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KOH + Br2 + Sb2O3 = H2O + KBr + KSbO3

Input interpretation

KOH potassium hydroxide + Br_2 bromine + Sb_2O_3 antimony trioxide ⟶ H_2O water + KBr potassium bromide + KSbO3
KOH potassium hydroxide + Br_2 bromine + Sb_2O_3 antimony trioxide ⟶ H_2O water + KBr potassium bromide + KSbO3

Balanced equation

Balance the chemical equation algebraically: KOH + Br_2 + Sb_2O_3 ⟶ H_2O + KBr + KSbO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 Br_2 + c_3 Sb_2O_3 ⟶ c_4 H_2O + c_5 KBr + c_6 KSbO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Br and Sb: H: | c_1 = 2 c_4 K: | c_1 = c_5 + c_6 O: | c_1 + 3 c_3 = c_4 + 3 c_6 Br: | 2 c_2 = c_5 Sb: | 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 = 6 c_2 = 2 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: | | 6 KOH + 2 Br_2 + Sb_2O_3 ⟶ 3 H_2O + 4 KBr + 2 KSbO3
Balance the chemical equation algebraically: KOH + Br_2 + Sb_2O_3 ⟶ H_2O + KBr + KSbO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 Br_2 + c_3 Sb_2O_3 ⟶ c_4 H_2O + c_5 KBr + c_6 KSbO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Br and Sb: H: | c_1 = 2 c_4 K: | c_1 = c_5 + c_6 O: | c_1 + 3 c_3 = c_4 + 3 c_6 Br: | 2 c_2 = c_5 Sb: | 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 = 6 c_2 = 2 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: | | 6 KOH + 2 Br_2 + Sb_2O_3 ⟶ 3 H_2O + 4 KBr + 2 KSbO3

Structures

+ + ⟶ + + KSbO3
+ + ⟶ + + KSbO3

Names

potassium hydroxide + bromine + antimony trioxide ⟶ water + potassium bromide + KSbO3
potassium hydroxide + bromine + antimony trioxide ⟶ water + potassium bromide + KSbO3

Equilibrium constant

Construct the equilibrium constant, K, expression for: KOH + Br_2 + Sb_2O_3 ⟶ H_2O + KBr + KSbO3 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: 6 KOH + 2 Br_2 + Sb_2O_3 ⟶ 3 H_2O + 4 KBr + 2 KSbO3 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 | 6 | -6 Br_2 | 2 | -2 Sb_2O_3 | 1 | -1 H_2O | 3 | 3 KBr | 4 | 4 KSbO3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 6 | -6 | ([KOH])^(-6) Br_2 | 2 | -2 | ([Br2])^(-2) Sb_2O_3 | 1 | -1 | ([Sb2O3])^(-1) H_2O | 3 | 3 | ([H2O])^3 KBr | 4 | 4 | ([KBr])^4 KSbO3 | 2 | 2 | ([KSbO3])^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])^(-6) ([Br2])^(-2) ([Sb2O3])^(-1) ([H2O])^3 ([KBr])^4 ([KSbO3])^2 = (([H2O])^3 ([KBr])^4 ([KSbO3])^2)/(([KOH])^6 ([Br2])^2 [Sb2O3])
Construct the equilibrium constant, K, expression for: KOH + Br_2 + Sb_2O_3 ⟶ H_2O + KBr + KSbO3 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: 6 KOH + 2 Br_2 + Sb_2O_3 ⟶ 3 H_2O + 4 KBr + 2 KSbO3 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 | 6 | -6 Br_2 | 2 | -2 Sb_2O_3 | 1 | -1 H_2O | 3 | 3 KBr | 4 | 4 KSbO3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 6 | -6 | ([KOH])^(-6) Br_2 | 2 | -2 | ([Br2])^(-2) Sb_2O_3 | 1 | -1 | ([Sb2O3])^(-1) H_2O | 3 | 3 | ([H2O])^3 KBr | 4 | 4 | ([KBr])^4 KSbO3 | 2 | 2 | ([KSbO3])^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])^(-6) ([Br2])^(-2) ([Sb2O3])^(-1) ([H2O])^3 ([KBr])^4 ([KSbO3])^2 = (([H2O])^3 ([KBr])^4 ([KSbO3])^2)/(([KOH])^6 ([Br2])^2 [Sb2O3])

