## 11.83 standard deltaG

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

cbercik
Posts: 30
Joined: Wed Sep 21, 2016 3:00 pm

### 11.83 standard deltaG

For 11.83, we are calculating deltaG for a reaction at two temperatures, 25C and 150C. For the first part, when T = 25C, I thought to use deltaG(rxn) = deltaG(products) - deltaG(reactants) with the standard free energies of formation in Appendix 2A because the temperature is at standard conditions. However, solutions say to use deltaG = deltaH - TdeltaS. Why do we need use this equation for the first part at standard temperature?

Dylan Davisson 2B
Posts: 50
Joined: Thu Jul 27, 2017 3:00 am
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### Re: 11.83 standard deltaG

Yes, you can use the deltaG(rxn) = deltaG(products) - deltaG(reactants) equation for the first part of the question because it is at 25 degrees Celsius. I think the only reason the book does it a different way is because it is more convenient when finding k at two different temperatures. They use this deltaG = deltaH - TdeltaS equation because in that equation all you would have to do to find a deltaG at a new temperature is plug in the new temperature, which is quicker to do because you have to use this equation anyways to solve the second half of the problem at 150 degrees Celsius.