KOH + KMnO4 + NO2 = H2O + MnO2 + KNO3

KOH potassium hydroxide + KMnO_4 potassium permanganate + NO_2 nitrogen dioxide ⟶ H_2O water + MnO_2 manganese dioxide + KNO_3 potassium nitrate

Balance the chemical equation algebraically: KOH + KMnO_4 + NO_2 ⟶ H_2O + MnO_2 + KNO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 KMnO_4 + c_3 NO_2 ⟶ c_4 H_2O + c_5 MnO_2 + c_6 KNO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Mn and N: H: | c_1 = 2 c_4 K: | c_1 + c_2 = c_6 O: | c_1 + 4 c_2 + 2 c_3 = c_4 + 2 c_5 + 3 c_6 Mn: | c_2 = c_5 N: | 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 = 2 c_2 = 1 c_3 = 3 c_4 = 1 c_5 = 1 c_6 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + KMnO_4 + 3 NO_2 ⟶ H_2O + MnO_2 + 3 KNO_3

+ + ⟶ + +

potassium hydroxide + potassium permanganate + nitrogen dioxide ⟶ water + manganese dioxide + potassium nitrate

| potassium hydroxide | potassium permanganate | nitrogen dioxide | water | manganese dioxide | potassium nitrate molecular free energy | -379.4 kJ/mol | -737.6 kJ/mol | 51.3 kJ/mol | -237.1 kJ/mol | -465.1 kJ/mol | -394.9 kJ/mol total free energy | -758.8 kJ/mol | -737.6 kJ/mol | 153.9 kJ/mol | -237.1 kJ/mol | -465.1 kJ/mol | -1185 kJ/mol | G_initial = -1343 kJ/mol | | | G_final = -1887 kJ/mol | | ΔG_rxn^0 | -1887 kJ/mol - -1343 kJ/mol = -544.4 kJ/mol (exergonic) | | | | |

Construct the equilibrium constant, K, expression for: KOH + KMnO_4 + NO_2 ⟶ H_2O + MnO_2 + KNO_3 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 KOH + KMnO_4 + 3 NO_2 ⟶ H_2O + MnO_2 + 3 KNO_3 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 KOH | 2 | -2 KMnO_4 | 1 | -1 NO_2 | 3 | -3 H_2O | 1 | 1 MnO_2 | 1 | 1 KNO_3 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 2 | -2 | ([KOH])^(-2) KMnO_4 | 1 | -1 | ([KMnO4])^(-1) NO_2 | 3 | -3 | ([NO2])^(-3) H_2O | 1 | 1 | [H2O] MnO_2 | 1 | 1 | [MnO2] KNO_3 | 3 | 3 | ([KNO3])^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 = ([KOH])^(-2) ([KMnO4])^(-1) ([NO2])^(-3) [H2O] [MnO2] ([KNO3])^3 = ([H2O] [MnO2] ([KNO3])^3)/(([KOH])^2 [KMnO4] ([NO2])^3)

Construct the rate of reaction expression for: KOH + KMnO_4 + NO_2 ⟶ H_2O + MnO_2 + KNO_3 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 KOH + KMnO_4 + 3 NO_2 ⟶ H_2O + MnO_2 + 3 KNO_3 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 KOH | 2 | -2 KMnO_4 | 1 | -1 NO_2 | 3 | -3 H_2O | 1 | 1 MnO_2 | 1 | 1 KNO_3 | 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 KOH | 2 | -2 | -1/2 (Δ[KOH])/(Δt) KMnO_4 | 1 | -1 | -(Δ[KMnO4])/(Δt) NO_2 | 3 | -3 | -1/3 (Δ[NO2])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) MnO_2 | 1 | 1 | (Δ[MnO2])/(Δt) KNO_3 | 3 | 3 | 1/3 (Δ[KNO3])/(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[KOH])/(Δt) = -(Δ[KMnO4])/(Δt) = -1/3 (Δ[NO2])/(Δt) = (Δ[H2O])/(Δt) = (Δ[MnO2])/(Δt) = 1/3 (Δ[KNO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium hydroxide | potassium permanganate | nitrogen dioxide | water | manganese dioxide | potassium nitrate formula | KOH | KMnO_4 | NO_2 | H_2O | MnO_2 | KNO_3 Hill formula | HKO | KMnO_4 | NO_2 | H_2O | MnO_2 | KNO_3 name | potassium hydroxide | potassium permanganate | nitrogen dioxide | water | manganese dioxide | potassium nitrate IUPAC name | potassium hydroxide | potassium permanganate | Nitrogen dioxide | water | dioxomanganese | potassium nitrate

| potassium hydroxide | potassium permanganate | nitrogen dioxide | water | manganese dioxide | potassium nitrate molar mass | 56.105 g/mol | 158.03 g/mol | 46.005 g/mol | 18.015 g/mol | 86.936 g/mol | 101.1 g/mol phase | solid (at STP) | solid (at STP) | gas (at STP) | liquid (at STP) | solid (at STP) | solid (at STP) melting point | 406 °C | 240 °C | -11 °C | 0 °C | 535 °C | 334 °C boiling point | 1327 °C | | 21 °C | 99.9839 °C | | density | 2.044 g/cm^3 | 1 g/cm^3 | 0.00188 g/cm^3 (at 25 °C) | 1 g/cm^3 | 5.03 g/cm^3 | solubility in water | soluble | | reacts | | insoluble | soluble surface tension | | | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | | 4.02×10^-4 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) | | odor | | odorless | | odorless | | odorless