HNO3 + Mg = H2O + NO + N2O + Mg(NO3)2

HNO_3 nitric acid + Mg magnesium ⟶ H_2O water + NO nitric oxide + N_2O nitrous oxide + Mg(NO_3)_2 magnesium nitrate

Balance the chemical equation algebraically: HNO_3 + Mg ⟶ H_2O + NO + N_2O + Mg(NO_3)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Mg ⟶ c_3 H_2O + c_4 NO + c_5 N_2O + c_6 Mg(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 Mg: H: | c_1 = 2 c_3 N: | c_1 = c_4 + 2 c_5 + 2 c_6 O: | 3 c_1 = c_3 + c_4 + c_5 + 6 c_6 Mg: | c_2 = c_6 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_2 = (2 c_1)/5 - 1/10 c_3 = c_1/2 c_4 = 1 c_5 = c_1/10 - 2/5 c_6 = (2 c_1)/5 - 1/10 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_2 = (2 c_1)/5 - 1/5 c_3 = c_1/2 c_4 = 2 c_5 = c_1/10 - 4/5 c_6 = (2 c_1)/5 - 1/5 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 38 and solve for the remaining coefficients: c_1 = 38 c_2 = 15 c_3 = 19 c_4 = 2 c_5 = 3 c_6 = 15 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 38 HNO_3 + 15 Mg ⟶ 19 H_2O + 2 NO + 3 N_2O + 15 Mg(NO_3)_2

+ ⟶ + + +

nitric acid + magnesium ⟶ water + nitric oxide + nitrous oxide + magnesium nitrate

Construct the equilibrium constant, K, expression for: HNO_3 + Mg ⟶ H_2O + NO + N_2O + Mg(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: 38 HNO_3 + 15 Mg ⟶ 19 H_2O + 2 NO + 3 N_2O + 15 Mg(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 | 38 | -38 Mg | 15 | -15 H_2O | 19 | 19 NO | 2 | 2 N_2O | 3 | 3 Mg(NO_3)_2 | 15 | 15 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 38 | -38 | ([HNO3])^(-38) Mg | 15 | -15 | ([Mg])^(-15) H_2O | 19 | 19 | ([H2O])^19 NO | 2 | 2 | ([NO])^2 N_2O | 3 | 3 | ([N2O])^3 Mg(NO_3)_2 | 15 | 15 | ([Mg(NO3)2])^15 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])^(-38) ([Mg])^(-15) ([H2O])^19 ([NO])^2 ([N2O])^3 ([Mg(NO3)2])^15 = (([H2O])^19 ([NO])^2 ([N2O])^3 ([Mg(NO3)2])^15)/(([HNO3])^38 ([Mg])^15)

Construct the rate of reaction expression for: HNO_3 + Mg ⟶ H_2O + NO + N_2O + Mg(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: 38 HNO_3 + 15 Mg ⟶ 19 H_2O + 2 NO + 3 N_2O + 15 Mg(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 | 38 | -38 Mg | 15 | -15 H_2O | 19 | 19 NO | 2 | 2 N_2O | 3 | 3 Mg(NO_3)_2 | 15 | 15 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 | 38 | -38 | -1/38 (Δ[HNO3])/(Δt) Mg | 15 | -15 | -1/15 (Δ[Mg])/(Δt) H_2O | 19 | 19 | 1/19 (Δ[H2O])/(Δt) NO | 2 | 2 | 1/2 (Δ[NO])/(Δt) N_2O | 3 | 3 | 1/3 (Δ[N2O])/(Δt) Mg(NO_3)_2 | 15 | 15 | 1/15 (Δ[Mg(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 = -1/38 (Δ[HNO3])/(Δt) = -1/15 (Δ[Mg])/(Δt) = 1/19 (Δ[H2O])/(Δt) = 1/2 (Δ[NO])/(Δt) = 1/3 (Δ[N2O])/(Δt) = 1/15 (Δ[Mg(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | magnesium | water | nitric oxide | nitrous oxide | magnesium nitrate formula | HNO_3 | Mg | H_2O | NO | N_2O | Mg(NO_3)_2 Hill formula | HNO_3 | Mg | H_2O | NO | N_2O | MgN_2O_6 name | nitric acid | magnesium | water | nitric oxide | nitrous oxide | magnesium nitrate IUPAC name | nitric acid | magnesium | water | nitric oxide | nitrous oxide | magnesium dinitrate