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PbS + PbO = SO2 + Pb

Input interpretation

PbS lead sulfide + PbO lead monoxide ⟶ SO_2 sulfur dioxide + Pb lead
PbS lead sulfide + PbO lead monoxide ⟶ SO_2 sulfur dioxide + Pb lead

Balanced equation

Balance the chemical equation algebraically: PbS + PbO ⟶ SO_2 + Pb Add stoichiometric coefficients, c_i, to the reactants and products: c_1 PbS + c_2 PbO ⟶ c_3 SO_2 + c_4 Pb Set the number of atoms in the reactants equal to the number of atoms in the products for Pb, S and O: Pb: | c_1 + c_2 = c_4 S: | c_1 = c_3 O: | c_2 = 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 = 2 c_3 = 1 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | PbS + 2 PbO ⟶ SO_2 + 3 Pb
Balance the chemical equation algebraically: PbS + PbO ⟶ SO_2 + Pb Add stoichiometric coefficients, c_i, to the reactants and products: c_1 PbS + c_2 PbO ⟶ c_3 SO_2 + c_4 Pb Set the number of atoms in the reactants equal to the number of atoms in the products for Pb, S and O: Pb: | c_1 + c_2 = c_4 S: | c_1 = c_3 O: | c_2 = 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 = 2 c_3 = 1 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | PbS + 2 PbO ⟶ SO_2 + 3 Pb

Structures

 + ⟶ +
+ ⟶ +

Names

lead sulfide + lead monoxide ⟶ sulfur dioxide + lead
lead sulfide + lead monoxide ⟶ sulfur dioxide + lead

Equilibrium constant

Construct the equilibrium constant, K, expression for: PbS + PbO ⟶ SO_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: PbS + 2 PbO ⟶ SO_2 + 3 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 PbS | 1 | -1 PbO | 2 | -2 SO_2 | 1 | 1 Pb | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression PbS | 1 | -1 | ([PbS])^(-1) PbO | 2 | -2 | ([PbO])^(-2) SO_2 | 1 | 1 | [SO2] Pb | 3 | 3 | ([Pb])^3 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 = ([PbS])^(-1) ([PbO])^(-2) [SO2] ([Pb])^3 = ([SO2] ([Pb])^3)/([PbS] ([PbO])^2)
Construct the equilibrium constant, K, expression for: PbS + PbO ⟶ SO_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: PbS + 2 PbO ⟶ SO_2 + 3 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 PbS | 1 | -1 PbO | 2 | -2 SO_2 | 1 | 1 Pb | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression PbS | 1 | -1 | ([PbS])^(-1) PbO | 2 | -2 | ([PbO])^(-2) SO_2 | 1 | 1 | [SO2] Pb | 3 | 3 | ([Pb])^3 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 = ([PbS])^(-1) ([PbO])^(-2) [SO2] ([Pb])^3 = ([SO2] ([Pb])^3)/([PbS] ([PbO])^2)

Rate of reaction

Construct the rate of reaction expression for: PbS + PbO ⟶ SO_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: PbS + 2 PbO ⟶ SO_2 + 3 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 PbS | 1 | -1 PbO | 2 | -2 SO_2 | 1 | 1 Pb | 3 | 3 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 PbS | 1 | -1 | -(Δ[PbS])/(Δt) PbO | 2 | -2 | -1/2 (Δ[PbO])/(Δt) SO_2 | 1 | 1 | (Δ[SO2])/(Δt) Pb | 3 | 3 | 1/3 (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[PbS])/(Δt) = -1/2 (Δ[PbO])/(Δt) = (Δ[SO2])/(Δt) = 1/3 (Δ[Pb])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: PbS + PbO ⟶ SO_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: PbS + 2 PbO ⟶ SO_2 + 3 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 PbS | 1 | -1 PbO | 2 | -2 SO_2 | 1 | 1 Pb | 3 | 3 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 PbS | 1 | -1 | -(Δ[PbS])/(Δt) PbO | 2 | -2 | -1/2 (Δ[PbO])/(Δt) SO_2 | 1 | 1 | (Δ[SO2])/(Δt) Pb | 3 | 3 | 1/3 (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[PbS])/(Δt) = -1/2 (Δ[PbO])/(Δt) = (Δ[SO2])/(Δt) = 1/3 (Δ[Pb])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | lead sulfide | lead monoxide | sulfur dioxide | lead formula | PbS | PbO | SO_2 | Pb Hill formula | PbS | OPb | O_2S | Pb name | lead sulfide | lead monoxide | sulfur dioxide | lead
| lead sulfide | lead monoxide | sulfur dioxide | lead formula | PbS | PbO | SO_2 | Pb Hill formula | PbS | OPb | O_2S | Pb name | lead sulfide | lead monoxide | sulfur dioxide | lead

Substance properties

 | lead sulfide | lead monoxide | sulfur dioxide | lead molar mass | 239.3 g/mol | 223.2 g/mol | 64.06 g/mol | 207.2 g/mol phase | solid (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | 1114 °C | 886 °C | -73 °C | 327.4 °C boiling point | 1344 °C | 1470 °C | -10 °C | 1740 °C density | 7.5 g/cm^3 | 9.5 g/cm^3 | 0.002619 g/cm^3 (at 25 °C) | 11.34 g/cm^3 solubility in water | insoluble | insoluble | | insoluble surface tension | | | 0.02859 N/m |  dynamic viscosity | | 1.45×10^-4 Pa s (at 1000 °C) | 1.282×10^-5 Pa s (at 25 °C) | 0.00183 Pa s (at 38 °C)
| lead sulfide | lead monoxide | sulfur dioxide | lead molar mass | 239.3 g/mol | 223.2 g/mol | 64.06 g/mol | 207.2 g/mol phase | solid (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | 1114 °C | 886 °C | -73 °C | 327.4 °C boiling point | 1344 °C | 1470 °C | -10 °C | 1740 °C density | 7.5 g/cm^3 | 9.5 g/cm^3 | 0.002619 g/cm^3 (at 25 °C) | 11.34 g/cm^3 solubility in water | insoluble | insoluble | | insoluble surface tension | | | 0.02859 N/m | dynamic viscosity | | 1.45×10^-4 Pa s (at 1000 °C) | 1.282×10^-5 Pa s (at 25 °C) | 0.00183 Pa s (at 38 °C)

Units