O2 + Pb = PbO2

O_2 (oxygen) + Pb (lead) ⟶ PbO_2 (lead dioxide)

Balance the chemical equation algebraically: O_2 + Pb ⟶ PbO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 Pb ⟶ c_3 PbO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for O and Pb: O: | 2 c_1 = 2 c_3 Pb: | c_2 = c_3 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 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | O_2 + Pb ⟶ PbO_2

+ ⟶

oxygen + lead ⟶ lead dioxide

| oxygen | lead | lead dioxide molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -277.4 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -277.4 kJ/mol | H_initial = 0 kJ/mol | | H_final = -277.4 kJ/mol ΔH_rxn^0 | -277.4 kJ/mol - 0 kJ/mol = -277.4 kJ/mol (exothermic) | |

| oxygen | lead | lead dioxide molecular entropy | 205 J/(mol K) | 65 J/(mol K) | 69 J/(mol K) total entropy | 205 J/(mol K) | 65 J/(mol K) | 69 J/(mol K) | S_initial = 270 J/(mol K) | | S_final = 69 J/(mol K) ΔS_rxn^0 | 69 J/(mol K) - 270 J/(mol K) = -201 J/(mol K) (exoentropic) | |

Construct the equilibrium constant, K, expression for: O_2 + Pb ⟶ PbO_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: O_2 + Pb ⟶ PbO_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 O_2 | 1 | -1 Pb | 1 | -1 PbO_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 1 | -1 | ([O2])^(-1) Pb | 1 | -1 | ([Pb])^(-1) PbO_2 | 1 | 1 | [PbO2] 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 = ([O2])^(-1) ([Pb])^(-1) [PbO2] = ([PbO2])/([O2] [Pb])

Construct the rate of reaction expression for: O_2 + Pb ⟶ PbO_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: O_2 + Pb ⟶ PbO_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 O_2 | 1 | -1 Pb | 1 | -1 PbO_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 O_2 | 1 | -1 | -(Δ[O2])/(Δt) Pb | 1 | -1 | -(Δ[Pb])/(Δt) PbO_2 | 1 | 1 | (Δ[PbO2])/(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 = -(Δ[O2])/(Δt) = -(Δ[Pb])/(Δt) = (Δ[PbO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | lead | lead dioxide formula | O_2 | Pb | PbO_2 Hill formula | O_2 | Pb | O_2Pb name | oxygen | lead | lead dioxide IUPAC name | molecular oxygen | lead |

| oxygen | lead | lead dioxide molar mass | 31.998 g/mol | 207.2 g/mol | 239.2 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) melting point | -218 °C | 327.4 °C | 290 °C boiling point | -183 °C | 1740 °C | density | 0.001429 g/cm^3 (at 0 °C) | 11.34 g/cm^3 | 9.58 g/cm^3 solubility in water | | insoluble | insoluble surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 0.00183 Pa s (at 38 °C) | odor | odorless | |