Enthelpy change of a vaporization rxn

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Discussion Overview

The discussion revolves around the calculation of thermodynamic quantities (q, w, delta H, and delta U) for a vaporization reaction involving a liquid with a known standard enthalpy of vaporization. The context includes homework-related inquiries about the assumptions and principles governing these calculations at a specified temperature and pressure.

Discussion Character

  • Homework-related, Technical explanation

Main Points Raised

  • One participant expresses confusion about the assumption that q equals delta H at constant pressure and questions how to determine when this assumption is valid.
  • Another participant explains that work is nonzero due to the volume increase during vaporization, highlighting the distinction between energy and enthalpy.
  • A third participant seeks clarification on how to calculate delta H, indicating they can handle the other calculations.
  • A later reply provides a brief explanation that delta H is the enthalpy of vaporization adjusted for the number of moles involved.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the assumptions regarding q and delta H, and there is ongoing clarification about the calculations involved, indicating that multiple views and uncertainties remain in the discussion.

Contextual Notes

Unresolved aspects include the specific conditions under which q equals delta H and the implications of work being nonzero in this context. The discussion also reflects varying levels of understanding among participants regarding thermodynamic principles.

yungwun22
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Homework Statement



A certain liquid has a standard enthalpy of vaporization of 32 KJ/mol. Calculate q, w, delta H and delta U when 0.75 mol is vaporized at 260 K and 765 Torr.



Homework Equations





The Attempt at a Solution



I was able to get delta H, but the solution states that q is equal to delta H. If delta H is equal to q at constant pressure, I don't know how I'm supposed to know when to make that assumption. Also, why does work not equal zero if the pressure is constant, is this additional work?
 
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The amount of work is nonzero because the volume increases and the system therefore does work on its surroundings. This is the big difference between energy and enthalpy; enthalpy takes into account that a system's volume tends to vary during processes such as heating and vaporization.
 
I have a similar question, just wondering how you solved ∆H? I can do the rest.

Thanks
 
Hi hinglis, welcome to PF. ∆H is the enthalpy of vaporization, adjusted to the number of moles in the system.
 

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