Van't Hoff Approach to Finding ∆H and ∆S

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SUMMARY

The discussion centers on the application of the Van't Hoff equation to determine the standard enthalpy change (∆H) and entropy change (∆S) for a reaction. The equation incorporates ∆Ho, which represents the enthalpy change for one mole of reactant at 298K. It is clarified that T1 does not always have to be 298K; however, to calculate ∆Ho, one must input 298K for T1 and the corresponding equilibrium constant (K) for that temperature. The assumption that ∆H remains constant with temperature is noted as a limitation of this approach.

PREREQUISITES
  • Understanding of the Van't Hoff equation
  • Knowledge of thermodynamic concepts such as enthalpy (∆H) and entropy (∆S)
  • Familiarity with the concept of equilibrium constants (K)
  • Basic calculus for integration and differentiation
NEXT STEPS
  • Study the derivation of the Van't Hoff equation in detail
  • Learn about the relationship between temperature and equilibrium constants
  • Explore the concept of standard enthalpy changes in thermodynamics
  • Investigate the limitations of assuming constant ∆H with temperature changes
USEFUL FOR

Chemistry students, researchers in thermodynamics, and professionals involved in chemical kinetics and equilibrium studies will benefit from this discussion.

ampakine
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Heres the question:
[PLAIN]http://img194.imageshack.us/img194/9042/vanthoff.png
and here's the Van't Hoff equation (the form I like to use):
[PLAIN]http://img703.imageshack.us/img703/1577/vtequation.png
I'm a bit confused about what's going on here. I see the formula includes ∆Ho, meaning the enthalpy change for 1 mole of reactant at 298K. Does that mean to use this formula T1 always has to be 298K? In other words to get the ∆Ho for the reaction would I just plug 298K into T1 and 4.25 x 10-7 into KT1 then plug any other pair of values from that table into T2 and KT2?
 
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The ∆H is always assumed to be constant with temp in this equation (standard enthalpy change), which is an obvious downside to the relation. If you derive it from
(d ln(K)) / (d 1/T) = -∆Ho/R you will find that from the integration.
 

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