## 9.91

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

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

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

ClaireHW
Posts: 60
Joined: Fri Sep 29, 2017 7:07 am
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### 9.91

Using values in Appendix 2A, calculate the standard Gibbs free energy for the vaporization of water at 25, 100, and 150 degrees C.
b) what should the value at 100 degrees C be?
c) Why is there a discrepancy?

(Claire Woolson Dis 1K)

Isabella Sanzi 2E
Posts: 54
Joined: Thu Jul 13, 2017 3:00 am

### Re: 9.91

So unless you have a more specific question, I can just go through the methodology of this problem.
For part a, you have to calculate the delta H and delta S for the vaporization of H2O (reaction: H2O(l) --> H2O(g)). To find delta H, use the enthalpy of formation values at 25 degrees C. Then do the same for the molar entropy values to find delta S. And then you can calculate delta G with the delta S and H you just solved for and the temperature using the equation delta G = delta H - T*delta S. There you have the values for 25 degrees C. To find the delta G for 100 and 150 degrees C, just plug in the temperature in the equation (delta G = delta H - T*delta S) using the values for delta S and delta H you solved for previously.
For part b, we would expect the value would be 0 since that is the boiling point of water.
For part c, there is a discrepancy because the enthalpy and entropy values are not consistent with temperature, so you would probably get more exact values if you used the actual enthalpy and entropy of vaporization from that specific temperature.
I hope this helps!