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KOH + PbO2 + Sb = H2O + PbO + KSbO2

Input interpretation

KOH potassium hydroxide + PbO_2 lead dioxide + Sb gray antimony ⟶ H_2O water + PbO lead monoxide + KSbO2
KOH potassium hydroxide + PbO_2 lead dioxide + Sb gray antimony ⟶ H_2O water + PbO lead monoxide + KSbO2

Balanced equation

Balance the chemical equation algebraically: KOH + PbO_2 + Sb ⟶ H_2O + PbO + KSbO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 PbO_2 + c_3 Sb ⟶ c_4 H_2O + c_5 PbO + c_6 KSbO2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Pb and Sb: H: | c_1 = 2 c_4 K: | c_1 = c_6 O: | c_1 + 2 c_2 = c_4 + c_5 + 2 c_6 Pb: | c_2 = c_5 Sb: | c_3 = c_6 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 2 c_4 = 1 c_5 = 3 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + 3 PbO_2 + 2 Sb ⟶ H_2O + 3 PbO + 2 KSbO2
Balance the chemical equation algebraically: KOH + PbO_2 + Sb ⟶ H_2O + PbO + KSbO2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 PbO_2 + c_3 Sb ⟶ c_4 H_2O + c_5 PbO + c_6 KSbO2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Pb and Sb: H: | c_1 = 2 c_4 K: | c_1 = c_6 O: | c_1 + 2 c_2 = c_4 + c_5 + 2 c_6 Pb: | c_2 = c_5 Sb: | c_3 = c_6 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 2 c_4 = 1 c_5 = 3 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + 3 PbO_2 + 2 Sb ⟶ H_2O + 3 PbO + 2 KSbO2

Structures

+ + ⟶ + + KSbO2
+ + ⟶ + + KSbO2

Names

potassium hydroxide + lead dioxide + gray antimony ⟶ water + lead monoxide + KSbO2
potassium hydroxide + lead dioxide + gray antimony ⟶ water + lead monoxide + KSbO2

Equilibrium constant

Construct the equilibrium constant, K, expression for: KOH + PbO_2 + Sb ⟶ H_2O + PbO + KSbO2 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 KOH + 3 PbO_2 + 2 Sb ⟶ H_2O + 3 PbO + 2 KSbO2 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 KOH | 2 | -2 PbO_2 | 3 | -3 Sb | 2 | -2 H_2O | 1 | 1 PbO | 3 | 3 KSbO2 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 2 | -2 | ([KOH])^(-2) PbO_2 | 3 | -3 | ([PbO2])^(-3) Sb | 2 | -2 | ([Sb])^(-2) H_2O | 1 | 1 | [H2O] PbO | 3 | 3 | ([PbO])^3 KSbO2 | 2 | 2 | ([KSbO2])^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 = ([KOH])^(-2) ([PbO2])^(-3) ([Sb])^(-2) [H2O] ([PbO])^3 ([KSbO2])^2 = ([H2O] ([PbO])^3 ([KSbO2])^2)/(([KOH])^2 ([PbO2])^3 ([Sb])^2)
Construct the equilibrium constant, K, expression for: KOH + PbO_2 + Sb ⟶ H_2O + PbO + KSbO2 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 KOH + 3 PbO_2 + 2 Sb ⟶ H_2O + 3 PbO + 2 KSbO2 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 KOH | 2 | -2 PbO_2 | 3 | -3 Sb | 2 | -2 H_2O | 1 | 1 PbO | 3 | 3 KSbO2 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 2 | -2 | ([KOH])^(-2) PbO_2 | 3 | -3 | ([PbO2])^(-3) Sb | 2 | -2 | ([Sb])^(-2) H_2O | 1 | 1 | [H2O] PbO | 3 | 3 | ([PbO])^3 KSbO2 | 2 | 2 | ([KSbO2])^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 = ([KOH])^(-2) ([PbO2])^(-3) ([Sb])^(-2) [H2O] ([PbO])^3 ([KSbO2])^2 = ([H2O] ([PbO])^3 ([KSbO2])^2)/(([KOH])^2 ([PbO2])^3 ([Sb])^2)

