NaBrO3 = O2 + NaBr

NaBrO_3 sodium bromate ⟶ O_2 oxygen + NaBr sodium bromide

Balance the chemical equation algebraically: NaBrO_3 ⟶ O_2 + NaBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaBrO_3 ⟶ c_2 O_2 + c_3 NaBr Set the number of atoms in the reactants equal to the number of atoms in the products for Br, Na and O: Br: | c_1 = c_3 Na: | c_1 = c_3 O: | 3 c_1 = 2 c_2 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 = 3/2 c_3 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 NaBrO_3 ⟶ 3 O_2 + 2 NaBr

⟶ +

sodium bromate ⟶ oxygen + sodium bromide

| sodium bromate | oxygen | sodium bromide molecular enthalpy | -334.1 kJ/mol | 0 kJ/mol | -361.1 kJ/mol total enthalpy | -668.2 kJ/mol | 0 kJ/mol | -722.2 kJ/mol | H_initial = -668.2 kJ/mol | H_final = -722.2 kJ/mol | ΔH_rxn^0 | -722.2 kJ/mol - -668.2 kJ/mol = -54 kJ/mol (exothermic) | |

| sodium bromate | oxygen | sodium bromide molecular free energy | -242.6 kJ/mol | 231.7 kJ/mol | -349 kJ/mol total free energy | -485.2 kJ/mol | 695.1 kJ/mol | -698 kJ/mol | G_initial = -485.2 kJ/mol | G_final = -2.9 kJ/mol | ΔG_rxn^0 | -2.9 kJ/mol - -485.2 kJ/mol = 482.3 kJ/mol (endergonic) | |

Construct the equilibrium constant, K, expression for: NaBrO_3 ⟶ O_2 + NaBr 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: 2 NaBrO_3 ⟶ 3 O_2 + 2 NaBr 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 | 2 | -2 O_2 | 3 | 3 NaBr | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaBrO_3 | 2 | -2 | ([NaBrO3])^(-2) O_2 | 3 | 3 | ([O2])^3 NaBr | 2 | 2 | ([NaBr])^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 = ([NaBrO3])^(-2) ([O2])^3 ([NaBr])^2 = (([O2])^3 ([NaBr])^2)/([NaBrO3])^2

Construct the rate of reaction expression for: NaBrO_3 ⟶ O_2 + NaBr 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: 2 NaBrO_3 ⟶ 3 O_2 + 2 NaBr 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 | 2 | -2 O_2 | 3 | 3 NaBr | 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 NaBrO_3 | 2 | -2 | -1/2 (Δ[NaBrO3])/(Δt) O_2 | 3 | 3 | 1/3 (Δ[O2])/(Δt) NaBr | 2 | 2 | 1/2 (Δ[NaBr])/(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 = -1/2 (Δ[NaBrO3])/(Δt) = 1/3 (Δ[O2])/(Δt) = 1/2 (Δ[NaBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium bromate | oxygen | sodium bromide formula | NaBrO_3 | O_2 | NaBr Hill formula | BrNaO_3 | O_2 | BrNa name | sodium bromate | oxygen | sodium bromide IUPAC name | sodium bromate | molecular oxygen | sodium bromide

| sodium bromate | oxygen | sodium bromide molar mass | 150.89 g/mol | 31.998 g/mol | 102.89 g/mol phase | solid (at STP) | gas (at STP) | solid (at STP) melting point | 381 °C | -218 °C | 755 °C boiling point | 1390 °C | -183 °C | 1396 °C density | 3.339 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | 3.2 g/cm^3 solubility in water | soluble | | soluble surface tension | | 0.01347 N/m | dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | odor | odorless | odorless |