HNO3 + Fe = H2 + Fe(NO3)2

HNO_3 nitric acid + Fe iron ⟶ H_2 hydrogen + Fe(NO_3)_2 iron(II) nitrate

Balance the chemical equation algebraically: HNO_3 + Fe ⟶ H_2 + Fe(NO_3)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Fe ⟶ c_3 H_2 + c_4 Fe(NO_3)_2 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 O: | 3 c_1 = 6 c_4 Fe: | 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 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HNO_3 + Fe ⟶ H_2 + Fe(NO_3)_2

+ ⟶ +

nitric acid + iron ⟶ hydrogen + iron(II) nitrate

Construct the equilibrium constant, K, expression for: HNO_3 + Fe ⟶ H_2 + Fe(NO_3)_2 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 HNO_3 + Fe ⟶ H_2 + Fe(NO_3)_2 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 | 2 | -2 Fe | 1 | -1 H_2 | 1 | 1 Fe(NO_3)_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 2 | -2 | ([HNO3])^(-2) Fe | 1 | -1 | ([Fe])^(-1) H_2 | 1 | 1 | [H2] Fe(NO_3)_2 | 1 | 1 | [Fe(NO3)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 = ([HNO3])^(-2) ([Fe])^(-1) [H2] [Fe(NO3)2] = ([H2] [Fe(NO3)2])/(([HNO3])^2 [Fe])

Construct the rate of reaction expression for: HNO_3 + Fe ⟶ H_2 + Fe(NO_3)_2 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 HNO_3 + Fe ⟶ H_2 + Fe(NO_3)_2 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 | 2 | -2 Fe | 1 | -1 H_2 | 1 | 1 Fe(NO_3)_2 | 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 HNO_3 | 2 | -2 | -1/2 (Δ[HNO3])/(Δt) Fe | 1 | -1 | -(Δ[Fe])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) Fe(NO_3)_2 | 1 | 1 | (Δ[Fe(NO3)2])/(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[HNO3])/(Δt) = -(Δ[Fe])/(Δt) = (Δ[H2])/(Δt) = (Δ[Fe(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | iron | hydrogen | iron(II) nitrate formula | HNO_3 | Fe | H_2 | Fe(NO_3)_2 Hill formula | HNO_3 | Fe | H_2 | FeN_2O_6 name | nitric acid | iron | hydrogen | iron(II) nitrate IUPAC name | nitric acid | iron | molecular hydrogen |

| nitric acid | iron | hydrogen | iron(II) nitrate molar mass | 63.012 g/mol | 55.845 g/mol | 2.016 g/mol | 179.85 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | melting point | -41.6 °C | 1535 °C | -259.2 °C | boiling point | 83 °C | 2750 °C | -252.8 °C | density | 1.5129 g/cm^3 | 7.874 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | solubility in water | miscible | insoluble | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-6 Pa s (at 25 °C) | odor | | | odorless |