C4H10 = H2 + C4H8

CH_3CH_2CH_2CH_3 butane ⟶ H_2 hydrogen + (CH_3)_2C=CH_2 isobutylene

Balance the chemical equation algebraically: CH_3CH_2CH_2CH_3 ⟶ H_2 + (CH_3)_2C=CH_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CH_3CH_2CH_2CH_3 ⟶ c_2 H_2 + c_3 (CH_3)_2C=CH_2 Set the number of atoms in the reactants equal to the number of atoms in the products for C and H: C: | 4 c_1 = 4 c_3 H: | 10 c_1 = 2 c_2 + 8 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | CH_3CH_2CH_2CH_3 ⟶ H_2 + (CH_3)_2C=CH_2

⟶ +

butane ⟶ hydrogen + isobutylene

| butane | hydrogen | isobutylene molecular enthalpy | -125.7 kJ/mol | 0 kJ/mol | -16.9 kJ/mol total enthalpy | -125.7 kJ/mol | 0 kJ/mol | -16.9 kJ/mol | H_initial = -125.7 kJ/mol | H_final = -16.9 kJ/mol | ΔH_rxn^0 | -16.9 kJ/mol - -125.7 kJ/mol = 108.8 kJ/mol (endothermic) | |

Construct the equilibrium constant, K, expression for: CH_3CH_2CH_2CH_3 ⟶ H_2 + (CH_3)_2C=CH_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: CH_3CH_2CH_2CH_3 ⟶ H_2 + (CH_3)_2C=CH_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_3CH_2CH_2CH_3 | 1 | -1 H_2 | 1 | 1 (CH_3)_2C=CH_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CH_3CH_2CH_2CH_3 | 1 | -1 | ([CH3CH2CH2CH3])^(-1) H_2 | 1 | 1 | [H2] (CH_3)_2C=CH_2 | 1 | 1 | [(CH3)2C=CH2] 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 = ([CH3CH2CH2CH3])^(-1) [H2] [(CH3)2C=CH2] = ([H2] [(CH3)2C=CH2])/([CH3CH2CH2CH3])

Construct the rate of reaction expression for: CH_3CH_2CH_2CH_3 ⟶ H_2 + (CH_3)_2C=CH_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: CH_3CH_2CH_2CH_3 ⟶ H_2 + (CH_3)_2C=CH_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_3CH_2CH_2CH_3 | 1 | -1 H_2 | 1 | 1 (CH_3)_2C=CH_2 | 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 CH_3CH_2CH_2CH_3 | 1 | -1 | -(Δ[CH3CH2CH2CH3])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) (CH_3)_2C=CH_2 | 1 | 1 | (Δ[(CH3)2C=CH2])/(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 = -(Δ[CH3CH2CH2CH3])/(Δt) = (Δ[H2])/(Δt) = (Δ[(CH3)2C=CH2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| butane | hydrogen | isobutylene formula | CH_3CH_2CH_2CH_3 | H_2 | (CH_3)_2C=CH_2 Hill formula | C_4H_10 | H_2 | C_4H_8 name | butane | hydrogen | isobutylene IUPAC name | butane | molecular hydrogen | 2-methylprop-1-ene

| butane | hydrogen | isobutylene molar mass | 58.12 g/mol | 2.016 g/mol | 56.11 g/mol phase | gas (at STP) | gas (at STP) | gas (at STP) melting point | -138.3 °C | -259.2 °C | -140 °C boiling point | -0.5 °C | -252.8 °C | -6.9 °C density | 0.00249343 g/cm^3 (at 20 °C) | 8.99×10^-5 g/cm^3 (at 0 °C) | 0.5942 g/cm^3 (at 20 °C) solubility in water | | | insoluble surface tension | 0.01487 N/m | | 0.0123 N/m dynamic viscosity | 7×10^-6 Pa s (at 25 °C) | 8.9×10^-6 Pa s (at 25 °C) | 8.084×10^-6 Pa s (at 25 °C) odor | | odorless |