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For 2-methylbutane, the AH° of vaporization is 25.22 kJ/mol and the

AS° of vaporization is 84.48 J/mol K. At 1.00 atm and 201 K, what is

the AG° of vaporization for 2-methylbutane, in kJ/mol?

Respuesta :

sciencegeekgirl2017

Answer:

ΔG° = ΔH° - TΔS°

ΔG° = 25.22-(201 )(0.08448)

ΔG° = 8.24 kJ/mol

pstnonsonjoku

The standard free energy is obtained as 8.2  kJ/mol.

We know that; ΔG° = ΔH° - TΔS° where;

ΔG° = standard free energy

ΔH° = standard enthalpy

T = temperature

ΔS° = standard entropy

Hence substituting the values;

ΔG° =25.22 kJ/mol - (201 K * 84.48 J/mol K)

ΔG° = 25.22 * 10^3 J/mol -   16980.5 J/mol

ΔG° = 8.2  kJ/mol

Learn more about free energy: https://brainly.com/question/1301963

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