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C + PbO2 = CO + Pb

Input interpretation

C activated charcoal + PbO_2 lead dioxide ⟶ CO carbon monoxide + Pb lead
C activated charcoal + PbO_2 lead dioxide ⟶ CO carbon monoxide + Pb lead

Balanced equation

Balance the chemical equation algebraically: C + PbO_2 ⟶ CO + Pb Add stoichiometric coefficients, c_i, to the reactants and products: c_1 C + c_2 PbO_2 ⟶ c_3 CO + 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: | 2 c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 2 C + PbO_2 ⟶ 2 CO + Pb
Balance the chemical equation algebraically: C + PbO_2 ⟶ CO + Pb Add stoichiometric coefficients, c_i, to the reactants and products: c_1 C + c_2 PbO_2 ⟶ c_3 CO + 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: | 2 c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 C + PbO_2 ⟶ 2 CO + Pb

Structures

 + ⟶ +
+ ⟶ +

Names

activated charcoal + lead dioxide ⟶ carbon monoxide + lead
activated charcoal + lead dioxide ⟶ carbon monoxide + lead

Equilibrium constant

Construct the equilibrium constant, K, expression for: C + PbO_2 ⟶ CO + 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: 2 C + PbO_2 ⟶ 2 CO + 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 | 2 | -2 PbO_2 | 1 | -1 CO | 2 | 2 Pb | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression C | 2 | -2 | ([C])^(-2) PbO_2 | 1 | -1 | ([PbO2])^(-1) CO | 2 | 2 | ([CO])^2 Pb | 1 | 1 | [Pb] 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])^(-2) ([PbO2])^(-1) ([CO])^2 [Pb] = (([CO])^2 [Pb])/(([C])^2 [PbO2])
Construct the equilibrium constant, K, expression for: C + PbO_2 ⟶ CO + 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: 2 C + PbO_2 ⟶ 2 CO + 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 | 2 | -2 PbO_2 | 1 | -1 CO | 2 | 2 Pb | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression C | 2 | -2 | ([C])^(-2) PbO_2 | 1 | -1 | ([PbO2])^(-1) CO | 2 | 2 | ([CO])^2 Pb | 1 | 1 | [Pb] 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])^(-2) ([PbO2])^(-1) ([CO])^2 [Pb] = (([CO])^2 [Pb])/(([C])^2 [PbO2])

Rate of reaction

Construct the rate of reaction expression for: C + PbO_2 ⟶ CO + 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: 2 C + PbO_2 ⟶ 2 CO + 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 | 2 | -2 PbO_2 | 1 | -1 CO | 2 | 2 Pb | 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 C | 2 | -2 | -1/2 (Δ[C])/(Δt) PbO_2 | 1 | -1 | -(Δ[PbO2])/(Δt) CO | 2 | 2 | 1/2 (Δ[CO])/(Δt) Pb | 1 | 1 | (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 = -1/2 (Δ[C])/(Δt) = -(Δ[PbO2])/(Δt) = 1/2 (Δ[CO])/(Δt) = (Δ[Pb])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: C + PbO_2 ⟶ CO + 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: 2 C + PbO_2 ⟶ 2 CO + 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 | 2 | -2 PbO_2 | 1 | -1 CO | 2 | 2 Pb | 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 C | 2 | -2 | -1/2 (Δ[C])/(Δt) PbO_2 | 1 | -1 | -(Δ[PbO2])/(Δt) CO | 2 | 2 | 1/2 (Δ[CO])/(Δt) Pb | 1 | 1 | (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 = -1/2 (Δ[C])/(Δt) = -(Δ[PbO2])/(Δt) = 1/2 (Δ[CO])/(Δt) = (Δ[Pb])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | activated charcoal | lead dioxide | carbon monoxide | lead formula | C | PbO_2 | CO | Pb Hill formula | C | O_2Pb | CO | Pb name | activated charcoal | lead dioxide | carbon monoxide | lead IUPAC name | carbon | | carbon monoxide | lead
| activated charcoal | lead dioxide | carbon monoxide | lead formula | C | PbO_2 | CO | Pb Hill formula | C | O_2Pb | CO | Pb name | activated charcoal | lead dioxide | carbon monoxide | lead IUPAC name | carbon | | carbon monoxide | lead

Substance properties

 | activated charcoal | lead dioxide | carbon monoxide | lead molar mass | 12.011 g/mol | 239.2 g/mol | 28.01 g/mol | 207.2 g/mol phase | solid (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | 3550 °C | 290 °C | -205 °C | 327.4 °C boiling point | 4027 °C | | -191.5 °C | 1740 °C density | 2.26 g/cm^3 | 9.58 g/cm^3 | 0.001145 g/cm^3 (at 25 °C) | 11.34 g/cm^3 solubility in water | insoluble | insoluble | | insoluble dynamic viscosity | | | 1.772×10^-5 Pa s (at 25 °C) | 0.00183 Pa s (at 38 °C) odor | | | odorless |
| activated charcoal | lead dioxide | carbon monoxide | lead molar mass | 12.011 g/mol | 239.2 g/mol | 28.01 g/mol | 207.2 g/mol phase | solid (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | 3550 °C | 290 °C | -205 °C | 327.4 °C boiling point | 4027 °C | | -191.5 °C | 1740 °C density | 2.26 g/cm^3 | 9.58 g/cm^3 | 0.001145 g/cm^3 (at 25 °C) | 11.34 g/cm^3 solubility in water | insoluble | insoluble | | insoluble dynamic viscosity | | | 1.772×10^-5 Pa s (at 25 °C) | 0.00183 Pa s (at 38 °C) odor | | | odorless |

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