octane + O2 -> water + CO2

CH_3(CH_2)_6CH_3 octane + O_2 oxygen ⟶ H_2O water + CO_2 carbon dioxide

Balance the chemical equation algebraically: CH_3(CH_2)_6CH_3 + O_2 ⟶ H_2O + CO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CH_3(CH_2)_6CH_3 + c_2 O_2 ⟶ 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 C, H and O: C: | 8 c_1 = c_4 H: | 18 c_1 = 2 c_3 O: | 2 c_2 = c_3 + 2 c_4 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 = 25/2 c_3 = 9 c_4 = 8 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 25 c_3 = 18 c_4 = 16 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 CH_3(CH_2)_6CH_3 + 25 O_2 ⟶ 18 H_2O + 16 CO_2

+ ⟶ +

octane + oxygen ⟶ water + carbon dioxide

Construct the equilibrium constant, K, expression for: CH_3(CH_2)_6CH_3 + O_2 ⟶ 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: 2 CH_3(CH_2)_6CH_3 + 25 O_2 ⟶ 18 H_2O + 16 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 CH_3(CH_2)_6CH_3 | 2 | -2 O_2 | 25 | -25 H_2O | 18 | 18 CO_2 | 16 | 16 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CH_3(CH_2)_6CH_3 | 2 | -2 | ([CH3(CH2)6CH3])^(-2) O_2 | 25 | -25 | ([O2])^(-25) H_2O | 18 | 18 | ([H2O])^18 CO_2 | 16 | 16 | ([CO2])^16 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 = ([CH3(CH2)6CH3])^(-2) ([O2])^(-25) ([H2O])^18 ([CO2])^16 = (([H2O])^18 ([CO2])^16)/(([CH3(CH2)6CH3])^2 ([O2])^25)

Construct the rate of reaction expression for: CH_3(CH_2)_6CH_3 + O_2 ⟶ 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: 2 CH_3(CH_2)_6CH_3 + 25 O_2 ⟶ 18 H_2O + 16 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 CH_3(CH_2)_6CH_3 | 2 | -2 O_2 | 25 | -25 H_2O | 18 | 18 CO_2 | 16 | 16 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_3(CH_2)_6CH_3 | 2 | -2 | -1/2 (Δ[CH3(CH2)6CH3])/(Δt) O_2 | 25 | -25 | -1/25 (Δ[O2])/(Δt) H_2O | 18 | 18 | 1/18 (Δ[H2O])/(Δt) CO_2 | 16 | 16 | 1/16 (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 = -1/2 (Δ[CH3(CH2)6CH3])/(Δt) = -1/25 (Δ[O2])/(Δt) = 1/18 (Δ[H2O])/(Δt) = 1/16 (Δ[CO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| octane | oxygen | water | carbon dioxide formula | CH_3(CH_2)_6CH_3 | O_2 | H_2O | CO_2 Hill formula | C_8H_18 | O_2 | H_2O | CO_2 name | octane | oxygen | water | carbon dioxide IUPAC name | octane | molecular oxygen | water | carbon dioxide