NaOH + Br2 + Na2SO3 = H2O + Na2SO4 + NaBr

NaOH sodium hydroxide + Br_2 bromine + Na_2SO_3 sodium sulfite ⟶ H_2O water + Na_2SO_4 sodium sulfate + NaBr sodium bromide

Balance the chemical equation algebraically: NaOH + Br_2 + Na_2SO_3 ⟶ H_2O + Na_2SO_4 + NaBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 Br_2 + c_3 Na_2SO_3 ⟶ c_4 H_2O + c_5 Na_2SO_4 + c_6 NaBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O, Br and S: H: | c_1 = 2 c_4 Na: | c_1 + 2 c_3 = 2 c_5 + c_6 O: | c_1 + 3 c_3 = c_4 + 4 c_5 Br: | 2 c_2 = c_6 S: | c_3 = c_5 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 = 1 c_4 = 1 c_5 = 1 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 NaOH + Br_2 + Na_2SO_3 ⟶ H_2O + Na_2SO_4 + 2 NaBr

+ + ⟶ + +

sodium hydroxide + bromine + sodium sulfite ⟶ water + sodium sulfate + sodium bromide

| sodium hydroxide | bromine | sodium sulfite | water | sodium sulfate | sodium bromide molecular enthalpy | -425.8 kJ/mol | 0 kJ/mol | -1101 kJ/mol | -285.8 kJ/mol | -1387 kJ/mol | -361.1 kJ/mol total enthalpy | -851.6 kJ/mol | 0 kJ/mol | -1101 kJ/mol | -285.8 kJ/mol | -1387 kJ/mol | -722.2 kJ/mol | H_initial = -1952 kJ/mol | | | H_final = -2395 kJ/mol | | ΔH_rxn^0 | -2395 kJ/mol - -1952 kJ/mol = -442.7 kJ/mol (exothermic) | | | | |

| sodium hydroxide | bromine | sodium sulfite | water | sodium sulfate | sodium bromide molecular free energy | -379.7 kJ/mol | 0 kJ/mol | -10125 kJ/mol | -237.1 kJ/mol | -1270 kJ/mol | -349 kJ/mol total free energy | -759.4 kJ/mol | 0 kJ/mol | -10125 kJ/mol | -237.1 kJ/mol | -1270 kJ/mol | -698 kJ/mol | G_initial = -10884 kJ/mol | | | G_final = -2205 kJ/mol | | ΔG_rxn^0 | -2205 kJ/mol - -10884 kJ/mol = 8679 kJ/mol (endergonic) | | | | |

Construct the equilibrium constant, K, expression for: NaOH + Br_2 + Na_2SO_3 ⟶ H_2O + Na_2SO_4 + 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 NaOH + Br_2 + Na_2SO_3 ⟶ H_2O + Na_2SO_4 + 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 NaOH | 2 | -2 Br_2 | 1 | -1 Na_2SO_3 | 1 | -1 H_2O | 1 | 1 Na_2SO_4 | 1 | 1 NaBr | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 2 | -2 | ([NaOH])^(-2) Br_2 | 1 | -1 | ([Br2])^(-1) Na_2SO_3 | 1 | -1 | ([Na2SO3])^(-1) H_2O | 1 | 1 | [H2O] Na_2SO_4 | 1 | 1 | [Na2SO4] 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 = ([NaOH])^(-2) ([Br2])^(-1) ([Na2SO3])^(-1) [H2O] [Na2SO4] ([NaBr])^2 = ([H2O] [Na2SO4] ([NaBr])^2)/(([NaOH])^2 [Br2] [Na2SO3])

Construct the rate of reaction expression for: NaOH + Br_2 + Na_2SO_3 ⟶ H_2O + Na_2SO_4 + 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 NaOH + Br_2 + Na_2SO_3 ⟶ H_2O + Na_2SO_4 + 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 NaOH | 2 | -2 Br_2 | 1 | -1 Na_2SO_3 | 1 | -1 H_2O | 1 | 1 Na_2SO_4 | 1 | 1 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 NaOH | 2 | -2 | -1/2 (Δ[NaOH])/(Δt) Br_2 | 1 | -1 | -(Δ[Br2])/(Δt) Na_2SO_3 | 1 | -1 | -(Δ[Na2SO3])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) Na_2SO_4 | 1 | 1 | (Δ[Na2SO4])/(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[NaOH])/(Δt) = -(Δ[Br2])/(Δt) = -(Δ[Na2SO3])/(Δt) = (Δ[H2O])/(Δt) = (Δ[Na2SO4])/(Δt) = 1/2 (Δ[NaBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium hydroxide | bromine | sodium sulfite | water | sodium sulfate | sodium bromide formula | NaOH | Br_2 | Na_2SO_3 | H_2O | Na_2SO_4 | NaBr Hill formula | HNaO | Br_2 | Na_2O_3S | H_2O | Na_2O_4S | BrNa name | sodium hydroxide | bromine | sodium sulfite | water | sodium sulfate | sodium bromide IUPAC name | sodium hydroxide | molecular bromine | disodium sulfite | water | disodium sulfate | sodium bromide