8.67

Moderators: Chem_Mod, Chem_Admin

Ridhi Ravichandran 1E
Posts: 35
Joined: Sat Jul 22, 2017 3:01 am

8.67

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?

Thanks!

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.


Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

Who is online

Users browsing this forum: No registered users and 2 guests