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Multiple Choice
Rank the following compounds in order of decreasing vapor pressure at room temperature: CH3CH2OH, CH3OCH3, CH3CH2CH3.
A
CH3OCH3 > CH3CH2CH3 > CH3CH2OH
B
CH3CH2CH3 > CH3OCH3 > CH3CH2OH
C
CH3CH2OH > CH3CH2CH3 > CH3OCH3
D
CH3CH2OH > CH3OCH3 > CH3CH2CH3
Verified step by step guidance
1
Identify the intermolecular forces present in each compound, as vapor pressure is inversely related to the strength of these forces. Stronger intermolecular forces result in lower vapor pressure.
For CH3CH2OH (ethanol), recognize that it can form hydrogen bonds due to the -OH group, which are strong intermolecular forces.
For CH3OCH3 (dimethyl ether), note that it has dipole-dipole interactions because of the polar C-O-C bond but cannot hydrogen bond as effectively as ethanol.
For CH3CH2CH3 (propane), understand that it is a nonpolar molecule with only London dispersion forces, which are the weakest intermolecular forces among the three.
Rank the compounds by comparing the strength of their intermolecular forces: weakest forces correspond to highest vapor pressure, so the order is propane (CH3CH2CH3) > dimethyl ether (CH3OCH3) > ethanol (CH3CH2OH).