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KOH + Br2 + Mn(OH)2 = H2O + KMnO4 + KBr

Input interpretation

KOH potassium hydroxide + Br_2 bromine + Mn(OH)_2 manganese hydroxide ⟶ H_2O water + KMnO_4 potassium permanganate + KBr potassium bromide
KOH potassium hydroxide + Br_2 bromine + Mn(OH)_2 manganese hydroxide ⟶ H_2O water + KMnO_4 potassium permanganate + KBr potassium bromide

Balanced equation

Balance the chemical equation algebraically: KOH + Br_2 + Mn(OH)_2 ⟶ H_2O + KMnO_4 + KBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 Br_2 + c_3 Mn(OH)_2 ⟶ c_4 H_2O + c_5 KMnO_4 + c_6 KBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Br and Mn: H: | c_1 + 2 c_3 = 2 c_4 K: | c_1 = c_5 + c_6 O: | c_1 + 2 c_3 = c_4 + 4 c_5 Br: | 2 c_2 = c_6 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 5/2 c_3 = 1 c_4 = 4 c_5 = 1 c_6 = 5 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 12 c_2 = 5 c_3 = 2 c_4 = 8 c_5 = 2 c_6 = 10 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 KOH + 5 Br_2 + 2 Mn(OH)_2 ⟶ 8 H_2O + 2 KMnO_4 + 10 KBr
Balance the chemical equation algebraically: KOH + Br_2 + Mn(OH)_2 ⟶ H_2O + KMnO_4 + KBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 Br_2 + c_3 Mn(OH)_2 ⟶ c_4 H_2O + c_5 KMnO_4 + c_6 KBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Br and Mn: H: | c_1 + 2 c_3 = 2 c_4 K: | c_1 = c_5 + c_6 O: | c_1 + 2 c_3 = c_4 + 4 c_5 Br: | 2 c_2 = c_6 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 5/2 c_3 = 1 c_4 = 4 c_5 = 1 c_6 = 5 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 12 c_2 = 5 c_3 = 2 c_4 = 8 c_5 = 2 c_6 = 10 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 KOH + 5 Br_2 + 2 Mn(OH)_2 ⟶ 8 H_2O + 2 KMnO_4 + 10 KBr

Structures

+ + ⟶ + +
+ + ⟶ + +

Names

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

Equilibrium constant

Construct the equilibrium constant, K, expression for: KOH + Br_2 + Mn(OH)_2 ⟶ H_2O + KMnO_4 + KBr 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: 12 KOH + 5 Br_2 + 2 Mn(OH)_2 ⟶ 8 H_2O + 2 KMnO_4 + 10 KBr 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 | 12 | -12 Br_2 | 5 | -5 Mn(OH)_2 | 2 | -2 H_2O | 8 | 8 KMnO_4 | 2 | 2 KBr | 10 | 10 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 12 | -12 | ([KOH])^(-12) Br_2 | 5 | -5 | ([Br2])^(-5) Mn(OH)_2 | 2 | -2 | ([Mn(OH)2])^(-2) H_2O | 8 | 8 | ([H2O])^8 KMnO_4 | 2 | 2 | ([KMnO4])^2 KBr | 10 | 10 | ([KBr])^10 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])^(-12) ([Br2])^(-5) ([Mn(OH)2])^(-2) ([H2O])^8 ([KMnO4])^2 ([KBr])^10 = (([H2O])^8 ([KMnO4])^2 ([KBr])^10)/(([KOH])^12 ([Br2])^5 ([Mn(OH)2])^2)
Construct the equilibrium constant, K, expression for: KOH + Br_2 + Mn(OH)_2 ⟶ H_2O + KMnO_4 + KBr 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: 12 KOH + 5 Br_2 + 2 Mn(OH)_2 ⟶ 8 H_2O + 2 KMnO_4 + 10 KBr 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 | 12 | -12 Br_2 | 5 | -5 Mn(OH)_2 | 2 | -2 H_2O | 8 | 8 KMnO_4 | 2 | 2 KBr | 10 | 10 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 12 | -12 | ([KOH])^(-12) Br_2 | 5 | -5 | ([Br2])^(-5) Mn(OH)_2 | 2 | -2 | ([Mn(OH)2])^(-2) H_2O | 8 | 8 | ([H2O])^8 KMnO_4 | 2 | 2 | ([KMnO4])^2 KBr | 10 | 10 | ([KBr])^10 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])^(-12) ([Br2])^(-5) ([Mn(OH)2])^(-2) ([H2O])^8 ([KMnO4])^2 ([KBr])^10 = (([H2O])^8 ([KMnO4])^2 ([KBr])^10)/(([KOH])^12 ([Br2])^5 ([Mn(OH)2])^2)

