azeotrope boiling point question

[mytoechondriac]

From Kaplan FL:

When the azeotrope is at the boiling point, the average kinetic energy of the molecules in the liquid is:

B)less than the average kinetic energy of the gas molecules.
C)the same as the average kinetic energy of the gas molecules.

The answer is B.

I thought that because at the boiling point, the vapor pressure of the liquid is the same as the local atmospheric pressure, and because the gas and the liquid phases are in EQUILIBRIUM, the KE of the gas molecules would be the same as the KE of the liquid. Especially since EK Chem defines temperature as KE per mole. Don't the gas and the liquid of the azeotrope have the same temperature at the boiling point??

Please help me. I'm kind of stupid right now, especially since my brain cells are fried at this point.

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[xIcewind]

An azeotrope is defined as a substance whose composition (by mole fraction) is the same in vapour as in solution. Gas molecules will always have more KE than liquids, will it not? I'm going by logic/memory, but if you were to ask for the kinetic energy of a perfect crystalline solid and a gas at the same temperature, would they be different?