

| nitric acid | iron | water | nitrogen | ferric nitrate molar mass | 63.012 g/mol | 55.845 g/mol | 18.015 g/mol | 28.014 g/mol | 241.86 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | solid (at STP) melting point | -41.6 °C | 1535 °C | 0 °C | -210 °C | 35 °C boiling point | 83 °C | 2750 °C | 99.9839 °C | -195.79 °C | density | 1.5129 g/cm^3 | 7.874 g/cm^3 | 1 g/cm^3 | 0.001251 g/cm^3 (at 0 °C) | 1.7 g/cm^3 solubility in water | miscible | insoluble | | insoluble | very soluble surface tension | | | 0.0728 N/m | 0.0066 N/m | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) | odor | | | odorless | odorless |