O2 + C5H12O = H2O + CO2

O_2 oxygen + (CH_3)_2CHCH_2CH_2OH isoamyl alcohol ⟶ H_2O water + CO_2 carbon dioxide

Balance the chemical equation algebraically: O_2 + (CH_3)_2CHCH_2CH_2OH ⟶ H_2O + CO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 (CH_3)_2CHCH_2CH_2OH ⟶ 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: | 5 c_2 = c_4 H: | 12 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 = 15/2 c_2 = 1 c_3 = 6 c_4 = 5 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 15 c_2 = 2 c_3 = 12 c_4 = 10 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 15 O_2 + 2 (CH_3)_2CHCH_2CH_2OH ⟶ 12 H_2O + 10 CO_2

+ ⟶ +

oxygen + isoamyl alcohol ⟶ water + carbon dioxide

Construct the equilibrium constant, K, expression for: O_2 + (CH_3)_2CHCH_2CH_2OH ⟶ 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: 15 O_2 + 2 (CH_3)_2CHCH_2CH_2OH ⟶ 12 H_2O + 10 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 | 15 | -15 (CH_3)_2CHCH_2CH_2OH | 2 | -2 H_2O | 12 | 12 CO_2 | 10 | 10 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 15 | -15 | ([O2])^(-15) (CH_3)_2CHCH_2CH_2OH | 2 | -2 | ([(CH3)2CHCH2CH2OH])^(-2) H_2O | 12 | 12 | ([H2O])^12 CO_2 | 10 | 10 | ([CO2])^10 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])^(-15) ([(CH3)2CHCH2CH2OH])^(-2) ([H2O])^12 ([CO2])^10 = (([H2O])^12 ([CO2])^10)/(([O2])^15 ([(CH3)2CHCH2CH2OH])^2)

Construct the rate of reaction expression for: O_2 + (CH_3)_2CHCH_2CH_2OH ⟶ 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: 15 O_2 + 2 (CH_3)_2CHCH_2CH_2OH ⟶ 12 H_2O + 10 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 | 15 | -15 (CH_3)_2CHCH_2CH_2OH | 2 | -2 H_2O | 12 | 12 CO_2 | 10 | 10 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 | 15 | -15 | -1/15 (Δ[O2])/(Δt) (CH_3)_2CHCH_2CH_2OH | 2 | -2 | -1/2 (Δ[(CH3)2CHCH2CH2OH])/(Δt) H_2O | 12 | 12 | 1/12 (Δ[H2O])/(Δt) CO_2 | 10 | 10 | 1/10 (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 = -1/15 (Δ[O2])/(Δt) = -1/2 (Δ[(CH3)2CHCH2CH2OH])/(Δt) = 1/12 (Δ[H2O])/(Δt) = 1/10 (Δ[CO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | isoamyl alcohol | water | carbon dioxide formula | O_2 | (CH_3)_2CHCH_2CH_2OH | H_2O | CO_2 Hill formula | O_2 | C_5H_12O | H_2O | CO_2 name | oxygen | isoamyl alcohol | water | carbon dioxide IUPAC name | molecular oxygen | 3-methylbutan-1-ol | water | carbon dioxide