KOH + MnSO4 + KBrO3 = H2O + K2SO4 + K2MnO4 + KBr

KOH potassium hydroxide + MnSO_4 manganese(II) sulfate + KBrO_3 potassium bromate ⟶ H_2O water + K_2SO_4 potassium sulfate + K_2MnO_4 potassium manganate + KBr potassium bromide

Balance the chemical equation algebraically: KOH + MnSO_4 + KBrO_3 ⟶ H_2O + K_2SO_4 + K_2MnO_4 + KBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 MnSO_4 + c_3 KBrO_3 ⟶ c_4 H_2O + c_5 K_2SO_4 + c_6 K_2MnO_4 + c_7 KBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Mn, S and Br: H: | c_1 = 2 c_4 K: | c_1 + c_3 = 2 c_5 + 2 c_6 + c_7 O: | c_1 + 4 c_2 + 3 c_3 = c_4 + 4 c_5 + 4 c_6 Mn: | c_2 = c_6 S: | c_2 = c_5 Br: | c_3 = c_7 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 = 3/2 c_3 = 1 c_4 = 3 c_5 = 3/2 c_6 = 3/2 c_7 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 12 c_2 = 3 c_3 = 2 c_4 = 6 c_5 = 3 c_6 = 3 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 KOH + 3 MnSO_4 + 2 KBrO_3 ⟶ 6 H_2O + 3 K_2SO_4 + 3 K_2MnO_4 + 2 KBr

+ + ⟶ + + +

potassium hydroxide + manganese(II) sulfate + potassium bromate ⟶ water + potassium sulfate + potassium manganate + potassium bromide

Construct the equilibrium constant, K, expression for: KOH + MnSO_4 + KBrO_3 ⟶ H_2O + K_2SO_4 + K_2MnO_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 + 3 MnSO_4 + 2 KBrO_3 ⟶ 6 H_2O + 3 K_2SO_4 + 3 K_2MnO_4 + 2 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 MnSO_4 | 3 | -3 KBrO_3 | 2 | -2 H_2O | 6 | 6 K_2SO_4 | 3 | 3 K_2MnO_4 | 3 | 3 KBr | 2 | 2 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) MnSO_4 | 3 | -3 | ([MnSO4])^(-3) KBrO_3 | 2 | -2 | ([KBrO3])^(-2) H_2O | 6 | 6 | ([H2O])^6 K_2SO_4 | 3 | 3 | ([K2SO4])^3 K_2MnO_4 | 3 | 3 | ([K2MnO4])^3 KBr | 2 | 2 | ([KBr])^2 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) ([MnSO4])^(-3) ([KBrO3])^(-2) ([H2O])^6 ([K2SO4])^3 ([K2MnO4])^3 ([KBr])^2 = (([H2O])^6 ([K2SO4])^3 ([K2MnO4])^3 ([KBr])^2)/(([KOH])^12 ([MnSO4])^3 ([KBrO3])^2)

Construct the rate of reaction expression for: KOH + MnSO_4 + KBrO_3 ⟶ H_2O + K_2SO_4 + K_2MnO_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 + 3 MnSO_4 + 2 KBrO_3 ⟶ 6 H_2O + 3 K_2SO_4 + 3 K_2MnO_4 + 2 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 MnSO_4 | 3 | -3 KBrO_3 | 2 | -2 H_2O | 6 | 6 K_2SO_4 | 3 | 3 K_2MnO_4 | 3 | 3 KBr | 2 | 2 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) MnSO_4 | 3 | -3 | -1/3 (Δ[MnSO4])/(Δt) KBrO_3 | 2 | -2 | -1/2 (Δ[KBrO3])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) K_2SO_4 | 3 | 3 | 1/3 (Δ[K2SO4])/(Δt) K_2MnO_4 | 3 | 3 | 1/3 (Δ[K2MnO4])/(Δt) KBr | 2 | 2 | 1/2 (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) = -1/3 (Δ[MnSO4])/(Δt) = -1/2 (Δ[KBrO3])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/3 (Δ[K2SO4])/(Δt) = 1/3 (Δ[K2MnO4])/(Δt) = 1/2 (Δ[KBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium hydroxide | manganese(II) sulfate | potassium bromate | water | potassium sulfate | potassium manganate | potassium bromide formula | KOH | MnSO_4 | KBrO_3 | H_2O | K_2SO_4 | K_2MnO_4 | KBr Hill formula | HKO | MnSO_4 | BrKO_3 | H_2O | K_2O_4S | K_2MnO_4 | BrK name | potassium hydroxide | manganese(II) sulfate | potassium bromate | water | potassium sulfate | potassium manganate | potassium bromide IUPAC name | potassium hydroxide | manganese(+2) cation sulfate | potassium bromate | water | dipotassium sulfate | dipotassium dioxido-dioxomanganese | potassium bromide