Rate of reaction

Construct the rate of reaction expression for: KOH + Br_2 + Mn(OH)_2 ⟶ H_2O + KMnO_4 + KBr 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: 12 KOH + 5 Br_2 + 2 Mn(OH)_2 ⟶ 8 H_2O + 2 KMnO_4 + 10 KBr 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 | 12 | -12 Br_2 | 5 | -5 Mn(OH)_2 | 2 | -2 H_2O | 8 | 8 KMnO_4 | 2 | 2 KBr | 10 | 10 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 | 12 | -12 | -1/12 (Δ[KOH])/(Δt) Br_2 | 5 | -5 | -1/5 (Δ[Br2])/(Δt) Mn(OH)_2 | 2 | -2 | -1/2 (Δ[Mn(OH)2])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) KMnO_4 | 2 | 2 | 1/2 (Δ[KMnO4])/(Δt) KBr | 10 | 10 | 1/10 (Δ[KBr])/(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 = -1/12 (Δ[KOH])/(Δt) = -1/5 (Δ[Br2])/(Δt) = -1/2 (Δ[Mn(OH)2])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/2 (Δ[KMnO4])/(Δt) = 1/10 (Δ[KBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KOH + Br_2 + Mn(OH)_2 ⟶ H_2O + KMnO_4 + KBr 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: 12 KOH + 5 Br_2 + 2 Mn(OH)_2 ⟶ 8 H_2O + 2 KMnO_4 + 10 KBr 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 | 12 | -12 Br_2 | 5 | -5 Mn(OH)_2 | 2 | -2 H_2O | 8 | 8 KMnO_4 | 2 | 2 KBr | 10 | 10 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 | 12 | -12 | -1/12 (Δ[KOH])/(Δt) Br_2 | 5 | -5 | -1/5 (Δ[Br2])/(Δt) Mn(OH)_2 | 2 | -2 | -1/2 (Δ[Mn(OH)2])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) KMnO_4 | 2 | 2 | 1/2 (Δ[KMnO4])/(Δt) KBr | 10 | 10 | 1/10 (Δ[KBr])/(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 = -1/12 (Δ[KOH])/(Δt) = -1/5 (Δ[Br2])/(Δt) = -1/2 (Δ[Mn(OH)2])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/2 (Δ[KMnO4])/(Δt) = 1/10 (Δ[KBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium hydroxide | bromine | manganese hydroxide | water | potassium permanganate | potassium bromide formula | KOH | Br_2 | Mn(OH)_2 | H_2O | KMnO_4 | KBr Hill formula | HKO | Br_2 | H_2MnO_2 | H_2O | KMnO_4 | BrK name | potassium hydroxide | bromine | manganese hydroxide | water | potassium permanganate | potassium bromide IUPAC name | potassium hydroxide | molecular bromine | manganous dihydroxide | water | potassium permanganate | potassium bromide
| potassium hydroxide | bromine | manganese hydroxide | water | potassium permanganate | potassium bromide formula | KOH | Br_2 | Mn(OH)_2 | H_2O | KMnO_4 | KBr Hill formula | HKO | Br_2 | H_2MnO_2 | H_2O | KMnO_4 | BrK name | potassium hydroxide | bromine | manganese hydroxide | water | potassium permanganate | potassium bromide IUPAC name | potassium hydroxide | molecular bromine | manganous dihydroxide | water | potassium permanganate | potassium bromide

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