$\ce{N2H4}$ has a boiling point of 114 °C, which is higher than the boiling point of water. Why is this so?

Both $\ce{N2H4}$ and $\ce{H2O}$ have two lone pairs, so the number of hydrogen bonds per molecule seems to be the same. Also, the $\ce{O...H}$ hydrogen bond (21 kJ/mol) is stronger than the $\ce{N...H}$ hydrogen bond (13 kJ/mol). So, given this information, why is it that $\ce{N2H4}$ has a higher boiling point?

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    $\begingroup$ Maybe partially due to the molecular weight difference? $\endgroup$ – Ed V Mar 3 '20 at 21:04
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    $\begingroup$ H2O isn't fully using it's lone pairs, some O atoms are getting only one, not two H-bonds $\endgroup$ – Mithoron Mar 3 '20 at 21:16
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    $\begingroup$ The reason is in the molar masses. For N2H4 it is 32 g/mol. For H2O, it is 18 g/mol. Heavier moles must get more energy for passing in the vapor phase. $\endgroup$ – Maurice Mar 3 '20 at 21:19
  • $\begingroup$ Being a larger molecule, i suspect you can excite more vibrational modes before the hydrogen bonds break. $\endgroup$ – Karl Mar 3 '20 at 23:04

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