O2 + C8H18 = H2O + CO

O_2 oxygen + CH_3(CH_2)_6CH_3 octane ⟶ H_2O water + CO carbon monoxide

Balance the chemical equation algebraically: O_2 + CH_3(CH_2)_6CH_3 ⟶ H_2O + CO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 CH_3(CH_2)_6CH_3 ⟶ c_3 H_2O + c_4 CO 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_3 + c_4 C: | 8 c_2 = c_4 H: | 18 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 = 17/2 c_2 = 1 c_3 = 9 c_4 = 8 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 17 c_2 = 2 c_3 = 18 c_4 = 16 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 17 O_2 + 2 CH_3(CH_2)_6CH_3 ⟶ 18 H_2O + 16 CO

+ ⟶ +

oxygen + octane ⟶ water + carbon monoxide

Construct the equilibrium constant, K, expression for: O_2 + CH_3(CH_2)_6CH_3 ⟶ H_2O + CO 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: 17 O_2 + 2 CH_3(CH_2)_6CH_3 ⟶ 18 H_2O + 16 CO 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 | 17 | -17 CH_3(CH_2)_6CH_3 | 2 | -2 H_2O | 18 | 18 CO | 16 | 16 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 17 | -17 | ([O2])^(-17) CH_3(CH_2)_6CH_3 | 2 | -2 | ([CH3(CH2)6CH3])^(-2) H_2O | 18 | 18 | ([H2O])^18 CO | 16 | 16 | ([CO])^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 = ([O2])^(-17) ([CH3(CH2)6CH3])^(-2) ([H2O])^18 ([CO])^16 = (([H2O])^18 ([CO])^16)/(([O2])^17 ([CH3(CH2)6CH3])^2)

Construct the rate of reaction expression for: O_2 + CH_3(CH_2)_6CH_3 ⟶ H_2O + CO 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: 17 O_2 + 2 CH_3(CH_2)_6CH_3 ⟶ 18 H_2O + 16 CO 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 | 17 | -17 CH_3(CH_2)_6CH_3 | 2 | -2 H_2O | 18 | 18 CO | 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 O_2 | 17 | -17 | -1/17 (Δ[O2])/(Δt) CH_3(CH_2)_6CH_3 | 2 | -2 | -1/2 (Δ[CH3(CH2)6CH3])/(Δt) H_2O | 18 | 18 | 1/18 (Δ[H2O])/(Δt) CO | 16 | 16 | 1/16 (Δ[CO])/(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 = -1/17 (Δ[O2])/(Δt) = -1/2 (Δ[CH3(CH2)6CH3])/(Δt) = 1/18 (Δ[H2O])/(Δt) = 1/16 (Δ[CO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

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