Input interpretation
CH_3CH_2OH ethanol ⟶ H_2O water + CH_2=CH_2 ethylene
Balanced equation
Balance the chemical equation algebraically: CH_3CH_2OH ⟶ H_2O + CH_2=CH_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 CH_3CH_2OH ⟶ c_2 H_2O + c_3 CH_2=CH_2 Set the number of atoms in the reactants equal to the number of atoms in the products for C, H and O: C: | 2 c_1 = 2 c_3 H: | 6 c_1 = 2 c_2 + 4 c_3 O: | c_1 = c_2 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_2OH ⟶ H_2O + CH_2=CH_2
Structures
⟶ +
Names
ethanol ⟶ water + ethylene
Reaction thermodynamics
Enthalpy
| ethanol | water | ethylene molecular enthalpy | -277.7 kJ/mol | -285.8 kJ/mol | 52.4 kJ/mol total enthalpy | -277.7 kJ/mol | -285.8 kJ/mol | 52.4 kJ/mol | H_initial = -277.7 kJ/mol | H_final = -233.4 kJ/mol | ΔH_rxn^0 | -233.4 kJ/mol - -277.7 kJ/mol = 44.26 kJ/mol (endothermic) | |
Gibbs free energy
| ethanol | water | ethylene molecular free energy | -174.8 kJ/mol | -237.1 kJ/mol | 68 kJ/mol total free energy | -174.8 kJ/mol | -237.1 kJ/mol | 68 kJ/mol | G_initial = -174.8 kJ/mol | G_final = -169.1 kJ/mol | ΔG_rxn^0 | -169.1 kJ/mol - -174.8 kJ/mol = 5.7 kJ/mol (endergonic) | |
Entropy
| ethanol | water | ethylene molecular entropy | 160.7 J/(mol K) | 69.91 J/(mol K) | 219 J/(mol K) total entropy | 160.7 J/(mol K) | 69.91 J/(mol K) | 219 J/(mol K) | S_initial = 160.7 J/(mol K) | S_final = 288.9 J/(mol K) | ΔS_rxn^0 | 288.9 J/(mol K) - 160.7 J/(mol K) = 128.2 J/(mol K) (endoentropic) | |
Equilibrium constant
Construct the equilibrium constant, K, expression for: CH_3CH_2OH ⟶ H_2O + CH_2=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_2OH ⟶ H_2O + CH_2=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_2OH | 1 | -1 H_2O | 1 | 1 CH_2=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_2OH | 1 | -1 | ([CH3CH2OH])^(-1) H_2O | 1 | 1 | [H2O] CH_2=CH_2 | 1 | 1 | [CH2=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 = ([CH3CH2OH])^(-1) [H2O] [CH2=CH2] = ([H2O] [CH2=CH2])/([CH3CH2OH])
Rate of reaction
Construct the rate of reaction expression for: CH_3CH_2OH ⟶ H_2O + CH_2=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_2OH ⟶ H_2O + CH_2=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_2OH | 1 | -1 H_2O | 1 | 1 CH_2=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_2OH | 1 | -1 | -(Δ[CH3CH2OH])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) CH_2=CH_2 | 1 | 1 | (Δ[CH2=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[CH3CH2OH])/(Δt) = (Δ[H2O])/(Δt) = (Δ[CH2=CH2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
| ethanol | water | ethylene formula | CH_3CH_2OH | H_2O | CH_2=CH_2 Hill formula | C_2H_6O | H_2O | C_2H_4 name | ethanol | water | ethylene