H2O + N2O3 = HNO3 + NO

H_2O water + N_2O_3 nitrogen trioxide ⟶ HNO_3 nitric acid + NO nitric oxide

Balance the chemical equation algebraically: H_2O + N_2O_3 ⟶ HNO_3 + NO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 N_2O_3 ⟶ c_3 HNO_3 + c_4 NO Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and N: H: | 2 c_1 = c_3 O: | c_1 + 3 c_2 = 3 c_3 + c_4 N: | 2 c_2 = c_3 + 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3 c_3 = 2 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O + 3 N_2O_3 ⟶ 2 HNO_3 + 4 NO

+ ⟶ +

water + nitrogen trioxide ⟶ nitric acid + nitric oxide

| water | nitrogen trioxide | nitric acid | nitric oxide molecular free energy | -237.1 kJ/mol | 142.4 kJ/mol | -80.7 kJ/mol | 87.6 kJ/mol total free energy | -237.1 kJ/mol | 427.2 kJ/mol | -161.4 kJ/mol | 350.4 kJ/mol | G_initial = 190.1 kJ/mol | | G_final = 189 kJ/mol | ΔG_rxn^0 | 189 kJ/mol - 190.1 kJ/mol = -1.1 kJ/mol (exergonic) | | |

Construct the equilibrium constant, K, expression for: H_2O + N_2O_3 ⟶ HNO_3 + NO 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: H_2O + 3 N_2O_3 ⟶ 2 HNO_3 + 4 NO 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 H_2O | 1 | -1 N_2O_3 | 3 | -3 HNO_3 | 2 | 2 NO | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 1 | -1 | ([H2O])^(-1) N_2O_3 | 3 | -3 | ([N2O3])^(-3) HNO_3 | 2 | 2 | ([HNO3])^2 NO | 4 | 4 | ([NO])^4 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 = ([H2O])^(-1) ([N2O3])^(-3) ([HNO3])^2 ([NO])^4 = (([HNO3])^2 ([NO])^4)/([H2O] ([N2O3])^3)

Construct the rate of reaction expression for: H_2O + N_2O_3 ⟶ HNO_3 + NO 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: H_2O + 3 N_2O_3 ⟶ 2 HNO_3 + 4 NO 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 H_2O | 1 | -1 N_2O_3 | 3 | -3 HNO_3 | 2 | 2 NO | 4 | 4 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 H_2O | 1 | -1 | -(Δ[H2O])/(Δt) N_2O_3 | 3 | -3 | -1/3 (Δ[N2O3])/(Δt) HNO_3 | 2 | 2 | 1/2 (Δ[HNO3])/(Δt) NO | 4 | 4 | 1/4 (Δ[NO])/(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 = -(Δ[H2O])/(Δt) = -1/3 (Δ[N2O3])/(Δt) = 1/2 (Δ[HNO3])/(Δt) = 1/4 (Δ[NO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | nitrogen trioxide | nitric acid | nitric oxide formula | H_2O | N_2O_3 | HNO_3 | NO name | water | nitrogen trioxide | nitric acid | nitric oxide IUPAC name | water | nitramide | nitric acid | nitric oxide