O2 + H2S = H2O + SO2

O_2 (oxygen) + H_2S (hydrogen sulfide) ⟶ H_2O (water) + SO_2 (sulfur dioxide)

Balance the chemical equation algebraically: O_2 + H_2S ⟶ H_2O + SO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 H_2S ⟶ c_3 H_2O + c_4 SO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for O, H and S: O: | 2 c_1 = c_3 + 2 c_4 H: | 2 c_2 = 2 c_3 S: | c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3/2 c_2 = 1 c_3 = 1 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 2 c_3 = 2 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 O_2 + 2 H_2S ⟶ 2 H_2O + 2 SO_2

+ ⟶ +

oxygen + hydrogen sulfide ⟶ water + sulfur dioxide

| oxygen | hydrogen sulfide | water | sulfur dioxide molecular enthalpy | 0 kJ/mol | -20.6 kJ/mol | -285.8 kJ/mol | -296.8 kJ/mol total enthalpy | 0 kJ/mol | -41.2 kJ/mol | -571.7 kJ/mol | -593.6 kJ/mol | H_initial = -41.2 kJ/mol | | H_final = -1165 kJ/mol | ΔH_rxn^0 | -1165 kJ/mol - -41.2 kJ/mol = -1124 kJ/mol (exothermic) | | |

| oxygen | hydrogen sulfide | water | sulfur dioxide molecular free energy | 231.7 kJ/mol | -33.4 kJ/mol | -237.1 kJ/mol | -300.1 kJ/mol total free energy | 695.1 kJ/mol | -66.8 kJ/mol | -474.2 kJ/mol | -600.2 kJ/mol | G_initial = 628.3 kJ/mol | | G_final = -1074 kJ/mol | ΔG_rxn^0 | -1074 kJ/mol - 628.3 kJ/mol = -1703 kJ/mol (exergonic) | | |

| oxygen | hydrogen sulfide | water | sulfur dioxide molecular entropy | 205 J/(mol K) | 206 J/(mol K) | 69.91 J/(mol K) | 248 J/(mol K) total entropy | 615 J/(mol K) | 412 J/(mol K) | 139.8 J/(mol K) | 496 J/(mol K) | S_initial = 1027 J/(mol K) | | S_final = 635.8 J/(mol K) | ΔS_rxn^0 | 635.8 J/(mol K) - 1027 J/(mol K) = -391.2 J/(mol K) (exoentropic) | | |

Construct the equilibrium constant, K, expression for: O_2 + H_2S ⟶ H_2O + SO_2 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 O_2 + 2 H_2S ⟶ 2 H_2O + 2 SO_2 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 O_2 | 3 | -3 H_2S | 2 | -2 H_2O | 2 | 2 SO_2 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 3 | -3 | ([O2])^(-3) H_2S | 2 | -2 | ([H2S])^(-2) H_2O | 2 | 2 | ([H2O])^2 SO_2 | 2 | 2 | ([SO2])^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 = ([O2])^(-3) ([H2S])^(-2) ([H2O])^2 ([SO2])^2 = (([H2O])^2 ([SO2])^2)/(([O2])^3 ([H2S])^2)

Construct the rate of reaction expression for: O_2 + H_2S ⟶ H_2O + SO_2 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 O_2 + 2 H_2S ⟶ 2 H_2O + 2 SO_2 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 O_2 | 3 | -3 H_2S | 2 | -2 H_2O | 2 | 2 SO_2 | 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 O_2 | 3 | -3 | -1/3 (Δ[O2])/(Δt) H_2S | 2 | -2 | -1/2 (Δ[H2S])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) SO_2 | 2 | 2 | 1/2 (Δ[SO2])/(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[O2])/(Δt) = -1/2 (Δ[H2S])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/2 (Δ[SO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | hydrogen sulfide | water | sulfur dioxide formula | O_2 | H_2S | H_2O | SO_2 Hill formula | O_2 | H_2S | H_2O | O_2S name | oxygen | hydrogen sulfide | water | sulfur dioxide IUPAC name | molecular oxygen | hydrogen sulfide | water | sulfur dioxide

| oxygen | hydrogen sulfide | water | sulfur dioxide molar mass | 31.998 g/mol | 34.08 g/mol | 18.015 g/mol | 64.06 g/mol phase | gas (at STP) | gas (at STP) | liquid (at STP) | gas (at STP) melting point | -218 °C | -85 °C | 0 °C | -73 °C boiling point | -183 °C | -60 °C | 99.9839 °C | -10 °C density | 0.001429 g/cm^3 (at 0 °C) | 0.001393 g/cm^3 (at 25 °C) | 1 g/cm^3 | 0.002619 g/cm^3 (at 25 °C) surface tension | 0.01347 N/m | | 0.0728 N/m | 0.02859 N/m dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 1.239×10^-5 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) | 1.282×10^-5 Pa s (at 25 °C) odor | odorless | | odorless |