High School

Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ 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.

Learn more about Normal Boiling Point of Ethanol here:

https://brainly.com/question/14656434

#SPJ4