HNO3 + Ba = H2O + N2 + Ba(NO3)2

HNO_3 nitric acid + Ba barium ⟶ H_2O water + N_2 nitrogen + Ba(NO_3)_2 barium nitrate

Balance the chemical equation algebraically: HNO_3 + Ba ⟶ H_2O + N_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 N_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 = 2 c_4 + 2 c_5 O: | 3 c_1 = c_3 + 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 12 c_2 = 5 c_3 = 6 c_4 = 1 c_5 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 HNO_3 + 5 Ba ⟶ 6 H_2O + N_2 + 5 Ba(NO_3)_2

+ ⟶ + +

nitric acid + barium ⟶ water + nitrogen + barium nitrate

Construct the equilibrium constant, K, expression for: HNO_3 + Ba ⟶ H_2O + N_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: 12 HNO_3 + 5 Ba ⟶ 6 H_2O + N_2 + 5 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 | 12 | -12 Ba | 5 | -5 H_2O | 6 | 6 N_2 | 1 | 1 Ba(NO_3)_2 | 5 | 5 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 12 | -12 | ([HNO3])^(-12) Ba | 5 | -5 | ([Ba])^(-5) H_2O | 6 | 6 | ([H2O])^6 N_2 | 1 | 1 | [N2] Ba(NO_3)_2 | 5 | 5 | ([Ba(NO3)2])^5 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])^(-12) ([Ba])^(-5) ([H2O])^6 [N2] ([Ba(NO3)2])^5 = (([H2O])^6 [N2] ([Ba(NO3)2])^5)/(([HNO3])^12 ([Ba])^5)

Construct the rate of reaction expression for: HNO_3 + Ba ⟶ H_2O + N_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: 12 HNO_3 + 5 Ba ⟶ 6 H_2O + N_2 + 5 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 | 12 | -12 Ba | 5 | -5 H_2O | 6 | 6 N_2 | 1 | 1 Ba(NO_3)_2 | 5 | 5 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 | 12 | -12 | -1/12 (Δ[HNO3])/(Δt) Ba | 5 | -5 | -1/5 (Δ[Ba])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) N_2 | 1 | 1 | (Δ[N2])/(Δt) Ba(NO_3)_2 | 5 | 5 | 1/5 (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 = -1/12 (Δ[HNO3])/(Δt) = -1/5 (Δ[Ba])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[N2])/(Δt) = 1/5 (Δ[Ba(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)