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Determine the normal boiling point of ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \)), given that its enthalpy of vaporization is 38.6 kJ/mol and its entropy of vaporization is 110 J/mol·K.

Answer :

Final answer:

The normal boiling point of ethanol is calculated by using the thermodynamic equation ΔG = ΔH - TΔS. At the boiling point, ΔG equals zero, which allows solving the equation for temperature (T), giving a normal boiling point of approximately 77.76 °C for ethanol.

Explanation:

To determine the normal boiling point of ethanol (CH3CH2OH), given its enthalpy of vaporization is 38.6 kJ/mol and its entropy of vaporization is 110 J/mol·K, we can use the thermodynamic equation that relates these quantities with temperature: ΔG = ΔH - TΔS.

At the boiling point, ΔG equals zero because the liquid and gas phases are in equilibrium. Therefore, we can set the equation to 0 = ΔH - TΔS to solve for T, which represents the normal boiling point.

The calculation would be as follows:

  1. ΔH (enthalpy of vaporization) = 38,600 J/mol
  2. ΔS (entropy of vaporization) = 110 J/mol·K
  3. Set ΔG = 0 and rearrange the equation for T: T = ΔH/ΔS
  4. Calculate T: T = 38,600 J/mol / 110 J/mol·K = 350.91 K
  5. Convert Kelvin to degrees Celsius: 350.91 K - 273.15 K = 77.76 °C

Therefore, the normal boiling point of ethanol is approximately 77.76 °C.

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