HBrO2 = H2O + O2 + Br2

HBrO2 ⟶ H_2O water + O_2 oxygen + Br_2 bromine

Balance the chemical equation algebraically: HBrO2 ⟶ H_2O + O_2 + Br_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBrO2 ⟶ c_2 H_2O + c_3 O_2 + c_4 Br_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Br and O: H: | c_1 = 2 c_2 Br: | c_1 = 2 c_4 O: | 2 c_1 = c_2 + 2 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 3/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 4 c_2 = 2 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HBrO2 ⟶ 2 H_2O + 3 O_2 + 2 Br_2

HBrO2 ⟶ + +

HBrO2 ⟶ water + oxygen + bromine

Construct the equilibrium constant, K, expression for: HBrO2 ⟶ H_2O + O_2 + Br_2 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: 4 HBrO2 ⟶ 2 H_2O + 3 O_2 + 2 Br_2 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 HBrO2 | 4 | -4 H_2O | 2 | 2 O_2 | 3 | 3 Br_2 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HBrO2 | 4 | -4 | ([HBrO2])^(-4) H_2O | 2 | 2 | ([H2O])^2 O_2 | 3 | 3 | ([O2])^3 Br_2 | 2 | 2 | ([Br2])^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 = ([HBrO2])^(-4) ([H2O])^2 ([O2])^3 ([Br2])^2 = (([H2O])^2 ([O2])^3 ([Br2])^2)/([HBrO2])^4

Construct the rate of reaction expression for: HBrO2 ⟶ H_2O + O_2 + Br_2 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: 4 HBrO2 ⟶ 2 H_2O + 3 O_2 + 2 Br_2 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 HBrO2 | 4 | -4 H_2O | 2 | 2 O_2 | 3 | 3 Br_2 | 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 HBrO2 | 4 | -4 | -1/4 (Δ[HBrO2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) O_2 | 3 | 3 | 1/3 (Δ[O2])/(Δt) Br_2 | 2 | 2 | 1/2 (Δ[Br2])/(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 = -1/4 (Δ[HBrO2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/3 (Δ[O2])/(Δt) = 1/2 (Δ[Br2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| HBrO2 | water | oxygen | bromine formula | HBrO2 | H_2O | O_2 | Br_2 name | | water | oxygen | bromine IUPAC name | | water | molecular oxygen | molecular bromine

| HBrO2 | water | oxygen | bromine molar mass | 112.91 g/mol | 18.015 g/mol | 31.998 g/mol | 159.81 g/mol phase | | liquid (at STP) | gas (at STP) | liquid (at STP) melting point | | 0 °C | -218 °C | -7.2 °C boiling point | | 99.9839 °C | -183 °C | 58.8 °C density | | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | 3.119 g/cm^3 solubility in water | | | | insoluble surface tension | | 0.0728 N/m | 0.01347 N/m | 0.0409 N/m dynamic viscosity | | 8.9×10^-4 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | 9.44×10^-4 Pa s (at 25 °C) odor | | odorless | odorless |