NaBrO = NaBr + NaBrO3

NaBrO ⟶ NaBr sodium bromide + NaBrO_3 sodium bromate

Balance the chemical equation algebraically: NaBrO ⟶ NaBr + NaBrO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaBrO ⟶ c_2 NaBr + c_3 NaBrO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Na, Br and O: Na: | 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 NaBrO ⟶ 2 NaBr + NaBrO_3

NaBrO ⟶ +

NaBrO ⟶ sodium bromide + sodium bromate

Construct the equilibrium constant, K, expression for: NaBrO ⟶ NaBr + NaBrO_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 NaBrO ⟶ 2 NaBr + NaBrO_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 NaBrO | 3 | -3 NaBr | 2 | 2 NaBrO_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaBrO | 3 | -3 | ([NaBrO])^(-3) NaBr | 2 | 2 | ([NaBr])^2 NaBrO_3 | 1 | 1 | [NaBrO3] 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 = ([NaBrO])^(-3) ([NaBr])^2 [NaBrO3] = (([NaBr])^2 [NaBrO3])/([NaBrO])^3

Construct the rate of reaction expression for: NaBrO ⟶ NaBr + NaBrO_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 NaBrO ⟶ 2 NaBr + NaBrO_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 NaBrO | 3 | -3 NaBr | 2 | 2 NaBrO_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 NaBrO | 3 | -3 | -1/3 (Δ[NaBrO])/(Δt) NaBr | 2 | 2 | 1/2 (Δ[NaBr])/(Δt) NaBrO_3 | 1 | 1 | (Δ[NaBrO3])/(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[NaBrO])/(Δt) = 1/2 (Δ[NaBr])/(Δt) = (Δ[NaBrO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| NaBrO | sodium bromide | sodium bromate formula | NaBrO | NaBr | NaBrO_3 Hill formula | BrNaO | BrNa | BrNaO_3 name | | sodium bromide | sodium bromate

| NaBrO | sodium bromide | sodium bromate molar mass | 118.89 g/mol | 102.89 g/mol | 150.89 g/mol phase | | solid (at STP) | solid (at STP) melting point | | 755 °C | 381 °C boiling point | | 1396 °C | 1390 °C density | | 3.2 g/cm^3 | 3.339 g/cm^3 solubility in water | | soluble | soluble odor | | | odorless