KBrO = KBr + KBrO3

KBrO ⟶ KBr potassium bromide + KBrO_3 potassium bromate

Balance the chemical equation algebraically: KBrO ⟶ KBr + KBrO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KBrO ⟶ c_2 KBr + c_3 KBrO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for K, Br and O: K: | c_1 = c_2 + c_3 Br: | c_1 = c_2 + c_3 O: | c_1 = 3 c_3 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 = 3 c_2 = 2 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 KBrO ⟶ 2 KBr + KBrO_3

KBrO ⟶ +

KBrO ⟶ potassium bromide + potassium bromate

Construct the equilibrium constant, K, expression for: KBrO ⟶ KBr + KBrO_3 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: 3 KBrO ⟶ 2 KBr + KBrO_3 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 KBrO | 3 | -3 KBr | 2 | 2 KBrO_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KBrO | 3 | -3 | ([KBrO])^(-3) KBr | 2 | 2 | ([KBr])^2 KBrO_3 | 1 | 1 | [KBrO3] 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 = ([KBrO])^(-3) ([KBr])^2 [KBrO3] = (([KBr])^2 [KBrO3])/([KBrO])^3

Construct the rate of reaction expression for: KBrO ⟶ KBr + KBrO_3 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: 3 KBrO ⟶ 2 KBr + KBrO_3 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 KBrO | 3 | -3 KBr | 2 | 2 KBrO_3 | 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 KBrO | 3 | -3 | -1/3 (Δ[KBrO])/(Δt) KBr | 2 | 2 | 1/2 (Δ[KBr])/(Δt) KBrO_3 | 1 | 1 | (Δ[KBrO3])/(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 = -1/3 (Δ[KBrO])/(Δt) = 1/2 (Δ[KBr])/(Δt) = (Δ[KBrO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| KBrO | potassium bromide | potassium bromate formula | KBrO | KBr | KBrO_3 Hill formula | BrKO | BrK | BrKO_3 name | | potassium bromide | potassium bromate

| KBrO | potassium bromide | potassium bromate molar mass | 135 g/mol | 119 g/mol | 167 g/mol phase | | solid (at STP) | solid (at STP) melting point | | 734 °C | 350 °C boiling point | | 1435 °C | density | | 2.75 g/cm^3 | 3.218 g/cm^3 solubility in water | | soluble |