Input interpretation
![H_2O water + CH_2=CH_2 ethylene ⟶ CH_3CH_2OH ethanol](../image_source/434feb8fe757282604869b0821a1dcc7.png)
H_2O water + CH_2=CH_2 ethylene ⟶ CH_3CH_2OH ethanol
Balanced equation
![Balance the chemical equation algebraically: H_2O + CH_2=CH_2 ⟶ CH_3CH_2OH Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 CH_2=CH_2 ⟶ c_3 CH_3CH_2OH Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and C: H: | 2 c_1 + 4 c_2 = 6 c_3 O: | c_1 = 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_2O + CH_2=CH_2 ⟶ CH_3CH_2OH](../image_source/56d75ba1f234cd4ae20e3741f0fc4d5f.png)
Balance the chemical equation algebraically: H_2O + CH_2=CH_2 ⟶ CH_3CH_2OH Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 CH_2=CH_2 ⟶ c_3 CH_3CH_2OH Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and C: H: | 2 c_1 + 4 c_2 = 6 c_3 O: | c_1 = 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_2O + CH_2=CH_2 ⟶ CH_3CH_2OH
Structures
![+ ⟶](../image_source/c5372eab3dea2e50433a45c656e56ecf.png)
+ ⟶
Names
![water + ethylene ⟶ ethanol](../image_source/656a3af79e5f76939e43123b7b6a09a3.png)
water + ethylene ⟶ ethanol
Reaction thermodynamics
Enthalpy
![| water | ethylene | ethanol molecular enthalpy | -285.8 kJ/mol | 52.4 kJ/mol | -277.7 kJ/mol total enthalpy | -285.8 kJ/mol | 52.4 kJ/mol | -277.7 kJ/mol | H_initial = -233.4 kJ/mol | | H_final = -277.7 kJ/mol ΔH_rxn^0 | -277.7 kJ/mol - -233.4 kJ/mol = -44.26 kJ/mol (exothermic) | |](../image_source/60f5e506d0d08df9c79ff3a4a296513f.png)
| water | ethylene | ethanol molecular enthalpy | -285.8 kJ/mol | 52.4 kJ/mol | -277.7 kJ/mol total enthalpy | -285.8 kJ/mol | 52.4 kJ/mol | -277.7 kJ/mol | H_initial = -233.4 kJ/mol | | H_final = -277.7 kJ/mol ΔH_rxn^0 | -277.7 kJ/mol - -233.4 kJ/mol = -44.26 kJ/mol (exothermic) | |
Gibbs free energy
![| water | ethylene | ethanol molecular free energy | -237.1 kJ/mol | 68 kJ/mol | -174.8 kJ/mol total free energy | -237.1 kJ/mol | 68 kJ/mol | -174.8 kJ/mol | G_initial = -169.1 kJ/mol | | G_final = -174.8 kJ/mol ΔG_rxn^0 | -174.8 kJ/mol - -169.1 kJ/mol = -5.7 kJ/mol (exergonic) | |](../image_source/d8e8b4d631049d36c276b33eb3e0b359.png)
| water | ethylene | ethanol molecular free energy | -237.1 kJ/mol | 68 kJ/mol | -174.8 kJ/mol total free energy | -237.1 kJ/mol | 68 kJ/mol | -174.8 kJ/mol | G_initial = -169.1 kJ/mol | | G_final = -174.8 kJ/mol ΔG_rxn^0 | -174.8 kJ/mol - -169.1 kJ/mol = -5.7 kJ/mol (exergonic) | |
Entropy
![| water | ethylene | ethanol molecular entropy | 69.91 J/(mol K) | 219 J/(mol K) | 160.7 J/(mol K) total entropy | 69.91 J/(mol K) | 219 J/(mol K) | 160.7 J/(mol K) | S_initial = 288.9 J/(mol K) | | S_final = 160.7 J/(mol K) ΔS_rxn^0 | 160.7 J/(mol K) - 288.9 J/(mol K) = -128.2 J/(mol K) (exoentropic) | |](../image_source/5c3b3fc4172510d583482d1b58cdf7f5.png)
