Vander Waal;s forces

• hoay

  24 May 15, 04:20

  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

  24 May 15, 10:24

  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

  24 May 15, 12:40

  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).