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O2 + KOH + MnO2 = H2O + KMnO4

Input interpretation

O_2 (oxygen) + KOH (potassium hydroxide) + MnO_2 (manganese dioxide) ⟶ H_2O (water) + KMnO_4 (potassium permanganate)
O_2 (oxygen) + KOH (potassium hydroxide) + MnO_2 (manganese dioxide) ⟶ H_2O (water) + KMnO_4 (potassium permanganate)

Balanced equation

Balance the chemical equation algebraically: O_2 + KOH + MnO_2 ⟶ H_2O + KMnO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 KOH + c_3 MnO_2 ⟶ c_4 H_2O + c_5 KMnO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for O, H, K and Mn: O: | 2 c_1 + c_2 + 2 c_3 = c_4 + 4 c_5 H: | c_2 = 2 c_4 K: | c_2 = c_5 Mn: | 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3/2 c_2 = 2 c_3 = 2 c_4 = 1 c_5 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 4 c_3 = 4 c_4 = 2 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 3 O_2 + 4 KOH + 4 MnO_2 ⟶ 2 H_2O + 4 KMnO_4
Balance the chemical equation algebraically: O_2 + KOH + MnO_2 ⟶ H_2O + KMnO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 KOH + c_3 MnO_2 ⟶ c_4 H_2O + c_5 KMnO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for O, H, K and Mn: O: | 2 c_1 + c_2 + 2 c_3 = c_4 + 4 c_5 H: | c_2 = 2 c_4 K: | c_2 = c_5 Mn: | 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3/2 c_2 = 2 c_3 = 2 c_4 = 1 c_5 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 4 c_3 = 4 c_4 = 2 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 O_2 + 4 KOH + 4 MnO_2 ⟶ 2 H_2O + 4 KMnO_4

Structures

 + + ⟶ +
+ + ⟶ +

Names

oxygen + potassium hydroxide + manganese dioxide ⟶ water + potassium permanganate
oxygen + potassium hydroxide + manganese dioxide ⟶ water + potassium permanganate

Reaction thermodynamics

Gibbs free energy

 | oxygen | potassium hydroxide | manganese dioxide | water | potassium permanganate molecular free energy | 231.7 kJ/mol | -379.4 kJ/mol | -465.1 kJ/mol | -237.1 kJ/mol | -737.6 kJ/mol total free energy | 695.1 kJ/mol | -1518 kJ/mol | -1860 kJ/mol | -474.2 kJ/mol | -2950 kJ/mol  | G_initial = -2683 kJ/mol | | | G_final = -3425 kJ/mol |  ΔG_rxn^0 | -3425 kJ/mol - -2683 kJ/mol = -741.7 kJ/mol (exergonic) | | | |
| oxygen | potassium hydroxide | manganese dioxide | water | potassium permanganate molecular free energy | 231.7 kJ/mol | -379.4 kJ/mol | -465.1 kJ/mol | -237.1 kJ/mol | -737.6 kJ/mol total free energy | 695.1 kJ/mol | -1518 kJ/mol | -1860 kJ/mol | -474.2 kJ/mol | -2950 kJ/mol | G_initial = -2683 kJ/mol | | | G_final = -3425 kJ/mol | ΔG_rxn^0 | -3425 kJ/mol - -2683 kJ/mol = -741.7 kJ/mol (exergonic) | | | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: O_2 + KOH + MnO_2 ⟶ H_2O + KMnO_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: 3 O_2 + 4 KOH + 4 MnO_2 ⟶ 2 H_2O + 4 KMnO_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 O_2 | 3 | -3 KOH | 4 | -4 MnO_2 | 4 | -4 H_2O | 2 | 2 KMnO_4 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 3 | -3 | ([O2])^(-3) KOH | 4 | -4 | ([KOH])^(-4) MnO_2 | 4 | -4 | ([MnO2])^(-4) H_2O | 2 | 2 | ([H2O])^2 KMnO_4 | 4 | 4 | ([KMnO4])^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 = ([O2])^(-3) ([KOH])^(-4) ([MnO2])^(-4) ([H2O])^2 ([KMnO4])^4 = (([H2O])^2 ([KMnO4])^4)/(([O2])^3 ([KOH])^4 ([MnO2])^4)
Construct the equilibrium constant, K, expression for: O_2 + KOH + MnO_2 ⟶ H_2O + KMnO_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: 3 O_2 + 4 KOH + 4 MnO_2 ⟶ 2 H_2O + 4 KMnO_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 O_2 | 3 | -3 KOH | 4 | -4 MnO_2 | 4 | -4 H_2O | 2 | 2 KMnO_4 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 3 | -3 | ([O2])^(-3) KOH | 4 | -4 | ([KOH])^(-4) MnO_2 | 4 | -4 | ([MnO2])^(-4) H_2O | 2 | 2 | ([H2O])^2 KMnO_4 | 4 | 4 | ([KMnO4])^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 = ([O2])^(-3) ([KOH])^(-4) ([MnO2])^(-4) ([H2O])^2 ([KMnO4])^4 = (([H2O])^2 ([KMnO4])^4)/(([O2])^3 ([KOH])^4 ([MnO2])^4)

