HNO3 + Cu = H2O + NO + CuO

HNO_3 nitric acid + Cu copper ⟶ H_2O water + NO nitric oxide + CuO cupric oxide

Balance the chemical equation algebraically: HNO_3 + Cu ⟶ H_2O + NO + CuO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Cu ⟶ c_3 H_2O + c_4 NO + c_5 CuO Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Cu: H: | c_1 = 2 c_3 N: | c_1 = c_4 O: | 3 c_1 = c_3 + c_4 + c_5 Cu: | 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 1 c_4 = 2 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HNO_3 + 3 Cu ⟶ H_2O + 2 NO + 3 CuO

+ ⟶ + +

nitric acid + copper ⟶ water + nitric oxide + cupric oxide

| nitric acid | copper | water | nitric oxide | cupric oxide molecular entropy | 156 J/(mol K) | 33 J/(mol K) | 69.91 J/(mol K) | 211 J/(mol K) | 43 J/(mol K) total entropy | 312 J/(mol K) | 99 J/(mol K) | 69.91 J/(mol K) | 422 J/(mol K) | 129 J/(mol K) | S_initial = 411 J/(mol K) | | S_final = 620.9 J/(mol K) | | ΔS_rxn^0 | 620.9 J/(mol K) - 411 J/(mol K) = 209.9 J/(mol K) (endoentropic) | | | |

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

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

| nitric acid | copper | water | nitric oxide | cupric oxide formula | HNO_3 | Cu | H_2O | NO | CuO name | nitric acid | copper | water | nitric oxide | cupric oxide