H2O + KMnO4 + KBr = KOH + MnO2 + Br2

H_2O water + KMnO_4 potassium permanganate + KBr potassium bromide ⟶ KOH potassium hydroxide + MnO_2 manganese dioxide + Br_2 bromine

Balance the chemical equation algebraically: H_2O + KMnO_4 + KBr ⟶ KOH + MnO_2 + Br_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 KMnO_4 + c_3 KBr ⟶ c_4 KOH + c_5 MnO_2 + c_6 Br_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, K, Mn and Br: H: | 2 c_1 = c_4 O: | c_1 + 4 c_2 = c_4 + 2 c_5 K: | c_2 + c_3 = c_4 Mn: | c_2 = c_5 Br: | c_3 = 2 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 = 2 c_2 = 1 c_3 = 3 c_4 = 4 c_5 = 1 c_6 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 4 c_2 = 2 c_3 = 6 c_4 = 8 c_5 = 2 c_6 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 H_2O + 2 KMnO_4 + 6 KBr ⟶ 8 KOH + 2 MnO_2 + 3 Br_2

+ + ⟶ + +

water + potassium permanganate + potassium bromide ⟶ potassium hydroxide + manganese dioxide + bromine

| water | potassium permanganate | potassium bromide | potassium hydroxide | manganese dioxide | bromine molecular free energy | -237.1 kJ/mol | -737.6 kJ/mol | -380.7 kJ/mol | -379.4 kJ/mol | -465.1 kJ/mol | 0 kJ/mol total free energy | -948.4 kJ/mol | -1475 kJ/mol | -2284 kJ/mol | -3035 kJ/mol | -930.2 kJ/mol | 0 kJ/mol | G_initial = -4708 kJ/mol | | | G_final = -3965 kJ/mol | | ΔG_rxn^0 | -3965 kJ/mol - -4708 kJ/mol = 742.4 kJ/mol (endergonic) | | | | |

Construct the equilibrium constant, K, expression for: H_2O + KMnO_4 + KBr ⟶ KOH + MnO_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 H_2O + 2 KMnO_4 + 6 KBr ⟶ 8 KOH + 2 MnO_2 + 3 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 H_2O | 4 | -4 KMnO_4 | 2 | -2 KBr | 6 | -6 KOH | 8 | 8 MnO_2 | 2 | 2 Br_2 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 4 | -4 | ([H2O])^(-4) KMnO_4 | 2 | -2 | ([KMnO4])^(-2) KBr | 6 | -6 | ([KBr])^(-6) KOH | 8 | 8 | ([KOH])^8 MnO_2 | 2 | 2 | ([MnO2])^2 Br_2 | 3 | 3 | ([Br2])^3 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 = ([H2O])^(-4) ([KMnO4])^(-2) ([KBr])^(-6) ([KOH])^8 ([MnO2])^2 ([Br2])^3 = (([KOH])^8 ([MnO2])^2 ([Br2])^3)/(([H2O])^4 ([KMnO4])^2 ([KBr])^6)

Construct the rate of reaction expression for: H_2O + KMnO_4 + KBr ⟶ KOH + MnO_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 H_2O + 2 KMnO_4 + 6 KBr ⟶ 8 KOH + 2 MnO_2 + 3 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 H_2O | 4 | -4 KMnO_4 | 2 | -2 KBr | 6 | -6 KOH | 8 | 8 MnO_2 | 2 | 2 Br_2 | 3 | 3 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 H_2O | 4 | -4 | -1/4 (Δ[H2O])/(Δt) KMnO_4 | 2 | -2 | -1/2 (Δ[KMnO4])/(Δt) KBr | 6 | -6 | -1/6 (Δ[KBr])/(Δt) KOH | 8 | 8 | 1/8 (Δ[KOH])/(Δt) MnO_2 | 2 | 2 | 1/2 (Δ[MnO2])/(Δt) Br_2 | 3 | 3 | 1/3 (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[H2O])/(Δt) = -1/2 (Δ[KMnO4])/(Δt) = -1/6 (Δ[KBr])/(Δt) = 1/8 (Δ[KOH])/(Δt) = 1/2 (Δ[MnO2])/(Δt) = 1/3 (Δ[Br2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | potassium permanganate | potassium bromide | potassium hydroxide | manganese dioxide | bromine formula | H_2O | KMnO_4 | KBr | KOH | MnO_2 | Br_2 Hill formula | H_2O | KMnO_4 | BrK | HKO | MnO_2 | Br_2 name | water | potassium permanganate | potassium bromide | potassium hydroxide | manganese dioxide | bromine IUPAC name | water | potassium permanganate | potassium bromide | potassium hydroxide | dioxomanganese | molecular bromine