Rate of reaction

Construct the rate of reaction expression for: KOH + PbO_2 + Sb ⟶ H_2O + PbO + KSbO2 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 KOH + 3 PbO_2 + 2 Sb ⟶ H_2O + 3 PbO + 2 KSbO2 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 KOH | 2 | -2 PbO_2 | 3 | -3 Sb | 2 | -2 H_2O | 1 | 1 PbO | 3 | 3 KSbO2 | 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 KOH | 2 | -2 | -1/2 (Δ[KOH])/(Δt) PbO_2 | 3 | -3 | -1/3 (Δ[PbO2])/(Δt) Sb | 2 | -2 | -1/2 (Δ[Sb])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) PbO | 3 | 3 | 1/3 (Δ[PbO])/(Δt) KSbO2 | 2 | 2 | 1/2 (Δ[KSbO2])/(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[KOH])/(Δt) = -1/3 (Δ[PbO2])/(Δt) = -1/2 (Δ[Sb])/(Δt) = (Δ[H2O])/(Δt) = 1/3 (Δ[PbO])/(Δt) = 1/2 (Δ[KSbO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KOH + PbO_2 + Sb ⟶ H_2O + PbO + KSbO2 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 KOH + 3 PbO_2 + 2 Sb ⟶ H_2O + 3 PbO + 2 KSbO2 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 KOH | 2 | -2 PbO_2 | 3 | -3 Sb | 2 | -2 H_2O | 1 | 1 PbO | 3 | 3 KSbO2 | 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 KOH | 2 | -2 | -1/2 (Δ[KOH])/(Δt) PbO_2 | 3 | -3 | -1/3 (Δ[PbO2])/(Δt) Sb | 2 | -2 | -1/2 (Δ[Sb])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) PbO | 3 | 3 | 1/3 (Δ[PbO])/(Δt) KSbO2 | 2 | 2 | 1/2 (Δ[KSbO2])/(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[KOH])/(Δt) = -1/3 (Δ[PbO2])/(Δt) = -1/2 (Δ[Sb])/(Δt) = (Δ[H2O])/(Δt) = 1/3 (Δ[PbO])/(Δt) = 1/2 (Δ[KSbO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium hydroxide | lead dioxide | gray antimony | water | lead monoxide | KSbO2 formula | KOH | PbO_2 | Sb | H_2O | PbO | KSbO2 Hill formula | HKO | O_2Pb | Sb | H_2O | OPb | KO2Sb name | potassium hydroxide | lead dioxide | gray antimony | water | lead monoxide | IUPAC name | potassium hydroxide | | antimony | water | |
| potassium hydroxide | lead dioxide | gray antimony | water | lead monoxide | KSbO2 formula | KOH | PbO_2 | Sb | H_2O | PbO | KSbO2 Hill formula | HKO | O_2Pb | Sb | H_2O | OPb | KO2Sb name | potassium hydroxide | lead dioxide | gray antimony | water | lead monoxide | IUPAC name | potassium hydroxide | | antimony | water | |

Substance properties

| potassium hydroxide | lead dioxide | gray antimony | water | lead monoxide | KSbO2 molar mass | 56.105 g/mol | 239.2 g/mol | 121.76 g/mol | 18.015 g/mol | 223.2 g/mol | 192.856 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | 406 °C | 290 °C | 630 °C | 0 °C | 886 °C | boiling point | 1327 °C | | 1587 °C | 99.9839 °C | 1470 °C | density | 2.044 g/cm^3 | 9.58 g/cm^3 | 6.69 g/cm^3 | 1 g/cm^3 | 9.5 g/cm^3 | solubility in water | soluble | insoluble | | | insoluble | surface tension | | | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | | | 8.9×10^-4 Pa s (at 25 °C) | 1.45×10^-4 Pa s (at 1000 °C) | odor | | | | odorless | |
| potassium hydroxide | lead dioxide | gray antimony | water | lead monoxide | KSbO2 molar mass | 56.105 g/mol | 239.2 g/mol | 121.76 g/mol | 18.015 g/mol | 223.2 g/mol | 192.856 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | 406 °C | 290 °C | 630 °C | 0 °C | 886 °C | boiling point | 1327 °C | | 1587 °C | 99.9839 °C | 1470 °C | density | 2.044 g/cm^3 | 9.58 g/cm^3 | 6.69 g/cm^3 | 1 g/cm^3 | 9.5 g/cm^3 | solubility in water | soluble | insoluble | | | insoluble | surface tension | | | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | | | 8.9×10^-4 Pa s (at 25 °C) | 1.45×10^-4 Pa s (at 1000 °C) | odor | | | | odorless | |

Units

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