Question

For 2-methylbutane, the ∆h° of vaporization is 25. 22 kj/mol and the ∆s° of vaporization is 84. 48 j/mol･k. At 1. 00 atm and 201 k, what is the ∆g° of vaporization for 2-methylbutane, in kj/mol?

Answer 1

The Gibbs free energy ΔG for the vaporisation of 2-methylbutane, in KJ/mol is 8.24 KJ/mol

Data obtained from the question

• Enthalpy change (ΔH) = 25. 22 KJ/mol
• Change in entropy (ΔS) = 84.48 J/Kmol = 84.48 / 1000 = 8.448×10¯² KJ/Kmol
• Temperature (T) = 201 K

• Gibbs free energy (ΔG) =?

How to determine the Gibbs free energy

The Gibbs free energy for the reaction can be obtained as follow:

ΔG = ΔH – TΔS

ΔG = 25.22 – (201 × 8.448×10¯²)

ΔG = 25.22 – 169.8048

ΔG = 8.24 KJ/mol

Thus, the ΔG for the reaction is 8.24 KJ/mol

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