KOH + HBrO = H2O + KBrO

KOH potassium hydroxide + HOBr hypobromous acid ⟶ H_2O water + KBrO

Balance the chemical equation algebraically: KOH + HOBr ⟶ H_2O + KBrO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 HOBr ⟶ c_3 H_2O + c_4 KBrO Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O and Br: H: | c_1 + c_2 = 2 c_3 K: | c_1 = c_4 O: | c_1 + c_2 = c_3 + c_4 Br: | c_2 = c_4 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | KOH + HOBr ⟶ H_2O + KBrO

+ ⟶ + KBrO

potassium hydroxide + hypobromous acid ⟶ water + KBrO

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

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

| potassium hydroxide | hypobromous acid | water | KBrO formula | KOH | HOBr | H_2O | KBrO Hill formula | HKO | BrHO | H_2O | BrKO name | potassium hydroxide | hypobromous acid | water |

| potassium hydroxide | hypobromous acid | water | KBrO molar mass | 56.105 g/mol | 96.91 g/mol | 18.015 g/mol | 135 g/mol phase | solid (at STP) | | liquid (at STP) | melting point | 406 °C | | 0 °C | boiling point | 1327 °C | | 99.9839 °C | density | 2.044 g/cm^3 | | 1 g/cm^3 | solubility in water | 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 |