H2 + C2H4 = C2H6

H_2 (hydrogen) + CH_2=CH_2 (ethylene) ⟶ CH_3CH_3 (ethane)

Balance the chemical equation algebraically: H_2 + CH_2=CH_2 ⟶ CH_3CH_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2 + c_2 CH_2=CH_2 ⟶ c_3 CH_3CH_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H and C: H: | 2 c_1 + 4 c_2 = 6 c_3 C: | 2 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_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: | | H_2 + CH_2=CH_2 ⟶ CH_3CH_3

+ ⟶

hydrogen + ethylene ⟶ ethane

| hydrogen | ethylene | ethane molecular enthalpy | 0 kJ/mol | 52.4 kJ/mol | -84 kJ/mol total enthalpy | 0 kJ/mol | 52.4 kJ/mol | -84 kJ/mol | H_initial = 52.4 kJ/mol | | H_final = -84 kJ/mol ΔH_rxn^0 | -84 kJ/mol - 52.4 kJ/mol = -136.4 kJ/mol (exothermic) | |

| hydrogen | ethylene | ethane molecular free energy | 0 kJ/mol | 68 kJ/mol | -32 kJ/mol total free energy | 0 kJ/mol | 68 kJ/mol | -32 kJ/mol | G_initial = 68 kJ/mol | | G_final = -32 kJ/mol ΔG_rxn^0 | -32 kJ/mol - 68 kJ/mol = -100 kJ/mol (exergonic) | |

| hydrogen | ethylene | ethane molecular entropy | 115 J/(mol K) | 219 J/(mol K) | 229.5 J/(mol K) total entropy | 115 J/(mol K) | 219 J/(mol K) | 229.5 J/(mol K) | S_initial = 334 J/(mol K) | | S_final = 229.5 J/(mol K) ΔS_rxn^0 | 229.5 J/(mol K) - 334 J/(mol K) = -104.5 J/(mol K) (exoentropic) | |

Construct the equilibrium constant, K, expression for: H_2 + CH_2=CH_2 ⟶ CH_3CH_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: H_2 + CH_2=CH_2 ⟶ CH_3CH_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 | 1 | -1 CH_2=CH_2 | 1 | -1 CH_3CH_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 | 1 | -1 | ([H2])^(-1) CH_2=CH_2 | 1 | -1 | ([CH2=CH2])^(-1) CH_3CH_3 | 1 | 1 | [CH3CH3] 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])^(-1) ([CH2=CH2])^(-1) [CH3CH3] = ([CH3CH3])/([H2] [CH2=CH2])

Construct the rate of reaction expression for: H_2 + CH_2=CH_2 ⟶ CH_3CH_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: H_2 + CH_2=CH_2 ⟶ CH_3CH_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 | 1 | -1 CH_2=CH_2 | 1 | -1 CH_3CH_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 | 1 | -1 | -(Δ[H2])/(Δt) CH_2=CH_2 | 1 | -1 | -(Δ[CH2=CH2])/(Δt) CH_3CH_3 | 1 | 1 | (Δ[CH3CH3])/(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[H2])/(Δt) = -(Δ[CH2=CH2])/(Δt) = (Δ[CH3CH3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen | ethylene | ethane formula | H_2 | CH_2=CH_2 | CH_3CH_3 Hill formula | H_2 | C_2H_4 | C_2H_6 name | hydrogen | ethylene | ethane IUPAC name | molecular hydrogen | ethylene | ethane