Rate of reaction

Construct the rate of reaction expression for: O_2 + KOH + MnO_2 ⟶ H_2O + KMnO_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: 3 O_2 + 4 KOH + 4 MnO_2 ⟶ 2 H_2O + 4 KMnO_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 O_2 | 3 | -3 KOH | 4 | -4 MnO_2 | 4 | -4 H_2O | 2 | 2 KMnO_4 | 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 O_2 | 3 | -3 | -1/3 (Δ[O2])/(Δt) KOH | 4 | -4 | -1/4 (Δ[KOH])/(Δt) MnO_2 | 4 | -4 | -1/4 (Δ[MnO2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) KMnO_4 | 4 | 4 | 1/4 (Δ[KMnO4])/(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 = -1/3 (Δ[O2])/(Δt) = -1/4 (Δ[KOH])/(Δt) = -1/4 (Δ[MnO2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/4 (Δ[KMnO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: O_2 + KOH + MnO_2 ⟶ H_2O + KMnO_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: 3 O_2 + 4 KOH + 4 MnO_2 ⟶ 2 H_2O + 4 KMnO_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 O_2 | 3 | -3 KOH | 4 | -4 MnO_2 | 4 | -4 H_2O | 2 | 2 KMnO_4 | 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 O_2 | 3 | -3 | -1/3 (Δ[O2])/(Δt) KOH | 4 | -4 | -1/4 (Δ[KOH])/(Δt) MnO_2 | 4 | -4 | -1/4 (Δ[MnO2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) KMnO_4 | 4 | 4 | 1/4 (Δ[KMnO4])/(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 = -1/3 (Δ[O2])/(Δt) = -1/4 (Δ[KOH])/(Δt) = -1/4 (Δ[MnO2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/4 (Δ[KMnO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | oxygen | potassium hydroxide | manganese dioxide | water | potassium permanganate formula | O_2 | KOH | MnO_2 | H_2O | KMnO_4 Hill formula | O_2 | HKO | MnO_2 | H_2O | KMnO_4 name | oxygen | potassium hydroxide | manganese dioxide | water | potassium permanganate IUPAC name | molecular oxygen | potassium hydroxide | dioxomanganese | water | potassium permanganate
| oxygen | potassium hydroxide | manganese dioxide | water | potassium permanganate formula | O_2 | KOH | MnO_2 | H_2O | KMnO_4 Hill formula | O_2 | HKO | MnO_2 | H_2O | KMnO_4 name | oxygen | potassium hydroxide | manganese dioxide | water | potassium permanganate IUPAC name | molecular oxygen | potassium hydroxide | dioxomanganese | water | potassium permanganate