NaOH + Br2 + MnCl2 = H2O + NaCl + MnO2 + NaBr

NaOH (sodium hydroxide) + Br_2 (bromine) + MnCl_2 (manganese(II) chloride) ⟶ H_2O (water) + NaCl (sodium chloride) + MnO_2 (manganese dioxide) + NaBr (sodium bromide)

Balance the chemical equation algebraically: NaOH + Br_2 + MnCl_2 ⟶ H_2O + NaCl + MnO_2 + NaBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 Br_2 + c_3 MnCl_2 ⟶ c_4 H_2O + c_5 NaCl + c_6 MnO_2 + c_7 NaBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O, Br, Cl and Mn: H: | c_1 = 2 c_4 Na: | c_1 = c_5 + c_7 O: | c_1 = c_4 + 2 c_6 Br: | 2 c_2 = c_7 Cl: | 2 c_3 = c_5 Mn: | c_3 = c_6 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 = 4 c_2 = 1 c_3 = 1 c_4 = 2 c_5 = 2 c_6 = 1 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 NaOH + Br_2 + MnCl_2 ⟶ 2 H_2O + 2 NaCl + MnO_2 + 2 NaBr

+ + ⟶ + + +

sodium hydroxide + bromine + manganese(II) chloride ⟶ water + sodium chloride + manganese dioxide + sodium bromide

| sodium hydroxide | bromine | manganese(II) chloride | water | sodium chloride | manganese dioxide | sodium bromide molecular enthalpy | -425.8 kJ/mol | 0 kJ/mol | -481.3 kJ/mol | -285.8 kJ/mol | -411.2 kJ/mol | -520 kJ/mol | -361.1 kJ/mol total enthalpy | -1703 kJ/mol | 0 kJ/mol | -481.3 kJ/mol | -571.7 kJ/mol | -822.4 kJ/mol | -520 kJ/mol | -722.2 kJ/mol | H_initial = -2185 kJ/mol | | | H_final = -2636 kJ/mol | | | ΔH_rxn^0 | -2636 kJ/mol - -2185 kJ/mol = -451.8 kJ/mol (exothermic) | | | | | |

| sodium hydroxide | bromine | manganese(II) chloride | water | sodium chloride | manganese dioxide | sodium bromide molecular free energy | -379.7 kJ/mol | 0 kJ/mol | -440.5 kJ/mol | -237.1 kJ/mol | -384.1 kJ/mol | -465.1 kJ/mol | -349 kJ/mol total free energy | -1519 kJ/mol | 0 kJ/mol | -440.5 kJ/mol | -474.2 kJ/mol | -768.2 kJ/mol | -465.1 kJ/mol | -698 kJ/mol | G_initial = -1959 kJ/mol | | | G_final = -2406 kJ/mol | | | ΔG_rxn^0 | -2406 kJ/mol - -1959 kJ/mol = -446.2 kJ/mol (exergonic) | | | | | |

Construct the equilibrium constant, K, expression for: NaOH + Br_2 + MnCl_2 ⟶ H_2O + NaCl + MnO_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: 4 NaOH + Br_2 + MnCl_2 ⟶ 2 H_2O + 2 NaCl + MnO_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 NaOH | 4 | -4 Br_2 | 1 | -1 MnCl_2 | 1 | -1 H_2O | 2 | 2 NaCl | 2 | 2 MnO_2 | 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 | 4 | -4 | ([NaOH])^(-4) Br_2 | 1 | -1 | ([Br2])^(-1) MnCl_2 | 1 | -1 | ([MnCl2])^(-1) H_2O | 2 | 2 | ([H2O])^2 NaCl | 2 | 2 | ([NaCl])^2 MnO_2 | 1 | 1 | [MnO2] 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])^(-4) ([Br2])^(-1) ([MnCl2])^(-1) ([H2O])^2 ([NaCl])^2 [MnO2] ([NaBr])^2 = (([H2O])^2 ([NaCl])^2 [MnO2] ([NaBr])^2)/(([NaOH])^4 [Br2] [MnCl2])

Construct the rate of reaction expression for: NaOH + Br_2 + MnCl_2 ⟶ H_2O + NaCl + MnO_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: 4 NaOH + Br_2 + MnCl_2 ⟶ 2 H_2O + 2 NaCl + MnO_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 NaOH | 4 | -4 Br_2 | 1 | -1 MnCl_2 | 1 | -1 H_2O | 2 | 2 NaCl | 2 | 2 MnO_2 | 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 | 4 | -4 | -1/4 (Δ[NaOH])/(Δt) Br_2 | 1 | -1 | -(Δ[Br2])/(Δt) MnCl_2 | 1 | -1 | -(Δ[MnCl2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) NaCl | 2 | 2 | 1/2 (Δ[NaCl])/(Δt) MnO_2 | 1 | 1 | (Δ[MnO2])/(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 = -1/4 (Δ[NaOH])/(Δt) = -(Δ[Br2])/(Δt) = -(Δ[MnCl2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/2 (Δ[NaCl])/(Δt) = (Δ[MnO2])/(Δt) = 1/2 (Δ[NaBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium hydroxide | bromine | manganese(II) chloride | water | sodium chloride | manganese dioxide | sodium bromide formula | NaOH | Br_2 | MnCl_2 | H_2O | NaCl | MnO_2 | NaBr Hill formula | HNaO | Br_2 | Cl_2Mn | H_2O | ClNa | MnO_2 | BrNa name | sodium hydroxide | bromine | manganese(II) chloride | water | sodium chloride | manganese dioxide | sodium bromide IUPAC name | sodium hydroxide | molecular bromine | dichloromanganese | water | sodium chloride | dioxomanganese | sodium bromide