C + PbO = CO2 + Pb

C (activated charcoal) + PbO (lead monoxide) ⟶ CO_2 (carbon dioxide) + Pb (lead)

Balance the chemical equation algebraically: C + PbO ⟶ CO_2 + Pb Add stoichiometric coefficients, c_i, to the reactants and products: c_1 C + c_2 PbO ⟶ c_3 CO_2 + c_4 Pb Set the number of atoms in the reactants equal to the number of atoms in the products for C, O and Pb: C: | c_1 = c_3 O: | c_2 = 2 c_3 Pb: | 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 = 2 c_3 = 1 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | C + 2 PbO ⟶ CO_2 + 2 Pb

+ ⟶ +

activated charcoal + lead monoxide ⟶ carbon dioxide + lead

Construct the equilibrium constant, K, expression for: C + PbO ⟶ CO_2 + Pb 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: C + 2 PbO ⟶ CO_2 + 2 Pb 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 C | 1 | -1 PbO | 2 | -2 CO_2 | 1 | 1 Pb | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression C | 1 | -1 | ([C])^(-1) PbO | 2 | -2 | ([PbO])^(-2) CO_2 | 1 | 1 | [CO2] Pb | 2 | 2 | ([Pb])^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 = ([C])^(-1) ([PbO])^(-2) [CO2] ([Pb])^2 = ([CO2] ([Pb])^2)/([C] ([PbO])^2)

Construct the rate of reaction expression for: C + PbO ⟶ CO_2 + Pb 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: C + 2 PbO ⟶ CO_2 + 2 Pb 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 C | 1 | -1 PbO | 2 | -2 CO_2 | 1 | 1 Pb | 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 C | 1 | -1 | -(Δ[C])/(Δt) PbO | 2 | -2 | -1/2 (Δ[PbO])/(Δt) CO_2 | 1 | 1 | (Δ[CO2])/(Δt) Pb | 2 | 2 | 1/2 (Δ[Pb])/(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[C])/(Δt) = -1/2 (Δ[PbO])/(Δt) = (Δ[CO2])/(Δt) = 1/2 (Δ[Pb])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| activated charcoal | lead monoxide | carbon dioxide | lead formula | C | PbO | CO_2 | Pb Hill formula | C | OPb | CO_2 | Pb name | activated charcoal | lead monoxide | carbon dioxide | lead IUPAC name | carbon | | carbon dioxide | lead

| activated charcoal | lead monoxide | carbon dioxide | lead molar mass | 12.011 g/mol | 223.2 g/mol | 44.009 g/mol | 207.2 g/mol phase | solid (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | 3550 °C | 886 °C | -56.56 °C (at triple point) | 327.4 °C boiling point | 4027 °C | 1470 °C | -78.5 °C (at sublimation point) | 1740 °C density | 2.26 g/cm^3 | 9.5 g/cm^3 | 0.00184212 g/cm^3 (at 20 °C) | 11.34 g/cm^3 solubility in water | insoluble | insoluble | | insoluble dynamic viscosity | | 1.45×10^-4 Pa s (at 1000 °C) | 1.491×10^-5 Pa s (at 25 °C) | 0.00183 Pa s (at 38 °C) odor | | | odorless |