[Enthalpy]

Hello nice people!

To compute the heat of formation of a solution of NO in N₂O₄, I have already the heat of formation of NO, N₂O₄, and still need the heat of dissolution.
(A) Would you know it? Or have some better method? Ask the solvation army maybe?

Yes, it's a rocket propellant, how did you guess...? 33%wt NO lowers the freezing point from -10°C down to -107°C, especially useful on Mars. Even better, it makes the solution endothermic, enabling my seducing scheme for pumping propellants, described there
http://saposjoint.net/Forum/viewtopic.php?f=66&t=2272#p27535

To compute the heat of dissolution, I decided to use the available pressure-temperature curve of NO in N₂O₄ exactly as is done with liquid-vapour equilibrium when inferring the heat of vaporization, by telling that P varies like exp(-H/RT).
http://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation#Applications
(B) Do you agree?

For instance at 30%wt between 298K and 328K, gas pressure increases from 7.3 to 22.4 atm, leading to 30.4kJ/mol heat of dissolution.
(C) Could you kindly check?

I feel this is a lot. Vaporization of water takes 40.5kJ/mol and of NO 13.8kJ/mol... But maybe this explains why NO dissolves in N₂O₄ while NO doesn't liquefy at 3.33 times that pressure.
(D) Do you feel 30.4kJ/mol is credible?

Comments, suggestions welcome!
Marc Schaefer, aka Enthalpy