## Calculating G if H and S are gven

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

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

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

Kimberly 1H
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Joined: Fri Sep 28, 2018 12:17 am
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### Calculating G if H and S are gven

Can someone please explain why we assume $\Delta$G is 0 when H and S are given?

whitney_2C
Posts: 74
Joined: Fri Sep 28, 2018 12:28 am

### Re: Calculating G if H and S are gven

We aren't assuming that the Gibbs Free Energy is equal to 0, rather we are putting in 0 for its value to see when $T\Delta S$ and $\Delta H$ are equal to one another in order to determine at what temperatures the Gibbs Free Energy would be negative (spontaneous reaction) or positive (non spontaneous reaction).

005199302
Posts: 108
Joined: Fri Sep 28, 2018 12:15 am

### Re: Calculating G if H and S are gven

If we have the value when the system is in equilibrium and delta G =0, then we can find the behavior (in terms of spontaneity) for any temperatures above or below that value.

annabel 2A
Posts: 67
Joined: Fri Sep 28, 2018 12:18 am

### Re: Calculating G if H and S are gven

Why can we use the same deltaH and deltaS to calculate deltaG at any temperature? (like in this problem: 9.53 Calculate the change in molar Gibbs free energy for the process NH3(l) --> NH3 (g) at 1atm and (a) 15.0 C; (b) 45. C (see Tables 8.3 and 9.1 for standard enthalpy and entropy of vaporization))

I thought that deltaH and deltaS would be different for different temperatures? Are the changes in enthalpy and entropy for a phase change the same at any temperature?