Vander Waal;s forces

hoay SG

In which change would only van der Waals’ forces have to be overcome?

A evaporation of ethanol C2H5OH(l) → C2H5OH(g)

B melting of ice H2O(s) → H2O(l)

C melting of solid carbon dioxide CO2(s) → CO2(l)

D solidification of butane C4H10(l) → C4H10(s)

In Both A amd B hydrogen bonding had to beovercome.

Now as for D the it is a condensation process so molecules have to come closer and the VDW forces must be lost or they would just lose kinetic energy i am not sure?..... In C (and in B also) melting of solid is occurring but are VDW forces present ins olid ice?

Metanoia SG

My take.

Options A, B and C involves overcoming of the intermolecular attraction forces, as they result in increased distances among the molecules.

Option A: VDW & H-bonding

Option B: VDW & H-bonding

Option C: VDW only.

UltimaOnline SG

Metanoia is correct. Van der Waals forces are always present regardless of proticity or polarity of the species.

For option D "solidification of butane", van der Waals forces are formed and/or strengthened (instead of overcome) during condensation and/or deposition, and kinetic energy is lost as heat (hence bond forming of any type is exothermic).