CH3COOH = H2O + CH2CO

CH_3CO_2H acetic acid ⟶ H_2O water + C_2H_2O ketene

Balance the chemical equation algebraically: CH_3CO_2H ⟶ H_2O + C_2H_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CH_3CO_2H ⟶ c_2 H_2O + c_3 C_2H_2O 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 = 2 c_3 H: | 4 c_1 = 2 c_2 + 2 c_3 O: | 2 c_1 = c_2 + c_3 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 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | CH_3CO_2H ⟶ H_2O + C_2H_2O

⟶ +

acetic acid ⟶ water + ketene

| acetic acid | water | ketene molecular free energy | -389.9 kJ/mol | -237.1 kJ/mol | -48.3 kJ/mol total free energy | -389.9 kJ/mol | -237.1 kJ/mol | -48.3 kJ/mol | G_initial = -389.9 kJ/mol | G_final = -285.4 kJ/mol | ΔG_rxn^0 | -285.4 kJ/mol - -389.9 kJ/mol = 104.5 kJ/mol (endergonic) | |

Construct the equilibrium constant, K, expression for: CH_3CO_2H ⟶ H_2O + C_2H_2O 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: CH_3CO_2H ⟶ H_2O + C_2H_2O 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_3CO_2H | 1 | -1 H_2O | 1 | 1 C_2H_2O | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CH_3CO_2H | 1 | -1 | ([CH3CO2H])^(-1) H_2O | 1 | 1 | [H2O] C_2H_2O | 1 | 1 | [C2H2O] 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 = ([CH3CO2H])^(-1) [H2O] [C2H2O] = ([H2O] [C2H2O])/([CH3CO2H])

Construct the rate of reaction expression for: CH_3CO_2H ⟶ H_2O + C_2H_2O 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: CH_3CO_2H ⟶ H_2O + C_2H_2O 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_3CO_2H | 1 | -1 H_2O | 1 | 1 C_2H_2O | 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_3CO_2H | 1 | -1 | -(Δ[CH3CO2H])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) C_2H_2O | 1 | 1 | (Δ[C2H2O])/(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[CH3CO2H])/(Δt) = (Δ[H2O])/(Δt) = (Δ[C2H2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| acetic acid | water | ketene formula | CH_3CO_2H | H_2O | C_2H_2O Hill formula | C_2H_4O_2 | H_2O | C_2H_2O name | acetic acid | water | ketene IUPAC name | acetic acid | water | ethenone

| acetic acid | water | ketene molar mass | 60.052 g/mol | 18.015 g/mol | 42.037 g/mol phase | liquid (at STP) | liquid (at STP) | gas (at STP) melting point | 16.2 °C | 0 °C | -134.1 °C boiling point | 117.5 °C | 99.9839 °C | -56 °C density | 1.049 g/cm^3 | 1 g/cm^3 | 0.65 g/cm^3 (at -60 °C) solubility in water | miscible | | very soluble surface tension | 0.0288 N/m | 0.0728 N/m | dynamic viscosity | 0.001056 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) | odor | vinegar-like | odorless |