HCl + PbO2 = H2O + Cl2 + PbCl2

HCl (hydrogen chloride) + PbO_2 (lead dioxide) ⟶ H_2O (water) + Cl_2 (chlorine) + PbCl_2 (lead(II) chloride)

Balance the chemical equation algebraically: HCl + PbO_2 ⟶ H_2O + Cl_2 + PbCl_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HCl + c_2 PbO_2 ⟶ c_3 H_2O + c_4 Cl_2 + c_5 PbCl_2 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, O and Pb: Cl: | c_1 = 2 c_4 + 2 c_5 H: | c_1 = 2 c_3 O: | 2 c_2 = c_3 Pb: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 2 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HCl + PbO_2 ⟶ 2 H_2O + Cl_2 + PbCl_2

+ ⟶ + +

hydrogen chloride + lead dioxide ⟶ water + chlorine + lead(II) chloride

| hydrogen chloride | lead dioxide | water | chlorine | lead(II) chloride molecular enthalpy | -92.3 kJ/mol | -277.4 kJ/mol | -285.8 kJ/mol | 0 kJ/mol | -359.4 kJ/mol total enthalpy | -369.2 kJ/mol | -277.4 kJ/mol | -571.7 kJ/mol | 0 kJ/mol | -359.4 kJ/mol | H_initial = -646.6 kJ/mol | | H_final = -931.1 kJ/mol | | ΔH_rxn^0 | -931.1 kJ/mol - -646.6 kJ/mol = -284.5 kJ/mol (exothermic) | | | |

| hydrogen chloride | lead dioxide | water | chlorine | lead(II) chloride molecular free energy | -95.3 kJ/mol | -217.3 kJ/mol | -237.1 kJ/mol | 0 kJ/mol | -314.1 kJ/mol total free energy | -381.2 kJ/mol | -217.3 kJ/mol | -474.2 kJ/mol | 0 kJ/mol | -314.1 kJ/mol | G_initial = -598.5 kJ/mol | | G_final = -788.3 kJ/mol | | ΔG_rxn^0 | -788.3 kJ/mol - -598.5 kJ/mol = -189.8 kJ/mol (exergonic) | | | |

Construct the equilibrium constant, K, expression for: HCl + PbO_2 ⟶ H_2O + Cl_2 + 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: 4 HCl + PbO_2 ⟶ 2 H_2O + Cl_2 + 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 | 4 | -4 PbO_2 | 1 | -1 H_2O | 2 | 2 Cl_2 | 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 | 4 | -4 | ([HCl])^(-4) PbO_2 | 1 | -1 | ([PbO2])^(-1) H_2O | 2 | 2 | ([H2O])^2 Cl_2 | 1 | 1 | [Cl2] 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])^(-4) ([PbO2])^(-1) ([H2O])^2 [Cl2] [PbCl2] = (([H2O])^2 [Cl2] [PbCl2])/(([HCl])^4 [PbO2])

Construct the rate of reaction expression for: HCl + PbO_2 ⟶ H_2O + Cl_2 + 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: 4 HCl + PbO_2 ⟶ 2 H_2O + Cl_2 + 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 | 4 | -4 PbO_2 | 1 | -1 H_2O | 2 | 2 Cl_2 | 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 | 4 | -4 | -1/4 (Δ[HCl])/(Δt) PbO_2 | 1 | -1 | -(Δ[PbO2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) Cl_2 | 1 | 1 | (Δ[Cl2])/(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 = -1/4 (Δ[HCl])/(Δt) = -(Δ[PbO2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[Cl2])/(Δt) = (Δ[PbCl2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen chloride | lead dioxide | water | chlorine | lead(II) chloride formula | HCl | PbO_2 | H_2O | Cl_2 | PbCl_2 Hill formula | ClH | O_2Pb | H_2O | Cl_2 | Cl_2Pb name | hydrogen chloride | lead dioxide | water | chlorine | lead(II) chloride IUPAC name | hydrogen chloride | | water | molecular chlorine | dichlorolead