HNO3 + Hg = H2O + NO + Hg2(NO3)4

HNO_3 nitric acid + Hg mercury ⟶ H_2O water + NO nitric oxide + Hg2(NO3)4

Balance the chemical equation algebraically: HNO_3 + Hg ⟶ H_2O + NO + Hg2(NO3)4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Hg ⟶ c_3 H_2O + c_4 NO + c_5 Hg2(NO3)4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Hg: H: | c_1 = 2 c_3 N: | c_1 = c_4 + 4 c_5 O: | 3 c_1 = c_3 + c_4 + 12 c_5 Hg: | c_2 = 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_5 = 1 and solve the system of equations for the remaining coefficients: c_1 = 16/3 c_2 = 2 c_3 = 8/3 c_4 = 4/3 c_5 = 1 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 16 c_2 = 6 c_3 = 8 c_4 = 4 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 16 HNO_3 + 6 Hg ⟶ 8 H_2O + 4 NO + 3 Hg2(NO3)4

+ ⟶ + + Hg2(NO3)4

nitric acid + mercury ⟶ water + nitric oxide + Hg2(NO3)4

Construct the equilibrium constant, K, expression for: HNO_3 + Hg ⟶ H_2O + NO + Hg2(NO3)4 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: 16 HNO_3 + 6 Hg ⟶ 8 H_2O + 4 NO + 3 Hg2(NO3)4 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 | 16 | -16 Hg | 6 | -6 H_2O | 8 | 8 NO | 4 | 4 Hg2(NO3)4 | 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 | 16 | -16 | ([HNO3])^(-16) Hg | 6 | -6 | ([Hg])^(-6) H_2O | 8 | 8 | ([H2O])^8 NO | 4 | 4 | ([NO])^4 Hg2(NO3)4 | 3 | 3 | ([Hg2(NO3)4])^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])^(-16) ([Hg])^(-6) ([H2O])^8 ([NO])^4 ([Hg2(NO3)4])^3 = (([H2O])^8 ([NO])^4 ([Hg2(NO3)4])^3)/(([HNO3])^16 ([Hg])^6)

Construct the rate of reaction expression for: HNO_3 + Hg ⟶ H_2O + NO + Hg2(NO3)4 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: 16 HNO_3 + 6 Hg ⟶ 8 H_2O + 4 NO + 3 Hg2(NO3)4 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 | 16 | -16 Hg | 6 | -6 H_2O | 8 | 8 NO | 4 | 4 Hg2(NO3)4 | 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 | 16 | -16 | -1/16 (Δ[HNO3])/(Δt) Hg | 6 | -6 | -1/6 (Δ[Hg])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) NO | 4 | 4 | 1/4 (Δ[NO])/(Δt) Hg2(NO3)4 | 3 | 3 | 1/3 (Δ[Hg2(NO3)4])/(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 = -1/16 (Δ[HNO3])/(Δt) = -1/6 (Δ[Hg])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/4 (Δ[NO])/(Δt) = 1/3 (Δ[Hg2(NO3)4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | mercury | water | nitric oxide | Hg2(NO3)4 formula | HNO_3 | Hg | H_2O | NO | Hg2(NO3)4 Hill formula | HNO_3 | Hg | H_2O | NO | Hg2N4O12 name | nitric acid | mercury | water | nitric oxide |