HNO3 + Ba = H2O + NO2 + Ba(NO3)2

HNO_3 nitric acid + Ba barium ⟶ H_2O water + NO_2 nitrogen dioxide + Ba(NO_3)_2 barium nitrate

Balance the chemical equation algebraically: HNO_3 + Ba ⟶ H_2O + NO_2 + Ba(NO_3)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Ba ⟶ c_3 H_2O + c_4 NO_2 + c_5 Ba(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 Ba: H: | c_1 = 2 c_3 N: | c_1 = c_4 + 2 c_5 O: | 3 c_1 = c_3 + 2 c_4 + 6 c_5 Ba: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 2 c_4 = 2 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HNO_3 + Ba ⟶ 2 H_2O + 2 NO_2 + Ba(NO_3)_2

+ ⟶ + +

nitric acid + barium ⟶ water + nitrogen dioxide + barium nitrate

Construct the equilibrium constant, K, expression for: HNO_3 + Ba ⟶ H_2O + NO_2 + Ba(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: 4 HNO_3 + Ba ⟶ 2 H_2O + 2 NO_2 + Ba(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 | 4 | -4 Ba | 1 | -1 H_2O | 2 | 2 NO_2 | 2 | 2 Ba(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 | 4 | -4 | ([HNO3])^(-4) Ba | 1 | -1 | ([Ba])^(-1) H_2O | 2 | 2 | ([H2O])^2 NO_2 | 2 | 2 | ([NO2])^2 Ba(NO_3)_2 | 1 | 1 | [Ba(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])^(-4) ([Ba])^(-1) ([H2O])^2 ([NO2])^2 [Ba(NO3)2] = (([H2O])^2 ([NO2])^2 [Ba(NO3)2])/(([HNO3])^4 [Ba])

Construct the rate of reaction expression for: HNO_3 + Ba ⟶ H_2O + NO_2 + Ba(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: 4 HNO_3 + Ba ⟶ 2 H_2O + 2 NO_2 + Ba(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 | 4 | -4 Ba | 1 | -1 H_2O | 2 | 2 NO_2 | 2 | 2 Ba(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 | 4 | -4 | -1/4 (Δ[HNO3])/(Δt) Ba | 1 | -1 | -(Δ[Ba])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) NO_2 | 2 | 2 | 1/2 (Δ[NO2])/(Δt) Ba(NO_3)_2 | 1 | 1 | (Δ[Ba(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 = -1/4 (Δ[HNO3])/(Δt) = -(Δ[Ba])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/2 (Δ[NO2])/(Δt) = (Δ[Ba(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | barium | water | nitrogen dioxide | barium nitrate formula | HNO_3 | Ba | H_2O | NO_2 | Ba(NO_3)_2 Hill formula | HNO_3 | Ba | H_2O | NO_2 | BaN_2O_6 name | nitric acid | barium | water | nitrogen dioxide | barium nitrate IUPAC name | nitric acid | barium | water | Nitrogen dioxide | barium(+2) cation dinitrate

| nitric acid | barium | water | nitrogen dioxide | barium nitrate molar mass | 63.012 g/mol | 137.327 g/mol | 18.015 g/mol | 46.005 g/mol | 261.34 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | solid (at STP) melting point | -41.6 °C | 725 °C | 0 °C | -11 °C | 592 °C boiling point | 83 °C | 1640 °C | 99.9839 °C | 21 °C | density | 1.5129 g/cm^3 | 3.6 g/cm^3 | 1 g/cm^3 | 0.00188 g/cm^3 (at 25 °C) | 3.23 g/cm^3 solubility in water | miscible | insoluble | | reacts | surface tension | | 0.224 N/m | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) | odor | | | odorless | |