H2 + CO = H2O + C8H18

H_2 hydrogen + CO carbon monoxide ⟶ H_2O water + CH_3(CH_2)_6CH_3 octane

Balance the chemical equation algebraically: H_2 + CO ⟶ H_2O + CH_3(CH_2)_6CH_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2 + c_2 CO ⟶ c_3 H_2O + c_4 CH_3(CH_2)_6CH_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, C and O: H: | 2 c_1 = 2 c_3 + 18 c_4 C: | c_2 = 8 c_4 O: | c_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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 17 c_2 = 8 c_3 = 8 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 17 H_2 + 8 CO ⟶ 8 H_2O + CH_3(CH_2)_6CH_3

+ ⟶ +

hydrogen + carbon monoxide ⟶ water + octane

Construct the equilibrium constant, K, expression for: H_2 + CO ⟶ H_2O + CH_3(CH_2)_6CH_3 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 H_2 + 8 CO ⟶ 8 H_2O + CH_3(CH_2)_6CH_3 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 H_2 | 17 | -17 CO | 8 | -8 H_2O | 8 | 8 CH_3(CH_2)_6CH_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2 | 17 | -17 | ([H2])^(-17) CO | 8 | -8 | ([CO])^(-8) H_2O | 8 | 8 | ([H2O])^8 CH_3(CH_2)_6CH_3 | 1 | 1 | [CH3(CH2)6CH3] 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 = ([H2])^(-17) ([CO])^(-8) ([H2O])^8 [CH3(CH2)6CH3] = (([H2O])^8 [CH3(CH2)6CH3])/(([H2])^17 ([CO])^8)

Construct the rate of reaction expression for: H_2 + CO ⟶ H_2O + CH_3(CH_2)_6CH_3 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 H_2 + 8 CO ⟶ 8 H_2O + CH_3(CH_2)_6CH_3 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 H_2 | 17 | -17 CO | 8 | -8 H_2O | 8 | 8 CH_3(CH_2)_6CH_3 | 1 | 1 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 H_2 | 17 | -17 | -1/17 (Δ[H2])/(Δt) CO | 8 | -8 | -1/8 (Δ[CO])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) CH_3(CH_2)_6CH_3 | 1 | 1 | (Δ[CH3(CH2)6CH3])/(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[H2])/(Δt) = -1/8 (Δ[CO])/(Δt) = 1/8 (Δ[H2O])/(Δt) = (Δ[CH3(CH2)6CH3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen | carbon monoxide | water | octane formula | H_2 | CO | H_2O | CH_3(CH_2)_6CH_3 Hill formula | H_2 | CO | H_2O | C_8H_18 name | hydrogen | carbon monoxide | water | octane IUPAC name | molecular hydrogen | carbon monoxide | water | octane