HCl + PbO2 + NaNO2 = H2O + NaNO3 + PbCl2

HCl hydrogen chloride + PbO_2 lead dioxide + NaNO_2 sodium nitrite ⟶ H_2O water + NaNO_3 sodium nitrate + PbCl_2 lead(II) chloride

Balance the chemical equation algebraically: HCl + PbO_2 + NaNO_2 ⟶ H_2O + NaNO_3 + PbCl_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HCl + c_2 PbO_2 + c_3 NaNO_2 ⟶ c_4 H_2O + c_5 NaNO_3 + c_6 PbCl_2 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, O, Pb, N and Na: Cl: | c_1 = 2 c_6 H: | c_1 = 2 c_4 O: | 2 c_2 + 2 c_3 = c_4 + 3 c_5 Pb: | c_2 = c_6 N: | c_3 = c_5 Na: | c_3 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 1 c_4 = 1 c_5 = 1 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HCl + PbO_2 + NaNO_2 ⟶ H_2O + NaNO_3 + PbCl_2

+ + ⟶ + +

hydrogen chloride + lead dioxide + sodium nitrite ⟶ water + sodium nitrate + lead(II) chloride

| hydrogen chloride | lead dioxide | sodium nitrite | water | sodium nitrate | lead(II) chloride molecular enthalpy | -92.3 kJ/mol | -277.4 kJ/mol | -359 kJ/mol | -285.8 kJ/mol | -467.9 kJ/mol | -359.4 kJ/mol total enthalpy | -184.6 kJ/mol | -277.4 kJ/mol | -359 kJ/mol | -285.8 kJ/mol | -467.9 kJ/mol | -359.4 kJ/mol | H_initial = -821 kJ/mol | | | H_final = -1113 kJ/mol | | ΔH_rxn^0 | -1113 kJ/mol - -821 kJ/mol = -292.1 kJ/mol (exothermic) | | | | |

| hydrogen chloride | lead dioxide | sodium nitrite | water | sodium nitrate | lead(II) chloride molecular free energy | -95.3 kJ/mol | -217.3 kJ/mol | -284.6 kJ/mol | -237.1 kJ/mol | -366 kJ/mol | -314.1 kJ/mol total free energy | -190.6 kJ/mol | -217.3 kJ/mol | -284.6 kJ/mol | -237.1 kJ/mol | -366 kJ/mol | -314.1 kJ/mol | G_initial = -692.5 kJ/mol | | | G_final = -917.2 kJ/mol | | ΔG_rxn^0 | -917.2 kJ/mol - -692.5 kJ/mol = -224.7 kJ/mol (exergonic) | | | | |

Construct the equilibrium constant, K, expression for: HCl + PbO_2 + NaNO_2 ⟶ H_2O + NaNO_3 + PbCl_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: 2 HCl + PbO_2 + NaNO_2 ⟶ H_2O + NaNO_3 + PbCl_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 HCl | 2 | -2 PbO_2 | 1 | -1 NaNO_2 | 1 | -1 H_2O | 1 | 1 NaNO_3 | 1 | 1 PbCl_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HCl | 2 | -2 | ([HCl])^(-2) PbO_2 | 1 | -1 | ([PbO2])^(-1) NaNO_2 | 1 | -1 | ([NaNO2])^(-1) H_2O | 1 | 1 | [H2O] NaNO_3 | 1 | 1 | [NaNO3] PbCl_2 | 1 | 1 | [PbCl2] 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 = ([HCl])^(-2) ([PbO2])^(-1) ([NaNO2])^(-1) [H2O] [NaNO3] [PbCl2] = ([H2O] [NaNO3] [PbCl2])/(([HCl])^2 [PbO2] [NaNO2])

Construct the rate of reaction expression for: HCl + PbO_2 + NaNO_2 ⟶ H_2O + NaNO_3 + PbCl_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: 2 HCl + PbO_2 + NaNO_2 ⟶ H_2O + NaNO_3 + PbCl_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 HCl | 2 | -2 PbO_2 | 1 | -1 NaNO_2 | 1 | -1 H_2O | 1 | 1 NaNO_3 | 1 | 1 PbCl_2 | 1 | 1 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 HCl | 2 | -2 | -1/2 (Δ[HCl])/(Δt) PbO_2 | 1 | -1 | -(Δ[PbO2])/(Δt) NaNO_2 | 1 | -1 | -(Δ[NaNO2])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NaNO_3 | 1 | 1 | (Δ[NaNO3])/(Δt) PbCl_2 | 1 | 1 | (Δ[PbCl2])/(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[HCl])/(Δt) = -(Δ[PbO2])/(Δt) = -(Δ[NaNO2])/(Δt) = (Δ[H2O])/(Δt) = (Δ[NaNO3])/(Δt) = (Δ[PbCl2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen chloride | lead dioxide | sodium nitrite | water | sodium nitrate | lead(II) chloride formula | HCl | PbO_2 | NaNO_2 | H_2O | NaNO_3 | PbCl_2 Hill formula | ClH | O_2Pb | NNaO_2 | H_2O | NNaO_3 | Cl_2Pb name | hydrogen chloride | lead dioxide | sodium nitrite | water | sodium nitrate | lead(II) chloride IUPAC name | hydrogen chloride | | sodium nitrite | water | sodium nitrate | dichlorolead