| water | ethylene | ethanol molecular entropy | 69.91 J/(mol K) | 219 J/(mol K) | 160.7 J/(mol K) total entropy | 69.91 J/(mol K) | 219 J/(mol K) | 160.7 J/(mol K) | S_initial = 288.9 J/(mol K) | | S_final = 160.7 J/(mol K) ΔS_rxn^0 | 160.7 J/(mol K) - 288.9 J/(mol K) = -128.2 J/(mol K) (exoentropic) | |
Equilibrium constant
![Construct the equilibrium constant, K, expression for: H_2O + CH_2=CH_2 ⟶ CH_3CH_2OH 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_2O + CH_2=CH_2 ⟶ CH_3CH_2OH 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_2O | 1 | -1 CH_2=CH_2 | 1 | -1 CH_3CH_2OH | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 1 | -1 | ([H2O])^(-1) CH_2=CH_2 | 1 | -1 | ([CH2=CH2])^(-1) CH_3CH_2OH | 1 | 1 | [CH3CH2OH] 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 = ([H2O])^(-1) ([CH2=CH2])^(-1) [CH3CH2OH] = ([CH3CH2OH])/([H2O] [CH2=CH2])](../image_source/99fa54636248e7eef212228dea06569b.png)
Construct the equilibrium constant, K, expression for: H_2O + CH_2=CH_2 ⟶ CH_3CH_2OH 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_2O + CH_2=CH_2 ⟶ CH_3CH_2OH 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_2O | 1 | -1 CH_2=CH_2 | 1 | -1 CH_3CH_2OH | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 1 | -1 | ([H2O])^(-1) CH_2=CH_2 | 1 | -1 | ([CH2=CH2])^(-1) CH_3CH_2OH | 1 | 1 | [CH3CH2OH] 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 = ([H2O])^(-1) ([CH2=CH2])^(-1) [CH3CH2OH] = ([CH3CH2OH])/([H2O] [CH2=CH2])
Rate of reaction
![Construct the rate of reaction expression for: H_2O + CH_2=CH_2 ⟶ CH_3CH_2OH 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_2O + CH_2=CH_2 ⟶ CH_3CH_2OH 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_2O | 1 | -1 CH_2=CH_2 | 1 | -1 CH_3CH_2OH | 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_2O | 1 | -1 | -(Δ[H2O])/(Δt) CH_2=CH_2 | 1 | -1 | -(Δ[CH2=CH2])/(Δt) CH_3CH_2OH | 1 | 1 | (Δ[CH3CH2OH])/(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[H2O])/(Δt) = -(Δ[CH2=CH2])/(Δt) = (Δ[CH3CH2OH])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/d9ba8c079246433a6a769802e823ef48.png)
Construct the rate of reaction expression for: H_2O + CH_2=CH_2 ⟶ CH_3CH_2OH 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_2O + CH_2=CH_2 ⟶ CH_3CH_2OH 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_2O | 1 | -1 CH_2=CH_2 | 1 | -1 CH_3CH_2OH | 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_2O | 1 | -1 | -(Δ[H2O])/(Δt) CH_2=CH_2 | 1 | -1 | -(Δ[CH2=CH2])/(Δt) CH_3CH_2OH | 1 | 1 | (Δ[CH3CH2OH])/(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[H2O])/(Δt) = -(Δ[CH2=CH2])/(Δt) = (Δ[CH3CH2OH])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| water | ethylene | ethanol formula | H_2O | CH_2=CH_2 | CH_3CH_2OH Hill formula | H_2O | C_2H_4 | C_2H_6O name | water | ethylene | ethanol](../image_source/f26df443c95b194c13ef569e94089893.png)
| water | ethylene | ethanol formula | H_2O | CH_2=CH_2 | CH_3CH_2OH Hill formula | H_2O | C_2H_4 | C_2H_6O name | water | ethylene | ethanol