O2 + C4H8 = H2O + CO2

O_2 (oxygen) + (CH_3)_2C=CH_2 (isobutylene) ⟶ H_2O (water) + CO_2 (carbon dioxide)

Balance the chemical equation algebraically: O_2 + (CH_3)_2C=CH_2 ⟶ H_2O + CO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 (CH_3)_2C=CH_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 O, C and H: O: | 2 c_1 = c_3 + 2 c_4 C: | 4 c_2 = c_4 H: | 8 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 = 6 c_2 = 1 c_3 = 4 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 O_2 + (CH_3)_2C=CH_2 ⟶ 4 H_2O + 4 CO_2

+ ⟶ +

oxygen + isobutylene ⟶ water + carbon dioxide

| oxygen | isobutylene | water | carbon dioxide molecular enthalpy | 0 kJ/mol | -16.9 kJ/mol | -285.8 kJ/mol | -393.5 kJ/mol total enthalpy | 0 kJ/mol | -16.9 kJ/mol | -1143 kJ/mol | -1574 kJ/mol | H_initial = -16.9 kJ/mol | | H_final = -2717 kJ/mol | ΔH_rxn^0 | -2717 kJ/mol - -16.9 kJ/mol = -2700 kJ/mol (exothermic) | | |

Construct the equilibrium constant, K, expression for: O_2 + (CH_3)_2C=CH_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: 6 O_2 + (CH_3)_2C=CH_2 ⟶ 4 H_2O + 4 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 | 6 | -6 (CH_3)_2C=CH_2 | 1 | -1 H_2O | 4 | 4 CO_2 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 6 | -6 | ([O2])^(-6) (CH_3)_2C=CH_2 | 1 | -1 | ([(CH3)2C=CH2])^(-1) H_2O | 4 | 4 | ([H2O])^4 CO_2 | 4 | 4 | ([CO2])^4 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])^(-6) ([(CH3)2C=CH2])^(-1) ([H2O])^4 ([CO2])^4 = (([H2O])^4 ([CO2])^4)/(([O2])^6 [(CH3)2C=CH2])

Construct the rate of reaction expression for: O_2 + (CH_3)_2C=CH_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: 6 O_2 + (CH_3)_2C=CH_2 ⟶ 4 H_2O + 4 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 | 6 | -6 (CH_3)_2C=CH_2 | 1 | -1 H_2O | 4 | 4 CO_2 | 4 | 4 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 | 6 | -6 | -1/6 (Δ[O2])/(Δt) (CH_3)_2C=CH_2 | 1 | -1 | -(Δ[(CH3)2C=CH2])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) CO_2 | 4 | 4 | 1/4 (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 = -1/6 (Δ[O2])/(Δt) = -(Δ[(CH3)2C=CH2])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/4 (Δ[CO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | isobutylene | water | carbon dioxide formula | O_2 | (CH_3)_2C=CH_2 | H_2O | CO_2 Hill formula | O_2 | C_4H_8 | H_2O | CO_2 name | oxygen | isobutylene | water | carbon dioxide IUPAC name | molecular oxygen | 2-methylprop-1-ene | water | carbon dioxide