4. The lower the intermolecular forces, the more easily a compound leaves the li

Question

4. The lower the intermolecular forces, the more easily a compound leaves the liquid phase, the faster the rate of evaporation, and thus the higher the vapor pressure. Rank the non-polar alkanes– pentane, hexane, heptane and octane - in terms of increasing vapor pressure. Explain your ranking in terms of the molar mass and intermolecular forces present.

5. Rank the polar molecules methanol, acetone and ethyl acetate in terms of increasing vapor pressure.   Explain your ranking in terms in terms of the intermolecular forces present, and the effects (if any) of molar mass.

6. How would you rank the relative strengths of the intermolecular forces present in the non-polarand polar molecules discussed in questions #4 and #5?

Explanation / Answer

4) AS the molar mass of a compound increases, the vander waals foreces of attraction also increase, thus increasing the intermolecular forces.

Arranging the molecules in increasing order of vapor presure is to arragne them in decreasing order of boiling point.

Of these octane will have the lowest vapor pressure and pentane has highest vapor pressure .

5) The increasing order of vapor pressure is ethylacetate, methanol , acetone.

Ethylacetate has highest molar mass of the three, thus less vanderwals foreces. However it bsing polar it also has stronger dipole-dipole interactions, which lower the vapor pressure further.

Methanol has lower molar mass than acetone still, has lower vapor pressure than former, due to the presence of strong intermolecular H-bonding present in it.

Acetone has highest vapor pressure as it has weakest inter molecular attaractions, onlu weak dipole-dipole interactions present.

6) The intermolecular forces discusssed here are dipole-dipole , vanderwaals' and H- onding.

Of these the strongest are H- bonded followed by dipole-dpole interactions and the weakest being vanderWaals' forces.

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