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Ridhi Ravichandran 1E
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Joined: Sat Jul 22, 2017 3:01 am


Postby Ridhi Ravichandran 1E » Mon Jan 22, 2018 1:49 am

For part b on 8.67, why do you have to include the heat released when methanol changes from a gas to a liquid? Isn't it already in the liquid phase? According to the solutions manual, the equation is C(gr) + 2H2 + (1/2)O2 ----> CH3OH(l). Doesn't this indicate that methanol is already a liquid?

Also, for c and d, are we expected to know what the structures look like with or without resonance?


Jonathan Tangonan 1E
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

Re: 8.67

Postby Jonathan Tangonan 1E » Mon Jan 22, 2018 9:21 am

Yes it does indicated that methanol is in a liquid phase, but bond enthalpies refer to gaseous substances. So any substance in a chemical reaction that isn't in a gaseous phase has to be converted from their previous phase into the gaseous phase. So what we are actually looking at is:

C(g) + 2H2 + ½ O2(g) -> CH3OH (g)

Then it goes from the gaseous phase to the liquid phase

CH3OH (g) -> CH3OH (l) so you would also have to take into account the negative standard enthalpy of vaporization for CH3OH because it's going from the gaseous phase to the liquid phase.

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