HNO3 + H2S = H2O + S + NO

HNO_3 (nitric acid) + H_2S (hydrogen sulfide) ⟶ H_2O (water) + S (mixed sulfur) + NO (nitric oxide)

Balance the chemical equation algebraically: HNO_3 + H_2S ⟶ H_2O + S + NO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 H_2S ⟶ c_3 H_2O + c_4 S + c_5 NO Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and S: H: | c_1 + 2 c_2 = 2 c_3 N: | c_1 = c_5 O: | 3 c_1 = c_3 + c_5 S: | c_2 = 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 = 3/2 c_3 = 2 c_4 = 3/2 c_5 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 4 c_4 = 3 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HNO_3 + 3 H_2S ⟶ 4 H_2O + 3 S + 2 NO

+ ⟶ + +

nitric acid + hydrogen sulfide ⟶ water + mixed sulfur + nitric oxide

Construct the equilibrium constant, K, expression for: HNO_3 + H_2S ⟶ H_2O + S + 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: 2 HNO_3 + 3 H_2S ⟶ 4 H_2O + 3 S + 2 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 HNO_3 | 2 | -2 H_2S | 3 | -3 H_2O | 4 | 4 S | 3 | 3 NO | 2 | 2 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) H_2S | 3 | -3 | ([H2S])^(-3) H_2O | 4 | 4 | ([H2O])^4 S | 3 | 3 | ([S])^3 NO | 2 | 2 | ([NO])^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])^(-2) ([H2S])^(-3) ([H2O])^4 ([S])^3 ([NO])^2 = (([H2O])^4 ([S])^3 ([NO])^2)/(([HNO3])^2 ([H2S])^3)

Construct the rate of reaction expression for: HNO_3 + H_2S ⟶ H_2O + S + 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: 2 HNO_3 + 3 H_2S ⟶ 4 H_2O + 3 S + 2 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 HNO_3 | 2 | -2 H_2S | 3 | -3 H_2O | 4 | 4 S | 3 | 3 NO | 2 | 2 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) H_2S | 3 | -3 | -1/3 (Δ[H2S])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) S | 3 | 3 | 1/3 (Δ[S])/(Δt) NO | 2 | 2 | 1/2 (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 = -1/2 (Δ[HNO3])/(Δt) = -1/3 (Δ[H2S])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/3 (Δ[S])/(Δt) = 1/2 (Δ[NO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | hydrogen sulfide | water | mixed sulfur | nitric oxide formula | HNO_3 | H_2S | H_2O | S | NO name | nitric acid | hydrogen sulfide | water | mixed sulfur | nitric oxide IUPAC name | nitric acid | hydrogen sulfide | water | sulfur | nitric oxide