H2SO4 + KBr = H2O + SO2 + Br2 + KHSO4

H_2SO_4 sulfuric acid + KBr potassium bromide ⟶ H_2O water + SO_2 sulfur dioxide + Br_2 bromine + KHSO_4 potassium bisulfate

Balance the chemical equation algebraically: H_2SO_4 + KBr ⟶ H_2O + SO_2 + Br_2 + KHSO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2SO_4 + c_2 KBr ⟶ c_3 H_2O + c_4 SO_2 + c_5 Br_2 + c_6 KHSO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, S, Br and K: H: | 2 c_1 = 2 c_3 + c_6 O: | 4 c_1 = c_3 + 2 c_4 + 4 c_6 S: | c_1 = c_4 + c_6 Br: | c_2 = 2 c_5 K: | c_2 = c_6 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 c_2 = 2 c_3 = 2 c_4 = 1 c_5 = 1 c_6 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 H_2SO_4 + 2 KBr ⟶ 2 H_2O + SO_2 + Br_2 + 2 KHSO_4

+ ⟶ + + +

sulfuric acid + potassium bromide ⟶ water + sulfur dioxide + bromine + potassium bisulfate

| sulfuric acid | potassium bromide | water | sulfur dioxide | bromine | potassium bisulfate molecular enthalpy | -814 kJ/mol | -393.8 kJ/mol | -285.8 kJ/mol | -296.8 kJ/mol | 0 kJ/mol | -1161 kJ/mol total enthalpy | -2442 kJ/mol | -787.6 kJ/mol | -571.7 kJ/mol | -296.8 kJ/mol | 0 kJ/mol | -2321 kJ/mol | H_initial = -3230 kJ/mol | | H_final = -3190 kJ/mol | | | ΔH_rxn^0 | -3190 kJ/mol - -3230 kJ/mol = 39.94 kJ/mol (endothermic) | | | | |

| sulfuric acid | potassium bromide | water | sulfur dioxide | bromine | potassium bisulfate molecular free energy | -690 kJ/mol | -380.7 kJ/mol | -237.1 kJ/mol | -300.1 kJ/mol | 0 kJ/mol | -1031 kJ/mol total free energy | -2070 kJ/mol | -761.4 kJ/mol | -474.2 kJ/mol | -300.1 kJ/mol | 0 kJ/mol | -2063 kJ/mol | G_initial = -2831 kJ/mol | | G_final = -2837 kJ/mol | | | ΔG_rxn^0 | -2837 kJ/mol - -2831 kJ/mol = -5.5 kJ/mol (exergonic) | | | | |

Construct the equilibrium constant, K, expression for: H_2SO_4 + KBr ⟶ H_2O + SO_2 + Br_2 + KHSO_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 H_2SO_4 + 2 KBr ⟶ 2 H_2O + SO_2 + Br_2 + 2 KHSO_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 H_2SO_4 | 3 | -3 KBr | 2 | -2 H_2O | 2 | 2 SO_2 | 1 | 1 Br_2 | 1 | 1 KHSO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2SO_4 | 3 | -3 | ([H2SO4])^(-3) KBr | 2 | -2 | ([KBr])^(-2) H_2O | 2 | 2 | ([H2O])^2 SO_2 | 1 | 1 | [SO2] Br_2 | 1 | 1 | [Br2] KHSO_4 | 2 | 2 | ([KHSO4])^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 = ([H2SO4])^(-3) ([KBr])^(-2) ([H2O])^2 [SO2] [Br2] ([KHSO4])^2 = (([H2O])^2 [SO2] [Br2] ([KHSO4])^2)/(([H2SO4])^3 ([KBr])^2)

Construct the rate of reaction expression for: H_2SO_4 + KBr ⟶ H_2O + SO_2 + Br_2 + KHSO_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 H_2SO_4 + 2 KBr ⟶ 2 H_2O + SO_2 + Br_2 + 2 KHSO_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 H_2SO_4 | 3 | -3 KBr | 2 | -2 H_2O | 2 | 2 SO_2 | 1 | 1 Br_2 | 1 | 1 KHSO_4 | 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 H_2SO_4 | 3 | -3 | -1/3 (Δ[H2SO4])/(Δt) KBr | 2 | -2 | -1/2 (Δ[KBr])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) SO_2 | 1 | 1 | (Δ[SO2])/(Δt) Br_2 | 1 | 1 | (Δ[Br2])/(Δt) KHSO_4 | 2 | 2 | 1/2 (Δ[KHSO4])/(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[H2SO4])/(Δt) = -1/2 (Δ[KBr])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[SO2])/(Δt) = (Δ[Br2])/(Δt) = 1/2 (Δ[KHSO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sulfuric acid | potassium bromide | water | sulfur dioxide | bromine | potassium bisulfate formula | H_2SO_4 | KBr | H_2O | SO_2 | Br_2 | KHSO_4 Hill formula | H_2O_4S | BrK | H_2O | O_2S | Br_2 | HKO_4S name | sulfuric acid | potassium bromide | water | sulfur dioxide | bromine | potassium bisulfate IUPAC name | sulfuric acid | potassium bromide | water | sulfur dioxide | molecular bromine | potassium hydrogen sulfate