H2O + KMnO4 + KCl = KOH + MnO2 + KClO4

H_2O water + KMnO_4 potassium permanganate + KCl potassium chloride ⟶ KOH potassium hydroxide + MnO_2 manganese dioxide + KClO_4 potassium perchlorate

Balance the chemical equation algebraically: H_2O + KMnO_4 + KCl ⟶ KOH + MnO_2 + KClO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 KMnO_4 + c_3 KCl ⟶ c_4 KOH + c_5 MnO_2 + c_6 KClO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, K, Mn and Cl: H: | 2 c_1 = c_4 O: | c_1 + 4 c_2 = c_4 + 2 c_5 + 4 c_6 K: | c_2 + c_3 = c_4 + c_6 Mn: | c_2 = c_5 Cl: | 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4/3 c_2 = 8/3 c_3 = 1 c_4 = 8/3 c_5 = 8/3 c_6 = 1 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 4 c_2 = 8 c_3 = 3 c_4 = 8 c_5 = 8 c_6 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 H_2O + 8 KMnO_4 + 3 KCl ⟶ 8 KOH + 8 MnO_2 + 3 KClO_4

+ + ⟶ + +

water + potassium permanganate + potassium chloride ⟶ potassium hydroxide + manganese dioxide + potassium perchlorate

| water | potassium permanganate | potassium chloride | potassium hydroxide | manganese dioxide | potassium perchlorate molecular free energy | -237.1 kJ/mol | -737.6 kJ/mol | -408.5 kJ/mol | -379.4 kJ/mol | -465.1 kJ/mol | -303.1 kJ/mol total free energy | -948.4 kJ/mol | -5901 kJ/mol | -1226 kJ/mol | -3035 kJ/mol | -3721 kJ/mol | -909.3 kJ/mol | G_initial = -8075 kJ/mol | | | G_final = -7665 kJ/mol | | ΔG_rxn^0 | -7665 kJ/mol - -8075 kJ/mol = 409.4 kJ/mol (endergonic) | | | | |

Construct the equilibrium constant, K, expression for: H_2O + KMnO_4 + KCl ⟶ KOH + MnO_2 + KClO_4 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 + 8 KMnO_4 + 3 KCl ⟶ 8 KOH + 8 MnO_2 + 3 KClO_4 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 | 8 | -8 KCl | 3 | -3 KOH | 8 | 8 MnO_2 | 8 | 8 KClO_4 | 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 | 8 | -8 | ([KMnO4])^(-8) KCl | 3 | -3 | ([KCl])^(-3) KOH | 8 | 8 | ([KOH])^8 MnO_2 | 8 | 8 | ([MnO2])^8 KClO_4 | 3 | 3 | ([KClO4])^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])^(-8) ([KCl])^(-3) ([KOH])^8 ([MnO2])^8 ([KClO4])^3 = (([KOH])^8 ([MnO2])^8 ([KClO4])^3)/(([H2O])^4 ([KMnO4])^8 ([KCl])^3)

Construct the rate of reaction expression for: H_2O + KMnO_4 + KCl ⟶ KOH + MnO_2 + KClO_4 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 + 8 KMnO_4 + 3 KCl ⟶ 8 KOH + 8 MnO_2 + 3 KClO_4 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 | 8 | -8 KCl | 3 | -3 KOH | 8 | 8 MnO_2 | 8 | 8 KClO_4 | 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 | 8 | -8 | -1/8 (Δ[KMnO4])/(Δt) KCl | 3 | -3 | -1/3 (Δ[KCl])/(Δt) KOH | 8 | 8 | 1/8 (Δ[KOH])/(Δt) MnO_2 | 8 | 8 | 1/8 (Δ[MnO2])/(Δt) KClO_4 | 3 | 3 | 1/3 (Δ[KClO4])/(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) = -1/8 (Δ[KMnO4])/(Δt) = -1/3 (Δ[KCl])/(Δt) = 1/8 (Δ[KOH])/(Δt) = 1/8 (Δ[MnO2])/(Δt) = 1/3 (Δ[KClO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | potassium permanganate | potassium chloride | potassium hydroxide | manganese dioxide | potassium perchlorate formula | H_2O | KMnO_4 | KCl | KOH | MnO_2 | KClO_4 Hill formula | H_2O | KMnO_4 | ClK | HKO | MnO_2 | ClKO_4 name | water | potassium permanganate | potassium chloride | potassium hydroxide | manganese dioxide | potassium perchlorate IUPAC name | water | potassium permanganate | potassium chloride | potassium hydroxide | dioxomanganese | potassium perchlorate