contestada

Given the values of So given below in J/mol K and the values of ΔHfo given in kJ/mol, calculate the value of ΔGo in kJ for the combustion of 1 mole of propane to form carbon dioxide and gaseous water at 298 K. S (C3H8(g)) = 271 S (O2(g)) = 204 S (CO2(g)) = 214 S (H2O(g)) = 184 ΔHfo (C3H8(g)) = -100 ΔHfo (CO2(g)) = -398 ΔHfo (H2O(g)) = -226

Respuesta :

IthaloAbreu

Answer:

-2024 kJ

Explanation:

The combustion of propane is

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

And, ΔG° = ΔH° - TΔS°

Where T is the temperature in K.

ΔH° = ∑n*H°f, reactants - ∑n*H°f, products

Where n is the number of moles in the stoichiometry reaction. H°f, O₂(g) = 0 because it's a substance formed by only one element.

ΔH° = [4*(-226) +3*(-398)] - [-100] = -1998 kJ

ΔS° = ∑n*S°, reactants - ∑n*S°, products

ΔS° = [4*(184) + 3*214] - [5*204 + 271] = 87 J/K = 0.087 kJ/K

So

ΔG° = -1998 - 298*0.087

ΔG° = -2024 kJ

Eduard22sly

The value of the Gibbs free energy, ΔG for the combustion of 1 mole of propane to form carbon dioxide and gaseous water at 298 K is –2023.926 KJ/mol

Balanced equation

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

How to determine the change in entropy ΔS

  • Entropy of C₃H₈ = 271 J/Kmol
  • Entropy of O₂ = 204 J/Kmol
  • Entropy of CO₂ = 214 J/Kmol
  • Entropy of H₂O(g) = 184 J/Kmol
  • Change in entropy (ΔS) =?

ΔS°rxn = ∑nS°(products) – ∑nS°(reactants)

ΔS°rxn = [3S°(CO₂) + 4S°(H₂O)] – [S°(C₃H₈) + 5S°(O₂)]

ΔS°rxn = [3(214) + 4(184)] – [271 + 5(204)]

ΔS°rxn = 1378 – 1291

ΔS°rxn = 87 J/Kmol

ΔS°rxn = 87 / 1000

ΔS°rxn = 0.087 KJ/Kmol

How to determine the change in enthalpy ΔH

  • Enthalpy of C₃H₈ = –100 KJ/Kmol
  • Enthalpy of O₂ = 0 KJ/Kmol
  • Enthalpy of CO₂ = –398 KJ/Kmol
  • Enthalpy of H₂O(g) = –226 KJ/Kmol
  • Enthalpy change (ΔH) =?

ΔH°rxn = ∑nH°(products) – ∑nH°(reactants)

ΔH°rxn = [3H°(CO₂) + 4H°(H₂O)] – [H°(C₃H₈) + 5H°(O₂)]

ΔH°rxn = [3(–398) + 4(–226)] – [–100 + 5(0)]

ΔH°rxn = [–2098] – [–100]

ΔH°rxn = –1998 KJ/Kmol

How to determine the Gibbs free energy

  • Change in entropy (ΔS) = 0.087 KJ/Kmol
  • Enthalpy change (ΔH) = –1998 KJ/Kmol
  • Temperature (T) = 298 K
  • Gibbs free energy (ΔG) =?

ΔG = ΔH – TΔS

ΔG = –1998 – (298 × 0.087)

ΔG = –1998 – 25.926

ΔG = –2023.926 KJ/mol

Thus, the ΔG for the reaction is –2023.926 KJ/mol

Learn more about Gibbs free energy:

https://brainly.com/question/9552459

Otras preguntas