HNO3 + Ac = H2O + N2O + Ac(NO3)3

HNO_3 nitric acid + Ac actinium ⟶ H_2O water + N_2O nitrous oxide + Ac(NO3)3

Balance the chemical equation algebraically: HNO_3 + Ac ⟶ H_2O + N_2O + Ac(NO3)3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Ac ⟶ c_3 H_2O + c_4 N_2O + c_5 Ac(NO3)3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Ac: H: | c_1 = 2 c_3 N: | c_1 = 2 c_4 + 3 c_5 O: | 3 c_1 = c_3 + c_4 + 9 c_5 Ac: | 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 = 10 c_2 = 8/3 c_3 = 5 c_4 = 1 c_5 = 8/3 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 30 c_2 = 8 c_3 = 15 c_4 = 3 c_5 = 8 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 30 HNO_3 + 8 Ac ⟶ 15 H_2O + 3 N_2O + 8 Ac(NO3)3

+ ⟶ + + Ac(NO3)3

nitric acid + actinium ⟶ water + nitrous oxide + Ac(NO3)3

Construct the equilibrium constant, K, expression for: HNO_3 + Ac ⟶ H_2O + N_2O + Ac(NO3)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: 30 HNO_3 + 8 Ac ⟶ 15 H_2O + 3 N_2O + 8 Ac(NO3)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 | 30 | -30 Ac | 8 | -8 H_2O | 15 | 15 N_2O | 3 | 3 Ac(NO3)3 | 8 | 8 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 30 | -30 | ([HNO3])^(-30) Ac | 8 | -8 | ([Ac])^(-8) H_2O | 15 | 15 | ([H2O])^15 N_2O | 3 | 3 | ([N2O])^3 Ac(NO3)3 | 8 | 8 | ([Ac(NO3)3])^8 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])^(-30) ([Ac])^(-8) ([H2O])^15 ([N2O])^3 ([Ac(NO3)3])^8 = (([H2O])^15 ([N2O])^3 ([Ac(NO3)3])^8)/(([HNO3])^30 ([Ac])^8)

Construct the rate of reaction expression for: HNO_3 + Ac ⟶ H_2O + N_2O + Ac(NO3)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: 30 HNO_3 + 8 Ac ⟶ 15 H_2O + 3 N_2O + 8 Ac(NO3)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 | 30 | -30 Ac | 8 | -8 H_2O | 15 | 15 N_2O | 3 | 3 Ac(NO3)3 | 8 | 8 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 | 30 | -30 | -1/30 (Δ[HNO3])/(Δt) Ac | 8 | -8 | -1/8 (Δ[Ac])/(Δt) H_2O | 15 | 15 | 1/15 (Δ[H2O])/(Δt) N_2O | 3 | 3 | 1/3 (Δ[N2O])/(Δt) Ac(NO3)3 | 8 | 8 | 1/8 (Δ[Ac(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 = -1/30 (Δ[HNO3])/(Δt) = -1/8 (Δ[Ac])/(Δt) = 1/15 (Δ[H2O])/(Δt) = 1/3 (Δ[N2O])/(Δt) = 1/8 (Δ[Ac(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | actinium | water | nitrous oxide | Ac(NO3)3 formula | HNO_3 | Ac | H_2O | N_2O | Ac(NO3)3 Hill formula | HNO_3 | Ac | H_2O | N_2O | AcN3O9 name | nitric acid | actinium | water | nitrous oxide |

| nitric acid | actinium | water | nitrous oxide | Ac(NO3)3 molar mass | 63.012 g/mol | 227 g/mol | 18.015 g/mol | 44.013 g/mol | 413 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 1050 °C | 0 °C | -91 °C | boiling point | 83 °C | 3200 °C | 99.9839 °C | -88 °C | density | 1.5129 g/cm^3 | 10.07 g/cm^3 | 1 g/cm^3 | 0.001799 g/cm^3 (at 25 °C) | solubility in water | miscible | | | | surface tension | | | 0.0728 N/m | 0.00175 N/m | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 1.491×10^-5 Pa s (at 25 °C) | odor | | | odorless | |