O2 + C2H3Br = H2O + CO + HBr

O_2 oxygen + CH_2=CHBr vinyl bromide ⟶ H_2O water + CO carbon monoxide + HBr hydrogen bromide

Balance the chemical equation algebraically: O_2 + CH_2=CHBr ⟶ H_2O + CO + HBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 CH_2=CHBr ⟶ c_3 H_2O + c_4 CO + c_5 HBr Set the number of atoms in the reactants equal to the number of atoms in the products for O, Br, C and H: O: | 2 c_1 = c_3 + c_4 Br: | c_2 = c_5 C: | 2 c_2 = c_4 H: | 3 c_2 = 2 c_3 + c_5 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 = 2 c_5 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 3 c_2 = 2 c_3 = 2 c_4 = 4 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 O_2 + 2 CH_2=CHBr ⟶ 2 H_2O + 4 CO + 2 HBr

+ ⟶ + +

oxygen + vinyl bromide ⟶ water + carbon monoxide + hydrogen bromide

| oxygen | vinyl bromide | water | carbon monoxide | hydrogen bromide molecular enthalpy | 0 kJ/mol | 79.2 kJ/mol | -285.8 kJ/mol | -110.5 kJ/mol | -36.3 kJ/mol total enthalpy | 0 kJ/mol | 158.4 kJ/mol | -571.7 kJ/mol | -442 kJ/mol | -72.6 kJ/mol | H_initial = 158.4 kJ/mol | | H_final = -1086 kJ/mol | | ΔH_rxn^0 | -1086 kJ/mol - 158.4 kJ/mol = -1245 kJ/mol (exothermic) | | | |

| oxygen | vinyl bromide | water | carbon monoxide | hydrogen bromide molecular free energy | 231.7 kJ/mol | 81.8 kJ/mol | -237.1 kJ/mol | -137 kJ/mol | -53.4 kJ/mol total free energy | 695.1 kJ/mol | 163.6 kJ/mol | -474.2 kJ/mol | -548 kJ/mol | -106.8 kJ/mol | G_initial = 858.7 kJ/mol | | G_final = -1129 kJ/mol | | ΔG_rxn^0 | -1129 kJ/mol - 858.7 kJ/mol = -1988 kJ/mol (exergonic) | | | |

Construct the equilibrium constant, K, expression for: O_2 + CH_2=CHBr ⟶ H_2O + CO + HBr 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 CH_2=CHBr ⟶ 2 H_2O + 4 CO + 2 HBr 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 CH_2=CHBr | 2 | -2 H_2O | 2 | 2 CO | 4 | 4 HBr | 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) CH_2=CHBr | 2 | -2 | ([CH2=CHBr])^(-2) H_2O | 2 | 2 | ([H2O])^2 CO | 4 | 4 | ([CO])^4 HBr | 2 | 2 | ([HBr])^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) ([CH2=CHBr])^(-2) ([H2O])^2 ([CO])^4 ([HBr])^2 = (([H2O])^2 ([CO])^4 ([HBr])^2)/(([O2])^3 ([CH2=CHBr])^2)

Construct the rate of reaction expression for: O_2 + CH_2=CHBr ⟶ H_2O + CO + HBr 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 CH_2=CHBr ⟶ 2 H_2O + 4 CO + 2 HBr 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 CH_2=CHBr | 2 | -2 H_2O | 2 | 2 CO | 4 | 4 HBr | 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) CH_2=CHBr | 2 | -2 | -1/2 (Δ[CH2=CHBr])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) CO | 4 | 4 | 1/4 (Δ[CO])/(Δt) HBr | 2 | 2 | 1/2 (Δ[HBr])/(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[CH2=CHBr])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/4 (Δ[CO])/(Δt) = 1/2 (Δ[HBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | vinyl bromide | water | carbon monoxide | hydrogen bromide formula | O_2 | CH_2=CHBr | H_2O | CO | HBr Hill formula | O_2 | C_2H_3Br | H_2O | CO | BrH name | oxygen | vinyl bromide | water | carbon monoxide | hydrogen bromide IUPAC name | molecular oxygen | bromoethylene | water | carbon monoxide | hydrogen bromide