C2H5OH = H2O + H2 + C4H6

CH_3CH_2OH ethanol ⟶ H_2O water + H_2 hydrogen + CH_2=CHCH=CH_2 1, 3-butadiene

Balance the chemical equation algebraically: CH_3CH_2OH ⟶ H_2O + H_2 + CH_2=CHCH=CH_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CH_3CH_2OH ⟶ c_2 H_2O + c_3 H_2 + c_4 CH_2=CHCH=CH_2 Set the number of atoms in the reactants equal to the number of atoms in the products for C, H and O: C: | 2 c_1 = 4 c_4 H: | 6 c_1 = 2 c_2 + 2 c_3 + 6 c_4 O: | c_1 = c_2 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 = 2 c_2 = 2 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 CH_3CH_2OH ⟶ 2 H_2O + H_2 + CH_2=CHCH=CH_2

⟶ + +

ethanol ⟶ water + hydrogen + 1, 3-butadiene

| ethanol | water | hydrogen | 1, 3-butadiene molecular enthalpy | -277.7 kJ/mol | -285.8 kJ/mol | 0 kJ/mol | -13.6 kJ/mol total enthalpy | -555.4 kJ/mol | -571.7 kJ/mol | 0 kJ/mol | -13.6 kJ/mol | H_initial = -555.4 kJ/mol | H_final = -585.3 kJ/mol | | ΔH_rxn^0 | -585.3 kJ/mol - -555.4 kJ/mol = -29.88 kJ/mol (exothermic) | | |

Construct the equilibrium constant, K, expression for: CH_3CH_2OH ⟶ H_2O + H_2 + CH_2=CHCH=CH_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: 2 CH_3CH_2OH ⟶ 2 H_2O + H_2 + CH_2=CHCH=CH_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 CH_3CH_2OH | 2 | -2 H_2O | 2 | 2 H_2 | 1 | 1 CH_2=CHCH=CH_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CH_3CH_2OH | 2 | -2 | ([CH3CH2OH])^(-2) H_2O | 2 | 2 | ([H2O])^2 H_2 | 1 | 1 | [H2] CH_2=CHCH=CH_2 | 1 | 1 | [CH2=CHCH=CH2] 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 = ([CH3CH2OH])^(-2) ([H2O])^2 [H2] [CH2=CHCH=CH2] = (([H2O])^2 [H2] [CH2=CHCH=CH2])/([CH3CH2OH])^2

Construct the rate of reaction expression for: CH_3CH_2OH ⟶ H_2O + H_2 + CH_2=CHCH=CH_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: 2 CH_3CH_2OH ⟶ 2 H_2O + H_2 + CH_2=CHCH=CH_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 CH_3CH_2OH | 2 | -2 H_2O | 2 | 2 H_2 | 1 | 1 CH_2=CHCH=CH_2 | 1 | 1 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 CH_3CH_2OH | 2 | -2 | -1/2 (Δ[CH3CH2OH])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) CH_2=CHCH=CH_2 | 1 | 1 | (Δ[CH2=CHCH=CH2])/(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 = -1/2 (Δ[CH3CH2OH])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[H2])/(Δt) = (Δ[CH2=CHCH=CH2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| ethanol | water | hydrogen | 1, 3-butadiene formula | CH_3CH_2OH | H_2O | H_2 | CH_2=CHCH=CH_2 Hill formula | C_2H_6O | H_2O | H_2 | C_4H_6 name | ethanol | water | hydrogen | 1, 3-butadiene IUPAC name | ethanol | water | molecular hydrogen | buta-1, 3-diene