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Cl2 + KOH + Bi2O3 = H2O + KCl + KBiO3

Input interpretation

Cl_2 (chlorine) + KOH (potassium hydroxide) + Bi_2O_3 (bismuth trioxide) ⟶ H_2O (water) + KCl (potassium chloride) + KBiO3
Cl_2 (chlorine) + KOH (potassium hydroxide) + Bi_2O_3 (bismuth trioxide) ⟶ H_2O (water) + KCl (potassium chloride) + KBiO3

Balanced equation

Balance the chemical equation algebraically: Cl_2 + KOH + Bi_2O_3 ⟶ H_2O + KCl + KBiO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Cl_2 + c_2 KOH + 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 Cl, H, K, O and Bi: Cl: | 2 c_1 = c_5 H: | c_2 = 2 c_4 K: | c_2 = c_5 + c_6 O: | c_2 + 3 c_3 = c_4 + 3 c_6 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 = 6 c_3 = 1 c_4 = 3 c_5 = 4 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Cl_2 + 6 KOH + Bi_2O_3 ⟶ 3 H_2O + 4 KCl + 2 KBiO3
Balance the chemical equation algebraically: Cl_2 + KOH + Bi_2O_3 ⟶ H_2O + KCl + KBiO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Cl_2 + c_2 KOH + 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 Cl, H, K, O and Bi: Cl: | 2 c_1 = c_5 H: | c_2 = 2 c_4 K: | c_2 = c_5 + c_6 O: | c_2 + 3 c_3 = c_4 + 3 c_6 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 = 6 c_3 = 1 c_4 = 3 c_5 = 4 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Cl_2 + 6 KOH + Bi_2O_3 ⟶ 3 H_2O + 4 KCl + 2 KBiO3

Structures

+ + ⟶ + + KBiO3
+ + ⟶ + + KBiO3

Names

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

Equilibrium constant

Construct the equilibrium constant, K, expression for: Cl_2 + KOH + 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 Cl_2 + 6 KOH + Bi_2O_3 ⟶ 3 H_2O + 4 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 Cl_2 | 2 | -2 KOH | 6 | -6 Bi_2O_3 | 1 | -1 H_2O | 3 | 3 KCl | 4 | 4 KBiO3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Cl_2 | 2 | -2 | ([Cl2])^(-2) KOH | 6 | -6 | ([KOH])^(-6) Bi_2O_3 | 1 | -1 | ([Bi2O3])^(-1) H_2O | 3 | 3 | ([H2O])^3 KCl | 4 | 4 | ([KCl])^4 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 = ([Cl2])^(-2) ([KOH])^(-6) ([Bi2O3])^(-1) ([H2O])^3 ([KCl])^4 ([KBiO3])^2 = (([H2O])^3 ([KCl])^4 ([KBiO3])^2)/(([Cl2])^2 ([KOH])^6 [Bi2O3])
Construct the equilibrium constant, K, expression for: Cl_2 + KOH + 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 Cl_2 + 6 KOH + Bi_2O_3 ⟶ 3 H_2O + 4 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 Cl_2 | 2 | -2 KOH | 6 | -6 Bi_2O_3 | 1 | -1 H_2O | 3 | 3 KCl | 4 | 4 KBiO3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Cl_2 | 2 | -2 | ([Cl2])^(-2) KOH | 6 | -6 | ([KOH])^(-6) Bi_2O_3 | 1 | -1 | ([Bi2O3])^(-1) H_2O | 3 | 3 | ([H2O])^3 KCl | 4 | 4 | ([KCl])^4 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 = ([Cl2])^(-2) ([KOH])^(-6) ([Bi2O3])^(-1) ([H2O])^3 ([KCl])^4 ([KBiO3])^2 = (([H2O])^3 ([KCl])^4 ([KBiO3])^2)/(([Cl2])^2 ([KOH])^6 [Bi2O3])

Rate of reaction

Construct the rate of reaction expression for: Cl_2 + KOH + 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 Cl_2 + 6 KOH + Bi_2O_3 ⟶ 3 H_2O + 4 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 Cl_2 | 2 | -2 KOH | 6 | -6 Bi_2O_3 | 1 | -1 H_2O | 3 | 3 KCl | 4 | 4 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 Cl_2 | 2 | -2 | -1/2 (Δ[Cl2])/(Δt) KOH | 6 | -6 | -1/6 (Δ[KOH])/(Δt) Bi_2O_3 | 1 | -1 | -(Δ[Bi2O3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) KCl | 4 | 4 | 1/4 (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[Cl2])/(Δt) = -1/6 (Δ[KOH])/(Δt) = -(Δ[Bi2O3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/4 (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: Cl_2 + KOH + 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 Cl_2 + 6 KOH + Bi_2O_3 ⟶ 3 H_2O + 4 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 Cl_2 | 2 | -2 KOH | 6 | -6 Bi_2O_3 | 1 | -1 H_2O | 3 | 3 KCl | 4 | 4 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 Cl_2 | 2 | -2 | -1/2 (Δ[Cl2])/(Δt) KOH | 6 | -6 | -1/6 (Δ[KOH])/(Δt) Bi_2O_3 | 1 | -1 | -(Δ[Bi2O3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) KCl | 4 | 4 | 1/4 (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[Cl2])/(Δt) = -1/6 (Δ[KOH])/(Δt) = -(Δ[Bi2O3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/4 (Δ[KCl])/(Δt) = 1/2 (Δ[KBiO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| chlorine | potassium hydroxide | bismuth trioxide | water | potassium chloride | KBiO3 formula | Cl_2 | KOH | Bi_2O_3 | H_2O | KCl | KBiO3 Hill formula | Cl_2 | HKO | Bi_2O_3 | H_2O | ClK | BiKO3 name | chlorine | potassium hydroxide | bismuth trioxide | water | potassium chloride | IUPAC name | molecular chlorine | potassium hydroxide | oxo-oxobismuthanyloxybismuthane | water | potassium chloride |
| chlorine | potassium hydroxide | bismuth trioxide | water | potassium chloride | KBiO3 formula | Cl_2 | KOH | Bi_2O_3 | H_2O | KCl | KBiO3 Hill formula | Cl_2 | HKO | Bi_2O_3 | H_2O | ClK | BiKO3 name | chlorine | potassium hydroxide | bismuth trioxide | water | potassium chloride | IUPAC name | molecular chlorine | potassium hydroxide | oxo-oxobismuthanyloxybismuthane | water | potassium chloride |

Substance properties

| chlorine | potassium hydroxide | bismuth trioxide | water | potassium chloride | KBiO3 molar mass | 70.9 g/mol | 56.105 g/mol | 465.958 g/mol | 18.015 g/mol | 74.55 g/mol | 296.076 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | -101 °C | 406 °C | 825 °C | 0 °C | 770 °C | boiling point | -34 °C | 1327 °C | 1890 °C | 99.9839 °C | 1420 °C | density | 0.003214 g/cm^3 (at 0 °C) | 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 |
| chlorine | potassium hydroxide | bismuth trioxide | water | potassium chloride | KBiO3 molar mass | 70.9 g/mol | 56.105 g/mol | 465.958 g/mol | 18.015 g/mol | 74.55 g/mol | 296.076 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | -101 °C | 406 °C | 825 °C | 0 °C | 770 °C | boiling point | -34 °C | 1327 °C | 1890 °C | 99.9839 °C | 1420 °C | density | 0.003214 g/cm^3 (at 0 °C) | 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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