Input interpretation
![O_2 (oxygen) + CH_3CH_3 (ethane) ⟶ H_2O (water) + CO_2 (carbon dioxide)](../image_source/248cce198e360e41dbca17e923bd1e00.png)
O_2 (oxygen) + CH_3CH_3 (ethane) ⟶ H_2O (water) + CO_2 (carbon dioxide)
Balanced equation
![Balance the chemical equation algebraically: O_2 + CH_3CH_3 ⟶ H_2O + CO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 CH_3CH_3 ⟶ c_3 H_2O + c_4 CO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for O, C and H: O: | 2 c_1 = c_3 + 2 c_4 C: | 2 c_2 = c_4 H: | 6 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 7/2 c_2 = 1 c_3 = 3 c_4 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 7 c_2 = 2 c_3 = 6 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 7 O_2 + 2 CH_3CH_3 ⟶ 6 H_2O + 4 CO_2](../image_source/af33fbc0565d278fc893811b3e252dea.png)
Balance the chemical equation algebraically: O_2 + CH_3CH_3 ⟶ H_2O + CO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 CH_3CH_3 ⟶ c_3 H_2O + c_4 CO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for O, C and H: O: | 2 c_1 = c_3 + 2 c_4 C: | 2 c_2 = c_4 H: | 6 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 7/2 c_2 = 1 c_3 = 3 c_4 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 7 c_2 = 2 c_3 = 6 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 7 O_2 + 2 CH_3CH_3 ⟶ 6 H_2O + 4 CO_2
Structures
![+ ⟶ +](../image_source/003b441d878fd3d16929626ee94d77a6.png)
+ ⟶ +
Names
![oxygen + ethane ⟶ water + carbon dioxide](../image_source/aad10ec647b4eaf8c50666c4c2c76022.png)
oxygen + ethane ⟶ water + carbon dioxide
Reaction thermodynamics
Enthalpy
![| oxygen | ethane | water | carbon dioxide molecular enthalpy | 0 kJ/mol | -84 kJ/mol | -285.8 kJ/mol | -393.5 kJ/mol total enthalpy | 0 kJ/mol | -168 kJ/mol | -1715 kJ/mol | -1574 kJ/mol | H_initial = -168 kJ/mol | | H_final = -3289 kJ/mol | ΔH_rxn^0 | -3289 kJ/mol - -168 kJ/mol = -3121 kJ/mol (exothermic) | | |](../image_source/002bb5315225b59b4e46048d8b541e22.png)
| oxygen | ethane | water | carbon dioxide molecular enthalpy | 0 kJ/mol | -84 kJ/mol | -285.8 kJ/mol | -393.5 kJ/mol total enthalpy | 0 kJ/mol | -168 kJ/mol | -1715 kJ/mol | -1574 kJ/mol | H_initial = -168 kJ/mol | | H_final = -3289 kJ/mol | ΔH_rxn^0 | -3289 kJ/mol - -168 kJ/mol = -3121 kJ/mol (exothermic) | | |
Gibbs free energy
![| oxygen | ethane | water | carbon dioxide molecular free energy | 231.7 kJ/mol | -32 kJ/mol | -237.1 kJ/mol | -394.4 kJ/mol total free energy | 1622 kJ/mol | -64 kJ/mol | -1423 kJ/mol | -1578 kJ/mol | G_initial = 1558 kJ/mol | | G_final = -3000 kJ/mol | ΔG_rxn^0 | -3000 kJ/mol - 1558 kJ/mol = -4558 kJ/mol (exergonic) | | |](../image_source/9ea189f78a5402926106d2395ef240ce.png)
| oxygen | ethane | water | carbon dioxide molecular free energy | 231.7 kJ/mol | -32 kJ/mol | -237.1 kJ/mol | -394.4 kJ/mol total free energy | 1622 kJ/mol | -64 kJ/mol | -1423 kJ/mol | -1578 kJ/mol | G_initial = 1558 kJ/mol | | G_final = -3000 kJ/mol | ΔG_rxn^0 | -3000 kJ/mol - 1558 kJ/mol = -4558 kJ/mol (exergonic) | | |
Entropy
![| oxygen | ethane | water | carbon dioxide molecular entropy | 205 J/(mol K) | 229.5 J/(mol K) | 69.91 J/(mol K) | 214 J/(mol K) total entropy | 1435 J/(mol K) | 459 J/(mol K) | 419.5 J/(mol K) | 856 J/(mol K) | S_initial = 1894 J/(mol K) | | S_final = 1275 J/(mol K) | ΔS_rxn^0 | 1275 J/(mol K) - 1894 J/(mol K) = -618.5 J/(mol K) (exoentropic) | | |](../image_source/d760642ad681478a5bd72ea3785310c4.png)
