SL Bonding Practice Exam Question

[Cherries]

The graph below illustrates how the vapour pressures of ethoxyethane, CH3CH2OCH2CH3, benzene, C6H6, and water, H2O, change with temperature.

Using data from the graph, explain the difference in vapour pressure of ethoxyethane, benzene and water at 30 °C.

ANSWER: water has hydrogen bonding; benzene has van der Waals’/London/dispersion forces; ethoxyethane has dipole–dipole forces (and van der Waals’/London/ dispersion) but they are weaker than benzene

I understand the forces of each of these substances BUT how does it relate to their positions on the graph?

 

Thanks!

Edited May 12, 2018 by Cherries

[kw0573]

Yeah this is one of those questions where there's no way to know what they are looking for. The idea is that high vapour pressure means more volatile species and lower boiling point. Low boiling point due to weak intermolecular forces. So it wants you to point out that the dipole-dipole in ethoxyethane is weaker than vdW in benzene and both weaker than hydrogen bonding.