Isobaric Process finding G using (deltaG-deltaGo)

Click For Summary
SUMMARY

The discussion focuses on the calculation of the change in Gibbs free energy (ΔG) for biological processes at body temperature (37°C) compared to standard conditions (25°C). It clarifies that ΔG(37°C) – ΔG(25°C) can be approximated using the equation ΔG(37°C) – ΔG(25°C) ≈ – (T – 298) (ΔS(25°C)), where ΔS(25°C) is assumed constant. The conversation also highlights that the differential change in standard free energy (dΔG) is influenced by temperature, allowing for integration to express ΔG as a function of temperature.

PREREQUISITES
  • Understanding of Gibbs free energy and its significance in thermodynamics.
  • Familiarity with the concept of entropy (ΔS) and its role in chemical reactions.
  • Basic knowledge of temperature effects on thermodynamic properties.
  • Ability to interpret differential equations in the context of physical chemistry.
NEXT STEPS
  • Study the derivation of the Gibbs-Helmholtz equation for a deeper understanding of ΔG.
  • Learn about the relationship between entropy and temperature in thermodynamic processes.
  • Explore the implications of temperature on reaction spontaneity and equilibrium.
  • Investigate the role of standard state conditions in thermodynamic calculations.
USEFUL FOR

Chemistry students, biochemists, and anyone studying thermodynamics in biological systems will benefit from this discussion, particularly those interested in the effects of temperature on Gibbs free energy.

mikece
Messages
1
Reaction score
0
Can anyone help me with understanding what is going on at page 11 of these notes?

http://www.chem.utoronto.ca/coursenotes/CHM223/Section%205B%20Fall%202010.pdf

Do you get why there is 2 delta Gs? I am really confused with what he
was trying to do there?
ΔG(37oC) – ΔG(25oC) ≈ – (T – 298) (ΔS(25oC))

Thanks
For biological processes at body temperature (37oC = 98.6oF)
ΔS(37oC) ≈ ΔS(25oC) ≈ constant
And thus:
ΔG(37oC) – ΔG(25oC) ≈ – (T – 298) (ΔS(25oC))
(b) dG is an exact differential
 
Science news on Phys.org
For some specified chemical reaction, the standard free energy change ##Delta G^0## is a function of temperature ##\Delta G^0(T)##. We know that the effect of temperature on ##Delta G^0## is given by $$d{\Delta G^0}=\Delta S^0dT$$. If ##\Delta S^0## is nearly constant over the temperature range of interest (between 298 K and T), then we can integrate to get $$\Delta G^0(T)-\Delta G^0(298)=\Delta S^0(298)(T-298)$$