What is the boiling point of methanol, if ∆H°for vaporization is 35.3 kJ/mol, and ∆S° for vaporization is 105 J/mol K?
1 answer:
Answer: 336.2K
Explanation:
The boiling point is obtained from the standard enthalpy and entropy of vaporization as shown in the image attached.
Standard enthalpy of vaporization=35.3KJmol-1=35.3×10^3 Jmol-1
Standard entropy of vaporization= 105 JMol-1K-1
The formula and details of the solution are shown in the image attached.
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