HI + PbO2 = H2O + I2 + PbI2

HI hydrogen iodide + PbO_2 lead dioxide ⟶ H_2O water + I_2 iodine + PbI_2 lead iodide

Balance the chemical equation algebraically: HI + PbO_2 ⟶ H_2O + I_2 + PbI_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HI + c_2 PbO_2 ⟶ c_3 H_2O + c_4 I_2 + c_5 PbI_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, I, O and Pb: H: | c_1 = 2 c_3 I: | c_1 = 2 c_4 + 2 c_5 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 HI + PbO_2 ⟶ 2 H_2O + I_2 + PbI_2

+ ⟶ + +

hydrogen iodide + lead dioxide ⟶ water + iodine + lead iodide

| hydrogen iodide | lead dioxide | water | iodine | lead iodide molecular enthalpy | 26.5 kJ/mol | -277.4 kJ/mol | -285.8 kJ/mol | 0 kJ/mol | -175.5 kJ/mol total enthalpy | 106 kJ/mol | -277.4 kJ/mol | -571.7 kJ/mol | 0 kJ/mol | -175.5 kJ/mol | H_initial = -171.4 kJ/mol | | H_final = -747.2 kJ/mol | | ΔH_rxn^0 | -747.2 kJ/mol - -171.4 kJ/mol = -575.8 kJ/mol (exothermic) | | | |

| hydrogen iodide | lead dioxide | water | iodine | lead iodide molecular free energy | 1.7 kJ/mol | -217.3 kJ/mol | -237.1 kJ/mol | 0 kJ/mol | -173.6 kJ/mol total free energy | 6.8 kJ/mol | -217.3 kJ/mol | -474.2 kJ/mol | 0 kJ/mol | -173.6 kJ/mol | G_initial = -210.5 kJ/mol | | G_final = -647.8 kJ/mol | | ΔG_rxn^0 | -647.8 kJ/mol - -210.5 kJ/mol = -437.3 kJ/mol (exergonic) | | | |

Construct the equilibrium constant, K, expression for: HI + PbO_2 ⟶ H_2O + I_2 + PbI_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 HI + PbO_2 ⟶ 2 H_2O + I_2 + PbI_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 HI | 4 | -4 PbO_2 | 1 | -1 H_2O | 2 | 2 I_2 | 1 | 1 PbI_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HI | 4 | -4 | ([HI])^(-4) PbO_2 | 1 | -1 | ([PbO2])^(-1) H_2O | 2 | 2 | ([H2O])^2 I_2 | 1 | 1 | [I2] PbI_2 | 1 | 1 | [PbI2] 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 = ([HI])^(-4) ([PbO2])^(-1) ([H2O])^2 [I2] [PbI2] = (([H2O])^2 [I2] [PbI2])/(([HI])^4 [PbO2])

Construct the rate of reaction expression for: HI + PbO_2 ⟶ H_2O + I_2 + PbI_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 HI + PbO_2 ⟶ 2 H_2O + I_2 + PbI_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 HI | 4 | -4 PbO_2 | 1 | -1 H_2O | 2 | 2 I_2 | 1 | 1 PbI_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 HI | 4 | -4 | -1/4 (Δ[HI])/(Δt) PbO_2 | 1 | -1 | -(Δ[PbO2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) I_2 | 1 | 1 | (Δ[I2])/(Δt) PbI_2 | 1 | 1 | (Δ[PbI2])/(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[HI])/(Δt) = -(Δ[PbO2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[I2])/(Δt) = (Δ[PbI2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen iodide | lead dioxide | water | iodine | lead iodide formula | HI | PbO_2 | H_2O | I_2 | PbI_2 Hill formula | HI | O_2Pb | H_2O | I_2 | I_2Pb name | hydrogen iodide | lead dioxide | water | iodine | lead iodide IUPAC name | hydrogen iodide | | water | molecular iodine |

| hydrogen iodide | lead dioxide | water | iodine | lead iodide molar mass | 127.912 g/mol | 239.2 g/mol | 18.015 g/mol | 253.80894 g/mol | 461 g/mol phase | gas (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | solid (at STP) melting point | -50.76 °C | 290 °C | 0 °C | 113 °C | 402 °C boiling point | -35.55 °C | | 99.9839 °C | 184 °C | 954 °C density | 0.005228 g/cm^3 (at 25 °C) | 9.58 g/cm^3 | 1 g/cm^3 | 4.94 g/cm^3 | 6.16 g/cm^3 solubility in water | very soluble | insoluble | | | surface tension | | | 0.0728 N/m | | dynamic viscosity | 0.001321 Pa s (at -39 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 0.00227 Pa s (at 116 °C) | odor | | | odorless | |