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KOH + KClO + Bi2O3 = H2O + KCl + KBiO3

Input interpretation

KOH potassium hydroxide + KClO + Bi_2O_3 bismuth trioxide ⟶ H_2O water + KCl potassium chloride + KBiO3
KOH potassium hydroxide + KClO + Bi_2O_3 bismuth trioxide ⟶ H_2O water + KCl potassium chloride + KBiO3

Balanced equation

Balance the chemical equation algebraically: KOH + KClO + Bi_2O_3 ⟶ H_2O + KCl + KBiO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 KClO + c_3 Bi_2O_3 ⟶ c_4 H_2O + c_5 KCl + c_6 KBiO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Cl and Bi: H: | c_1 = 2 c_4 K: | c_1 + c_2 = c_5 + c_6 O: | c_1 + c_2 + 3 c_3 = c_4 + 3 c_6 Cl: | c_2 = c_5 Bi: | 2 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 2 c_3 = 1 c_4 = 1 c_5 = 2 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + 2 KClO + Bi_2O_3 ⟶ H_2O + 2 KCl + 2 KBiO3
Balance the chemical equation algebraically: KOH + KClO + Bi_2O_3 ⟶ H_2O + KCl + KBiO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 KClO + c_3 Bi_2O_3 ⟶ c_4 H_2O + c_5 KCl + c_6 KBiO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Cl and Bi: H: | c_1 = 2 c_4 K: | c_1 + c_2 = c_5 + c_6 O: | c_1 + c_2 + 3 c_3 = c_4 + 3 c_6 Cl: | c_2 = c_5 Bi: | 2 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 2 c_3 = 1 c_4 = 1 c_5 = 2 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + 2 KClO + Bi_2O_3 ⟶ H_2O + 2 KCl + 2 KBiO3

Structures

+ KClO + ⟶ + + KBiO3
+ KClO + ⟶ + + KBiO3

Names

potassium hydroxide + KClO + bismuth trioxide ⟶ water + potassium chloride + KBiO3
potassium hydroxide + KClO + bismuth trioxide ⟶ water + potassium chloride + KBiO3

Equilibrium constant

Construct the equilibrium constant, K, expression for: KOH + KClO + Bi_2O_3 ⟶ H_2O + KCl + KBiO3 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 + 2 KClO + Bi_2O_3 ⟶ H_2O + 2 KCl + 2 KBiO3 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 KClO | 2 | -2 Bi_2O_3 | 1 | -1 H_2O | 1 | 1 KCl | 2 | 2 KBiO3 | 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) KClO | 2 | -2 | ([KClO])^(-2) Bi_2O_3 | 1 | -1 | ([Bi2O3])^(-1) H_2O | 1 | 1 | [H2O] KCl | 2 | 2 | ([KCl])^2 KBiO3 | 2 | 2 | ([KBiO3])^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) ([KClO])^(-2) ([Bi2O3])^(-1) [H2O] ([KCl])^2 ([KBiO3])^2 = ([H2O] ([KCl])^2 ([KBiO3])^2)/(([KOH])^2 ([KClO])^2 [Bi2O3])
Construct the equilibrium constant, K, expression for: KOH + KClO + Bi_2O_3 ⟶ H_2O + KCl + KBiO3 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 + 2 KClO + Bi_2O_3 ⟶ H_2O + 2 KCl + 2 KBiO3 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 KClO | 2 | -2 Bi_2O_3 | 1 | -1 H_2O | 1 | 1 KCl | 2 | 2 KBiO3 | 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) KClO | 2 | -2 | ([KClO])^(-2) Bi_2O_3 | 1 | -1 | ([Bi2O3])^(-1) H_2O | 1 | 1 | [H2O] KCl | 2 | 2 | ([KCl])^2 KBiO3 | 2 | 2 | ([KBiO3])^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) ([KClO])^(-2) ([Bi2O3])^(-1) [H2O] ([KCl])^2 ([KBiO3])^2 = ([H2O] ([KCl])^2 ([KBiO3])^2)/(([KOH])^2 ([KClO])^2 [Bi2O3])

