8.67

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Virpal Gill 1B
Posts: 32
Joined: Thu Jul 27, 2017 3:00 am

8.67

Postby Virpal Gill 1B » Mon Mar 12, 2018 10:15 am

The problem states: "Use the information in Tables 8.3, 8.6, and 8.7 to estimate the enthalpy of formation of each of the following compounds in the liquid state. The standard enthalpy of sublimation of carbon is 717 kJ/mol. (a) H2O; (b) methanol, CH3OH; (c) benzene, C6H6 (without resonance); (d) benzene, C6H6 (with resonance)." Is this problem assuming that each reaction is taking place at its boiling point because the solutions manual just subtracts the enthalpy of vaporization from the total calculated using bond enthalpies? If this is the case and the temperature was changed, how would this problem be solved?

Justin Lau 1D
Posts: 51
Joined: Sat Jul 22, 2017 3:00 am

Re: 8.67

Postby Justin Lau 1D » Mon Mar 12, 2018 11:15 am

The enthalpy off vaporization must be subtracted because you're looking for the liquid state. We know that all bond enthalpies are given in the gas state. So we need to account for the extra enthalpy gained during phase change.

McKenna disc 1C
Posts: 60
Joined: Fri Sep 29, 2017 7:04 am

Re: 8.67

Postby McKenna disc 1C » Wed Mar 14, 2018 2:56 pm

I was also wondering why we always subtract the enthalpy of vaporization. Some of the equations for the formation of the species have the species in its liquid state as a product; since we're concerned about the liquid state of the species, why do we still subtract the enthalpy of vaporization? And do we reverse the sign? I'm confused! Thanks :)

Kyung_Jin_Kim_1H
Posts: 53
Joined: Thu Jul 27, 2017 3:00 am

Re: 8.67

Postby Kyung_Jin_Kim_1H » Fri Mar 16, 2018 10:22 pm

deltaH(vap) gives you the heat for a substance going from liquid to gas (vaporization). In this case, we are going from gas to liquid, so we'll reverse the sign. Then we add the two deltaHs.

Think back to Hess's law problems-- we would make a positive enthalpy value negative (or vice versa) when we are reversing the reaction.


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