H2S + HBrO3 = H2O + H2SO4 + Br2

hydrogen sulfide + bromic acid ⟶ water + sulfuric acid + bromine

Balance the chemical equation algebraically: + ⟶ + + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 + c_5 Set the number of atoms in the reactants equal to the number of atoms in the products for H, S, Br and O: H: | 2 c_1 + c_2 = 2 c_3 + 2 c_4 S: | c_1 = c_4 Br: | c_2 = 2 c_5 O: | 3 c_2 = c_3 + 4 c_4 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 = 5/4 c_2 = 2 c_3 = 1 c_4 = 5/4 c_5 = 1 Multiply by the least common denominator, 4, to eliminate fractional coefficients: c_1 = 5 c_2 = 8 c_3 = 4 c_4 = 5 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 + 8 ⟶ 4 + 5 + 4

+ ⟶ + +

hydrogen sulfide + bromic acid ⟶ water + sulfuric acid + bromine

K_c = ([H2O]^4 [H2SO4]^5 [Br2]^4)/([H2S]^5 [H1O3Br1]^8)

rate = -1/5 (Δ[H2S])/(Δt) = -1/8 (Δ[H1O3Br1])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/5 (Δ[H2SO4])/(Δt) = 1/4 (Δ[Br2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen sulfide | bromic acid | water | sulfuric acid | bromine Hill formula | H_2S | BrHO_3 | H_2O | H_2O_4S | Br_2 name | hydrogen sulfide | bromic acid | water | sulfuric acid | bromine IUPAC name | hydrogen sulfide | bromic acid | water | sulfuric acid | molecular bromine