O3 + C4H10 = H2O + CO2

O_3 ozone + CH_3CH_2CH_2CH_3 butane ⟶ H_2O water + CO_2 carbon dioxide

Balance the chemical equation algebraically: O_3 + CH_3CH_2CH_2CH_3 ⟶ H_2O + CO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_3 + c_2 CH_3CH_2CH_2CH_3 ⟶ 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: | 3 c_1 = c_3 + 2 c_4 C: | 4 c_2 = c_4 H: | 10 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 = 13/3 c_2 = 1 c_3 = 5 c_4 = 4 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 13 c_2 = 3 c_3 = 15 c_4 = 12 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 13 O_3 + 3 CH_3CH_2CH_2CH_3 ⟶ 15 H_2O + 12 CO_2

+ ⟶ +

ozone + butane ⟶ water + carbon dioxide

| ozone | butane | water | carbon dioxide molecular enthalpy | 142.7 kJ/mol | -125.7 kJ/mol | -285.8 kJ/mol | -393.5 kJ/mol total enthalpy | 1855 kJ/mol | -377.1 kJ/mol | -4287 kJ/mol | -4722 kJ/mol | H_initial = 1478 kJ/mol | | H_final = -9009 kJ/mol | ΔH_rxn^0 | -9009 kJ/mol - 1478 kJ/mol = -10487 kJ/mol (exothermic) | | |

Construct the equilibrium constant, K, expression for: O_3 + CH_3CH_2CH_2CH_3 ⟶ 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: 13 O_3 + 3 CH_3CH_2CH_2CH_3 ⟶ 15 H_2O + 12 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_3 | 13 | -13 CH_3CH_2CH_2CH_3 | 3 | -3 H_2O | 15 | 15 CO_2 | 12 | 12 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_3 | 13 | -13 | ([O3])^(-13) CH_3CH_2CH_2CH_3 | 3 | -3 | ([CH3CH2CH2CH3])^(-3) H_2O | 15 | 15 | ([H2O])^15 CO_2 | 12 | 12 | ([CO2])^12 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 = ([O3])^(-13) ([CH3CH2CH2CH3])^(-3) ([H2O])^15 ([CO2])^12 = (([H2O])^15 ([CO2])^12)/(([O3])^13 ([CH3CH2CH2CH3])^3)

Construct the rate of reaction expression for: O_3 + CH_3CH_2CH_2CH_3 ⟶ 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: 13 O_3 + 3 CH_3CH_2CH_2CH_3 ⟶ 15 H_2O + 12 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_3 | 13 | -13 CH_3CH_2CH_2CH_3 | 3 | -3 H_2O | 15 | 15 CO_2 | 12 | 12 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_3 | 13 | -13 | -1/13 (Δ[O3])/(Δt) CH_3CH_2CH_2CH_3 | 3 | -3 | -1/3 (Δ[CH3CH2CH2CH3])/(Δt) H_2O | 15 | 15 | 1/15 (Δ[H2O])/(Δt) CO_2 | 12 | 12 | 1/12 (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 = -1/13 (Δ[O3])/(Δt) = -1/3 (Δ[CH3CH2CH2CH3])/(Δt) = 1/15 (Δ[H2O])/(Δt) = 1/12 (Δ[CO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| ozone | butane | water | carbon dioxide formula | O_3 | CH_3CH_2CH_2CH_3 | H_2O | CO_2 Hill formula | O_3 | C_4H_10 | H_2O | CO_2 name | ozone | butane | water | carbon dioxide