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The Gibbs free energy ΔG for the vaporisation of 2-methylbutane, in KJ/mol is 8.24 KJ/mol
<h3>Data obtained from the question </h3>
- 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) =?
<h3>How to determine the Gibbs free energy </h3>
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
Learn more about Gibbs free energy:
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