| oxygen | ethane | water | carbon dioxide molecular entropy | 205 J/(mol K) | 229.5 J/(mol K) | 69.91 J/(mol K) | 214 J/(mol K) total entropy | 1435 J/(mol K) | 459 J/(mol K) | 419.5 J/(mol K) | 856 J/(mol K) | S_initial = 1894 J/(mol K) | | S_final = 1275 J/(mol K) | ΔS_rxn^0 | 1275 J/(mol K) - 1894 J/(mol K) = -618.5 J/(mol K) (exoentropic) | | |
Equilibrium constant
![Construct the equilibrium constant, K, expression for: O_2 + CH_3CH_3 ⟶ H_2O + CO_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: 7 O_2 + 2 CH_3CH_3 ⟶ 6 H_2O + 4 CO_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 O_2 | 7 | -7 CH_3CH_3 | 2 | -2 H_2O | 6 | 6 CO_2 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 7 | -7 | ([O2])^(-7) CH_3CH_3 | 2 | -2 | ([CH3CH3])^(-2) H_2O | 6 | 6 | ([H2O])^6 CO_2 | 4 | 4 | ([CO2])^4 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 = ([O2])^(-7) ([CH3CH3])^(-2) ([H2O])^6 ([CO2])^4 = (([H2O])^6 ([CO2])^4)/(([O2])^7 ([CH3CH3])^2)](../image_source/ebcb8c45a952df9ccda4b16b876cb2f1.png)
Construct the equilibrium constant, K, expression for: O_2 + CH_3CH_3 ⟶ H_2O + CO_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: 7 O_2 + 2 CH_3CH_3 ⟶ 6 H_2O + 4 CO_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 O_2 | 7 | -7 CH_3CH_3 | 2 | -2 H_2O | 6 | 6 CO_2 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 7 | -7 | ([O2])^(-7) CH_3CH_3 | 2 | -2 | ([CH3CH3])^(-2) H_2O | 6 | 6 | ([H2O])^6 CO_2 | 4 | 4 | ([CO2])^4 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 = ([O2])^(-7) ([CH3CH3])^(-2) ([H2O])^6 ([CO2])^4 = (([H2O])^6 ([CO2])^4)/(([O2])^7 ([CH3CH3])^2)
Rate of reaction
![Construct the rate of reaction expression for: O_2 + CH_3CH_3 ⟶ H_2O + CO_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: 7 O_2 + 2 CH_3CH_3 ⟶ 6 H_2O + 4 CO_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 O_2 | 7 | -7 CH_3CH_3 | 2 | -2 H_2O | 6 | 6 CO_2 | 4 | 4 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 O_2 | 7 | -7 | -1/7 (Δ[O2])/(Δt) CH_3CH_3 | 2 | -2 | -1/2 (Δ[CH3CH3])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) CO_2 | 4 | 4 | 1/4 (Δ[CO2])/(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 = -1/7 (Δ[O2])/(Δt) = -1/2 (Δ[CH3CH3])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/4 (Δ[CO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/9506bd221f590092c5039a4e8d8661a7.png)
Construct the rate of reaction expression for: O_2 + CH_3CH_3 ⟶ H_2O + CO_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: 7 O_2 + 2 CH_3CH_3 ⟶ 6 H_2O + 4 CO_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 O_2 | 7 | -7 CH_3CH_3 | 2 | -2 H_2O | 6 | 6 CO_2 | 4 | 4 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 O_2 | 7 | -7 | -1/7 (Δ[O2])/(Δt) CH_3CH_3 | 2 | -2 | -1/2 (Δ[CH3CH3])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) CO_2 | 4 | 4 | 1/4 (Δ[CO2])/(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 = -1/7 (Δ[O2])/(Δt) = -1/2 (Δ[CH3CH3])/(Δt) = 1/6 (Δ[H2O])/(Δt) = 1/4 (Δ[CO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| oxygen | ethane | water | carbon dioxide formula | O_2 | CH_3CH_3 | H_2O | CO_2 Hill formula | O_2 | C_2H_6 | H_2O | CO_2 name | oxygen | ethane | water | carbon dioxide IUPAC name | molecular oxygen | ethane | water | carbon dioxide](../image_source/876fe4ada6b0aed80627588ae996aa43.png)
| oxygen | ethane | water | carbon dioxide formula | O_2 | CH_3CH_3 | H_2O | CO_2 Hill formula | O_2 | C_2H_6 | H_2O | CO_2 name | oxygen | ethane | water | carbon dioxide IUPAC name | molecular oxygen | ethane | water | carbon dioxide