O2 + C7H15OH = H2O + CO2

O_2 oxygen + CH_3(CH_2)_6OH 1-heptanol ⟶ H_2O water + CO_2 carbon dioxide

Balance the chemical equation algebraically: O_2 + CH_3(CH_2)_6OH ⟶ H_2O + CO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 CH_3(CH_2)_6OH ⟶ c_3 H_2O + c_4 CO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for O, C and H: O: | 2 c_1 + c_2 = c_3 + 2 c_4 C: | 7 c_2 = c_4 H: | 16 c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 21/2 c_2 = 1 c_3 = 8 c_4 = 7 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 21 c_2 = 2 c_3 = 16 c_4 = 14 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 21 O_2 + 2 CH_3(CH_2)_6OH ⟶ 16 H_2O + 14 CO_2

+ ⟶ +

oxygen + 1-heptanol ⟶ water + carbon dioxide

Construct the equilibrium constant, K, expression for: O_2 + CH_3(CH_2)_6OH ⟶ H_2O + CO_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: 21 O_2 + 2 CH_3(CH_2)_6OH ⟶ 16 H_2O + 14 CO_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 | 21 | -21 CH_3(CH_2)_6OH | 2 | -2 H_2O | 16 | 16 CO_2 | 14 | 14 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 21 | -21 | ([O2])^(-21) CH_3(CH_2)_6OH | 2 | -2 | ([CH3(CH2)6OH])^(-2) H_2O | 16 | 16 | ([H2O])^16 CO_2 | 14 | 14 | ([CO2])^14 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])^(-21) ([CH3(CH2)6OH])^(-2) ([H2O])^16 ([CO2])^14 = (([H2O])^16 ([CO2])^14)/(([O2])^21 ([CH3(CH2)6OH])^2)

Construct the rate of reaction expression for: O_2 + CH_3(CH_2)_6OH ⟶ H_2O + CO_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: 21 O_2 + 2 CH_3(CH_2)_6OH ⟶ 16 H_2O + 14 CO_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 | 21 | -21 CH_3(CH_2)_6OH | 2 | -2 H_2O | 16 | 16 CO_2 | 14 | 14 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 | 21 | -21 | -1/21 (Δ[O2])/(Δt) CH_3(CH_2)_6OH | 2 | -2 | -1/2 (Δ[CH3(CH2)6OH])/(Δt) H_2O | 16 | 16 | 1/16 (Δ[H2O])/(Δt) CO_2 | 14 | 14 | 1/14 (Δ[CO2])/(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 = -1/21 (Δ[O2])/(Δt) = -1/2 (Δ[CH3(CH2)6OH])/(Δt) = 1/16 (Δ[H2O])/(Δt) = 1/14 (Δ[CO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | 1-heptanol | water | carbon dioxide formula | O_2 | CH_3(CH_2)_6OH | H_2O | CO_2 Hill formula | O_2 | C_7H_16O | H_2O | CO_2 name | oxygen | 1-heptanol | water | carbon dioxide IUPAC name | molecular oxygen | heptan-1-ol | water | carbon dioxide