Rate of reaction

Construct the rate of reaction expression for: KOH + KClO + Bi_2O_3 ⟶ H_2O + KCl + KBiO3 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 + 2 KClO + Bi_2O_3 ⟶ H_2O + 2 KCl + 2 KBiO3 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 KClO | 2 | -2 Bi_2O_3 | 1 | -1 H_2O | 1 | 1 KCl | 2 | 2 KBiO3 | 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) KClO | 2 | -2 | -1/2 (Δ[KClO])/(Δt) Bi_2O_3 | 1 | -1 | -(Δ[Bi2O3])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) KCl | 2 | 2 | 1/2 (Δ[KCl])/(Δt) KBiO3 | 2 | 2 | 1/2 (Δ[KBiO3])/(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) = -1/2 (Δ[KClO])/(Δt) = -(Δ[Bi2O3])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[KCl])/(Δt) = 1/2 (Δ[KBiO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KOH + KClO + Bi_2O_3 ⟶ H_2O + KCl + KBiO3 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 + 2 KClO + Bi_2O_3 ⟶ H_2O + 2 KCl + 2 KBiO3 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 KClO | 2 | -2 Bi_2O_3 | 1 | -1 H_2O | 1 | 1 KCl | 2 | 2 KBiO3 | 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) KClO | 2 | -2 | -1/2 (Δ[KClO])/(Δt) Bi_2O_3 | 1 | -1 | -(Δ[Bi2O3])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) KCl | 2 | 2 | 1/2 (Δ[KCl])/(Δt) KBiO3 | 2 | 2 | 1/2 (Δ[KBiO3])/(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) = -1/2 (Δ[KClO])/(Δt) = -(Δ[Bi2O3])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[KCl])/(Δt) = 1/2 (Δ[KBiO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium hydroxide | KClO | bismuth trioxide | water | potassium chloride | KBiO3 formula | KOH | KClO | Bi_2O_3 | H_2O | KCl | KBiO3 Hill formula | HKO | ClKO | Bi_2O_3 | H_2O | ClK | BiKO3 name | potassium hydroxide | | bismuth trioxide | water | potassium chloride | IUPAC name | potassium hydroxide | | oxo-oxobismuthanyloxybismuthane | water | potassium chloride |
| potassium hydroxide | KClO | bismuth trioxide | water | potassium chloride | KBiO3 formula | KOH | KClO | Bi_2O_3 | H_2O | KCl | KBiO3 Hill formula | HKO | ClKO | Bi_2O_3 | H_2O | ClK | BiKO3 name | potassium hydroxide | | bismuth trioxide | water | potassium chloride | IUPAC name | potassium hydroxide | | oxo-oxobismuthanyloxybismuthane | water | potassium chloride |

Substance properties

| potassium hydroxide | KClO | bismuth trioxide | water | potassium chloride | KBiO3 molar mass | 56.105 g/mol | 90.55 g/mol | 465.958 g/mol | 18.015 g/mol | 74.55 g/mol | 296.076 g/mol phase | solid (at STP) | | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | 406 °C | | 825 °C | 0 °C | 770 °C | boiling point | 1327 °C | | 1890 °C | 99.9839 °C | 1420 °C | density | 2.044 g/cm^3 | | 8.9 g/cm^3 | 1 g/cm^3 | 1.98 g/cm^3 | solubility in water | soluble | | decomposes | | soluble | surface tension | | | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | odorless | odorless |
| potassium hydroxide | KClO | bismuth trioxide | water | potassium chloride | KBiO3 molar mass | 56.105 g/mol | 90.55 g/mol | 465.958 g/mol | 18.015 g/mol | 74.55 g/mol | 296.076 g/mol phase | solid (at STP) | | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | 406 °C | | 825 °C | 0 °C | 770 °C | boiling point | 1327 °C | | 1890 °C | 99.9839 °C | 1420 °C | density | 2.044 g/cm^3 | | 8.9 g/cm^3 | 1 g/cm^3 | 1.98 g/cm^3 | solubility in water | soluble | | decomposes | | soluble | surface tension | | | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | odorless | odorless |

Units

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