I'm trying to solve for the gibbs energy of vaporization of ethanol at a temperature lower than its boiling point. I am given the heat capacities of liquid and gaseous ethanol, the standard enthalpy of vaporization, and the standard boiling point of ethanol.

I start with the form:


Following integration, I get the form:

$\ce{\Delta H=\Delta H_{vap} +\Delta T(C_{p,g}-C_{p,l})}$

This form gives me the new enthalpy of vaporization at a non-standard temperature. However, When I try to enter this into the form:

$\ce{\Delta G=\Delta H - T\Delta S}$, with $\ce{\Delta S=\frac{\Delta H}{T}}$

This solution always results in an answer of zero (equilibrium). Intuitively, I understand that phase changes are typically reversible, however using another representation:

$\ce{\Delta G=-RTln(K)}$, with $\ce{K=\frac{P}{P_{0}}}$, the ratio of vapor pressure to standard pressure of the liquid (liquid activity = 1).

My question is, why do these formulas apparently conflate?


1 Answer 1


Alright, this is kind of fast but after thinking about it I think the issue with the problem is the definition of $\ce{\Delta S}$. My second try to compute this value involves starting with the form:

$\ce{dS = \frac{C_{p,g}-C_{p,l}}{T}*dT}$

Following integration, I arrive at the form:

$\ce{\Delta S = (C_{p,g}-C_{p,l})ln\frac{T}{T_b}+\frac{\Delta H}{T}}$ where $\ce{\Delta H}$ is determined in the way as in the question above.

Placing everything into the form $\ce{\Delta G = \Delta H-T\Delta S}$:

$\ce{\Delta G = \Delta H_{vap} +(T-T_b)(C_{p,g}-C_{p,l}) - T((C_{p,g}-C_{p,l})*ln\frac{T}{T_b}+\frac{\Delta H}{T})}$ where $\ce{\Delta H}$ is the non-standard enthalpy of vaporization.

  • $\begingroup$ Well, don't thank in posts or ask if they are ok. You can do it in comments, but as one can up/downvote answer or comment it then it could be said that all of them are waiting for review. As for your Q&A, for starters heat capacities don't change linearly. $\endgroup$
    – Mithoron
    Commented Mar 1, 2017 at 22:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.