As2O3 + HBrO3 = HBr + As2O5

Input interpretation

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As_2O_3 arsenic trioxide + HO_3Br bromic acid ⟶ HBr hydrogen bromide + As_2O_5 arsenic pentoxide

Balanced equation

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$Balance the chemical equation algebraically: As_2O_3 + HO_3Br ⟶ HBr + As_2O_5 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 As_2O_3 + c_2 HO_3Br ⟶ c_3 HBr + c_4 As_2O_5 Set the number of atoms in the reactants equal to the number of atoms in the products for As, O, Br and H: As: | 2 c_1 = 2 c_4 O: | 3 c_1 + 3 c_2 = 5 c_4 Br: | c_2 = c_3 H: | c_2 = c_3 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3/2 c_2 = 1 c_3 = 1 c_4 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 2 c_3 = 2 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 As_2O_3 + 2 HO_3Br ⟶ 2 HBr + 3 As_2O_5$

Balance the chemical equation algebraically: As_2O_3 + HO_3Br ⟶ HBr + As_2O_5 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 As_2O_3 + c_2 HO_3Br ⟶ c_3 HBr + c_4 As_2O_5 Set the number of atoms in the reactants equal to the number of atoms in the products for As, O, Br and H: As: | 2 c_1 = 2 c_4 O: | 3 c_1 + 3 c_2 = 5 c_4 Br: | c_2 = c_3 H: | c_2 = c_3 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 = 3/2 c_2 = 1 c_3 = 1 c_4 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 2 c_3 = 2 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 As_2O_3 + 2 HO_3Br ⟶ 2 HBr + 3 As_2O_5

+ ⟶ +

Names

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arsenic trioxide + bromic acid ⟶ hydrogen bromide + arsenic pentoxide

Equilibrium constant

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Construct the equilibrium constant, K, expression for: As_2O_3 + HO_3Br ⟶ HBr + As_2O_5 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 As_2O_3 + 2 HO_3Br ⟶ 2 HBr + 3 As_2O_5 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 As_2O_3 | 3 | -3 HO_3Br | 2 | -2 HBr | 2 | 2 As_2O_5 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression As_2O_3 | 3 | -3 | ([As2O3])^(-3) HO_3Br | 2 | -2 | ([H1O3Br1])^(-2) HBr | 2 | 2 | ([HBr])^2 As_2O_5 | 3 | 3 | ([As2O5])^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 = ([As2O3])^(-3) ([H1O3Br1])^(-2) ([HBr])^2 ([As2O5])^3 = (([HBr])^2 ([As2O5])^3)/(([As2O3])^3 ([H1O3Br1])^2)

Rate of reaction

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Construct the rate of reaction expression for: As_2O_3 + HO_3Br ⟶ HBr + As_2O_5 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 As_2O_3 + 2 HO_3Br ⟶ 2 HBr + 3 As_2O_5 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 As_2O_3 | 3 | -3 HO_3Br | 2 | -2 HBr | 2 | 2 As_2O_5 | 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 As_2O_3 | 3 | -3 | -1/3 (Δ[As2O3])/(Δt) HO_3Br | 2 | -2 | -1/2 (Δ[H1O3Br1])/(Δt) HBr | 2 | 2 | 1/2 (Δ[HBr])/(Δt) As_2O_5 | 3 | 3 | 1/3 (Δ[As2O5])/(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[As2O3])/(Δt) = -1/2 (Δ[H1O3Br1])/(Δt) = 1/2 (Δ[HBr])/(Δt) = 1/3 (Δ[As2O5])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)