# How to calculate ∆rGº from entropy and ∆fGº in different temperatures? [closed]

"The total oxidation of glucose occurs according to the following chemical equation:

C6H12O6 (s) + 6O2 (g) -> 6CO2 (g) + 6H2O (l)

The following table gives us the free energies of standard formation and the standard molar entropies of compounds involved in the previous reaction.

Compound ∆fGº(298 K) kJ/mol Smº (298 K) J/K/mol
C6H12O6 -917,2 212,10
O2 0 205,14
CO2 -394,36 213,14
H20 -273,13 69,91

Based on the previous data, determine the ∆rGº for glucose oxidation at 308 K"

It's a question of a doctoral selection process and I want to figure out if it is formulated wrong or if there is a way to an answer.

• Yes, it is answerable. From the individual free energies and entropies of formation, you can get delta_G and delta_S for the overall reaction at 298K, from which you can then get delta_H (from delta_G=delta_H – T delta_S). delta_H in turn gives you the temperature-dependence of the reaction, from which you can get delta_G at 308K. Now that you have a roadmap, try working through as much as possible on your own. – theorist Dec 15 '20 at 4:35
• Should not it be rather $\Delta G^{\circ}_f$ and $\Delta G^{\circ}_r$ ? – Poutnik Dec 15 '20 at 6:31

For a small temperature change like that, you can make use of dG=-SdT. Determine $$\Delta rG^0$$ and $$\Delta r S^0$$ at 298, multiply the latter by 10 , and subtract it from the former to get $$\Delta rG^0$$ at 308K.