Question

Place the following compounds in order of increasing vapor pressure. methane, CH4; pentane, CH3CH2CH2CH2CH3; 1-propanol, CH3CH2CH2OH; propanone, CH3COCH3

Answer 1

Explanation:

A compound that contains more stronger bonds will need more amount of heat in order to break the bonds so that it changes into vapor state.

In 1-propanol, there is hydrogen bonding and it is stronger in nature. As a result, more amount of heat is required to break the bonds between molecules of 1-propanol.

Whereas in propanone, there will be dipole-dipole interactions which are less stronger than hydrogen bonding. Hence, propanone molecule will need less amount of energy than 1-propanol.

On the other hand, pentane will need more amount of heat as it has longer chain of carbon atoms as compared to methane.

Thus, we can conclude that given compounds are arranged in order of increasing vapor pressure as follows.

         methane < pentane < propanone < 1-propanol

Answer 2

Answer:

1-propanol --> propanone --> pentane --> methane

Explanation:

Hello,

Vapor pressure is referred to the pressure exerted by the vapor that in equilibrium with a substance liquid. In organic compounds, such vapor pressure is higher as the number of carbons is lower since the intermolecular forces become weaker, nevertheless, the presence of functional groups modify this behavior by means of the intermolecular forces magnitudes (weaker intermolecular forces imply higher vapor pressures as the particles tend to separate to each other), thus, by searching on the NIST webpage for their vapor pressures at 25°C, one finds:

$Methane=large\\1-propanol=0.0282 bar\\Propanone=0.306 bar\\Pentane=0.683 bar$

In such a way, the lowest vapor pressure belongs to 1-propanol, next propanone, then pentane and finally methane.

Best regards.