Isobaric Process finding G using (deltaG-deltaGo)

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Can anyone help me with understanding what is going on at page 11 of these notes?

http://www.chem.utoronto.ca/coursenotes/CHM223/Section 5B Fall 2010.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
 

Answers and Replies

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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 $$\Delta G^0(T)-\Delta G^0(298)=\Delta S^0(298)(T-298)$$
 

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