H2O + I2 + Na2SO3 = Na2SO4 + HI

H_2O (water) + I_2 (iodine) + Na_2SO_3 (sodium sulfite) ⟶ Na_2SO_4 (sodium sulfate) + HI (hydrogen iodide)

Balance the chemical equation algebraically: H_2O + I_2 + Na_2SO_3 ⟶ Na_2SO_4 + HI Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 I_2 + c_3 Na_2SO_3 ⟶ c_4 Na_2SO_4 + c_5 HI Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, I, Na and S: H: | 2 c_1 = c_5 O: | c_1 + 3 c_3 = 4 c_4 I: | 2 c_2 = c_5 Na: | 2 c_3 = 2 c_4 S: | c_3 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 c_4 = 1 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O + I_2 + Na_2SO_3 ⟶ Na_2SO_4 + 2 HI

+ + ⟶ +

water + iodine + sodium sulfite ⟶ sodium sulfate + hydrogen iodide

| water | iodine | sodium sulfite | sodium sulfate | hydrogen iodide molecular enthalpy | -285.8 kJ/mol | 0 kJ/mol | -1101 kJ/mol | -1387 kJ/mol | 26.5 kJ/mol total enthalpy | -285.8 kJ/mol | 0 kJ/mol | -1101 kJ/mol | -1387 kJ/mol | 53 kJ/mol | H_initial = -1387 kJ/mol | | | H_final = -1334 kJ/mol | ΔH_rxn^0 | -1334 kJ/mol - -1387 kJ/mol = 52.53 kJ/mol (endothermic) | | | |

| water | iodine | sodium sulfite | sodium sulfate | hydrogen iodide molecular free energy | -237.1 kJ/mol | 0 kJ/mol | -10125 kJ/mol | -1270 kJ/mol | 1.7 kJ/mol total free energy | -237.1 kJ/mol | 0 kJ/mol | -10125 kJ/mol | -1270 kJ/mol | 3.4 kJ/mol | G_initial = -10362 kJ/mol | | | G_final = -1267 kJ/mol | ΔG_rxn^0 | -1267 kJ/mol - -10362 kJ/mol = 9095 kJ/mol (endergonic) | | | |

Construct the equilibrium constant, K, expression for: H_2O + I_2 + Na_2SO_3 ⟶ Na_2SO_4 + HI 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: H_2O + I_2 + Na_2SO_3 ⟶ Na_2SO_4 + 2 HI 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_2O | 1 | -1 I_2 | 1 | -1 Na_2SO_3 | 1 | -1 Na_2SO_4 | 1 | 1 HI | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 1 | -1 | ([H2O])^(-1) I_2 | 1 | -1 | ([I2])^(-1) Na_2SO_3 | 1 | -1 | ([Na2SO3])^(-1) Na_2SO_4 | 1 | 1 | [Na2SO4] HI | 2 | 2 | ([HI])^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 = ([H2O])^(-1) ([I2])^(-1) ([Na2SO3])^(-1) [Na2SO4] ([HI])^2 = ([Na2SO4] ([HI])^2)/([H2O] [I2] [Na2SO3])

Construct the rate of reaction expression for: H_2O + I_2 + Na_2SO_3 ⟶ Na_2SO_4 + HI 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: H_2O + I_2 + Na_2SO_3 ⟶ Na_2SO_4 + 2 HI 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_2O | 1 | -1 I_2 | 1 | -1 Na_2SO_3 | 1 | -1 Na_2SO_4 | 1 | 1 HI | 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_2O | 1 | -1 | -(Δ[H2O])/(Δt) I_2 | 1 | -1 | -(Δ[I2])/(Δt) Na_2SO_3 | 1 | -1 | -(Δ[Na2SO3])/(Δt) Na_2SO_4 | 1 | 1 | (Δ[Na2SO4])/(Δt) HI | 2 | 2 | 1/2 (Δ[HI])/(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 = -(Δ[H2O])/(Δt) = -(Δ[I2])/(Δt) = -(Δ[Na2SO3])/(Δt) = (Δ[Na2SO4])/(Δt) = 1/2 (Δ[HI])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | iodine | sodium sulfite | sodium sulfate | hydrogen iodide formula | H_2O | I_2 | Na_2SO_3 | Na_2SO_4 | HI Hill formula | H_2O | I_2 | Na_2O_3S | Na_2O_4S | HI name | water | iodine | sodium sulfite | sodium sulfate | hydrogen iodide IUPAC name | water | molecular iodine | disodium sulfite | disodium sulfate | hydrogen iodide