KMnO4 + C2H5OH = H2O + CO2 + KOH + MnO2

KMnO_4 (potassium permanganate) + CH_3CH_2OH (ethanol) ⟶ H_2O (water) + CO_2 (carbon dioxide) + KOH (potassium hydroxide) + MnO_2 (manganese dioxide)

Balance the chemical equation algebraically: KMnO_4 + CH_3CH_2OH ⟶ H_2O + CO_2 + KOH + MnO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KMnO_4 + c_2 CH_3CH_2OH ⟶ c_3 H_2O + c_4 CO_2 + c_5 KOH + c_6 MnO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for K, Mn, O, C and H: K: | c_1 = c_5 Mn: | c_1 = c_6 O: | 4 c_1 + c_2 = c_3 + 2 c_4 + c_5 + 2 c_6 C: | 2 c_2 = c_4 H: | 6 c_2 = 2 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 = 4 c_2 = 1 c_3 = 1 c_4 = 2 c_5 = 4 c_6 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 KMnO_4 + CH_3CH_2OH ⟶ H_2O + 2 CO_2 + 4 KOH + 4 MnO_2

+ ⟶ + + +

potassium permanganate + ethanol ⟶ water + carbon dioxide + potassium hydroxide + manganese dioxide

| potassium permanganate | ethanol | water | carbon dioxide | potassium hydroxide | manganese dioxide molecular free energy | -737.6 kJ/mol | -174.8 kJ/mol | -237.1 kJ/mol | -394.4 kJ/mol | -379.4 kJ/mol | -465.1 kJ/mol total free energy | -2950 kJ/mol | -174.8 kJ/mol | -237.1 kJ/mol | -788.8 kJ/mol | -1518 kJ/mol | -1860 kJ/mol | G_initial = -3125 kJ/mol | | G_final = -4404 kJ/mol | | | ΔG_rxn^0 | -4404 kJ/mol - -3125 kJ/mol = -1279 kJ/mol (exergonic) | | | | |

Construct the equilibrium constant, K, expression for: KMnO_4 + CH_3CH_2OH ⟶ H_2O + CO_2 + KOH + MnO_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 KMnO_4 + CH_3CH_2OH ⟶ H_2O + 2 CO_2 + 4 KOH + 4 MnO_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 KMnO_4 | 4 | -4 CH_3CH_2OH | 1 | -1 H_2O | 1 | 1 CO_2 | 2 | 2 KOH | 4 | 4 MnO_2 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KMnO_4 | 4 | -4 | ([KMnO4])^(-4) CH_3CH_2OH | 1 | -1 | ([CH3CH2OH])^(-1) H_2O | 1 | 1 | [H2O] CO_2 | 2 | 2 | ([CO2])^2 KOH | 4 | 4 | ([KOH])^4 MnO_2 | 4 | 4 | ([MnO2])^4 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 = ([KMnO4])^(-4) ([CH3CH2OH])^(-1) [H2O] ([CO2])^2 ([KOH])^4 ([MnO2])^4 = ([H2O] ([CO2])^2 ([KOH])^4 ([MnO2])^4)/(([KMnO4])^4 [CH3CH2OH])

Construct the rate of reaction expression for: KMnO_4 + CH_3CH_2OH ⟶ H_2O + CO_2 + KOH + MnO_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 KMnO_4 + CH_3CH_2OH ⟶ H_2O + 2 CO_2 + 4 KOH + 4 MnO_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 KMnO_4 | 4 | -4 CH_3CH_2OH | 1 | -1 H_2O | 1 | 1 CO_2 | 2 | 2 KOH | 4 | 4 MnO_2 | 4 | 4 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 KMnO_4 | 4 | -4 | -1/4 (Δ[KMnO4])/(Δt) CH_3CH_2OH | 1 | -1 | -(Δ[CH3CH2OH])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) CO_2 | 2 | 2 | 1/2 (Δ[CO2])/(Δt) KOH | 4 | 4 | 1/4 (Δ[KOH])/(Δt) MnO_2 | 4 | 4 | 1/4 (Δ[MnO2])/(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[KMnO4])/(Δt) = -(Δ[CH3CH2OH])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[CO2])/(Δt) = 1/4 (Δ[KOH])/(Δt) = 1/4 (Δ[MnO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium permanganate | ethanol | water | carbon dioxide | potassium hydroxide | manganese dioxide formula | KMnO_4 | CH_3CH_2OH | H_2O | CO_2 | KOH | MnO_2 Hill formula | KMnO_4 | C_2H_6O | H_2O | CO_2 | HKO | MnO_2 name | potassium permanganate | ethanol | water | carbon dioxide | potassium hydroxide | manganese dioxide IUPAC name | potassium permanganate | ethanol | water | carbon dioxide | potassium hydroxide | dioxomanganese