To determine which compound will evaporate most easily, we need to analyze and compare the type and strength of intermolecular forces present in each compound. The easier it is for a compound to evaporate, the weaker its intermolecular forces generally are.
Let's evaluate each compound:
1. [tex]$CH_3CH_2CH_3$[/tex] (Propane):
- This is a non-polar molecule.
- The dominant intermolecular forces are van der Waals forces (also known as dispersion forces).
- Dispersion forces are generally the weakest type of intermolecular force.
2. [tex]$CH_3CH_2OH$[/tex] (Ethanol):
- This molecule has a hydroxyl group (-OH), making it polar.
- Besides van der Waals forces, it can form hydrogen bonds due to the presence of the hydroxyl group.
- Hydrogen bonds are significantly stronger than van der Waals forces.
3. [tex]$CH_3CH_2NH_2$[/tex] (Ethylamine):
- This molecule has an amine group (-NH_2), making it polar.
- Like ethanol, ethylamine can form hydrogen bonds due to the presence of the amine group.
- However, the hydrogen bonds in amines are generally weaker than those in alcohols because the nitrogen-hydrogen bond is less polar than the oxygen-hydrogen bond.
Comparing the three compounds:
- Propane ([tex]$CH_3CH_2CH_3$[/tex]) has only weak van der Waals forces.
- Ethanol ([tex]$CH_3CH_2OH$[/tex]) has strong hydrogen bonds.
- Ethylamine ([tex]$CH_3CH_2NH_2$[/tex]) has hydrogen bonds, which are generally weaker than those in ethanol but still stronger than van der Waals forces.
Since van der Waals forces are weaker than hydrogen bonds, propane will have the weakest intermolecular forces among the three compounds, making it the compound that will evaporate most easily.
Therefore, the compound that will evaporate most easily is:
[tex]$CH_3CH_2CH_3$[/tex] (Propane).