Rate of reaction

Construct the rate of reaction expression for: KOH + Br_2 + Sb_2O_3 ⟶ H_2O + KBr + KSbO3 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: 6 KOH + 2 Br_2 + Sb_2O_3 ⟶ 3 H_2O + 4 KBr + 2 KSbO3 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 | 6 | -6 Br_2 | 2 | -2 Sb_2O_3 | 1 | -1 H_2O | 3 | 3 KBr | 4 | 4 KSbO3 | 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 | 6 | -6 | -1/6 (Δ[KOH])/(Δt) Br_2 | 2 | -2 | -1/2 (Δ[Br2])/(Δt) Sb_2O_3 | 1 | -1 | -(Δ[Sb2O3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) KBr | 4 | 4 | 1/4 (Δ[KBr])/(Δt) KSbO3 | 2 | 2 | 1/2 (Δ[KSbO3])/(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 = -1/6 (Δ[KOH])/(Δt) = -1/2 (Δ[Br2])/(Δt) = -(Δ[Sb2O3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/4 (Δ[KBr])/(Δt) = 1/2 (Δ[KSbO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KOH + Br_2 + Sb_2O_3 ⟶ H_2O + KBr + KSbO3 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: 6 KOH + 2 Br_2 + Sb_2O_3 ⟶ 3 H_2O + 4 KBr + 2 KSbO3 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 | 6 | -6 Br_2 | 2 | -2 Sb_2O_3 | 1 | -1 H_2O | 3 | 3 KBr | 4 | 4 KSbO3 | 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 | 6 | -6 | -1/6 (Δ[KOH])/(Δt) Br_2 | 2 | -2 | -1/2 (Δ[Br2])/(Δt) Sb_2O_3 | 1 | -1 | -(Δ[Sb2O3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) KBr | 4 | 4 | 1/4 (Δ[KBr])/(Δt) KSbO3 | 2 | 2 | 1/2 (Δ[KSbO3])/(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 = -1/6 (Δ[KOH])/(Δt) = -1/2 (Δ[Br2])/(Δt) = -(Δ[Sb2O3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/4 (Δ[KBr])/(Δt) = 1/2 (Δ[KSbO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium hydroxide | bromine | antimony trioxide | water | potassium bromide | KSbO3 formula | KOH | Br_2 | Sb_2O_3 | H_2O | KBr | KSbO3 Hill formula | HKO | Br_2 | O_3Sb_2 | H_2O | BrK | KO3Sb name | potassium hydroxide | bromine | antimony trioxide | water | potassium bromide | IUPAC name | potassium hydroxide | molecular bromine | oxo-oxostibanyloxystibane | water | potassium bromide |
| potassium hydroxide | bromine | antimony trioxide | water | potassium bromide | KSbO3 formula | KOH | Br_2 | Sb_2O_3 | H_2O | KBr | KSbO3 Hill formula | HKO | Br_2 | O_3Sb_2 | H_2O | BrK | KO3Sb name | potassium hydroxide | bromine | antimony trioxide | water | potassium bromide | IUPAC name | potassium hydroxide | molecular bromine | oxo-oxostibanyloxystibane | water | potassium bromide |

Substance properties

| potassium hydroxide | bromine | antimony trioxide | water | potassium bromide | KSbO3 molar mass | 56.105 g/mol | 159.81 g/mol | 291.517 g/mol | 18.015 g/mol | 119 g/mol | 208.855 g/mol phase | solid (at STP) | liquid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | 406 °C | -7.2 °C | 655 °C | 0 °C | 734 °C | boiling point | 1327 °C | 58.8 °C | 1550 °C | 99.9839 °C | 1435 °C | density | 2.044 g/cm^3 | 3.119 g/cm^3 | 5.2 g/cm^3 | 1 g/cm^3 | 2.75 g/cm^3 | solubility in water | soluble | insoluble | insoluble | | soluble | surface tension | | 0.0409 N/m | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | 9.44×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | | odorless | |
| potassium hydroxide | bromine | antimony trioxide | water | potassium bromide | KSbO3 molar mass | 56.105 g/mol | 159.81 g/mol | 291.517 g/mol | 18.015 g/mol | 119 g/mol | 208.855 g/mol phase | solid (at STP) | liquid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | 406 °C | -7.2 °C | 655 °C | 0 °C | 734 °C | boiling point | 1327 °C | 58.8 °C | 1550 °C | 99.9839 °C | 1435 °C | density | 2.044 g/cm^3 | 3.119 g/cm^3 | 5.2 g/cm^3 | 1 g/cm^3 | 2.75 g/cm^3 | solubility in water | soluble | insoluble | insoluble | | soluble | surface tension | | 0.0409 N/m | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | 9.44×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | | odorless | |

Units

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