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If ΔH vaporization of water (H2O) is 40.7 kJ/mol,
and the ΔS for vaporization of H2O is 109 J/mol K,
what is ΔG for water vaporization at 350K?
Include units.

Respuesta :

dsdrajlin

Answer:

2.6 kJ/mol

Explanation:

Step 1: Given data

  • Standard enthalpy of vaporization of water (ΔH°): 40.7 kJ/mol
  • Standard entropy of vaporization of water (ΔS°): 109 J/mol.K
  • Absolute temperature (T): 350 K
  • Standard Gibbs free energy of vaporization of water (ΔG°): ?

Step 2: Calculate ΔG°

We can calculate ΔG° using the following expression.

ΔG° = ΔH° - T × ΔS°

ΔG° = 40.7 kJ/mol - 350 K × 0.109 kJ/mol.K

ΔG° = 2.6 kJ/mol

Eduard22sly

The change in Gibbs free energy, ΔG for the water is 2.55 KJ/mol

Data obtained from the question

From the question given above, the following data were obtained:

  • Standard enthalpy of vaporization of water (ΔH°) = 40.7 kJ/mol
  • Standard entropy of vaporization of water (ΔS°) = 109 J/Kmol = 109 / 1000 = 0.109 KJ/Kmol
  • Absolute temperature (T) = 350 K
  • Standard Gibbs free energy (ΔG°) =?

How to determine the Gibbs free energy, ΔG°

The Gibbs free energy, ΔG° can be obtained as illustrated below:

ΔG° = ΔH° – TΔS°

ΔG° = 40.7 – (350 × 0.109)

ΔG° = 40.7 – 38.15

ΔG° = 2.55 KJ/mol

Learn more about Gibbs free energy:

https://brainly.com/question/9256736

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