## ΔG= ΔH -TΔS

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

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

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

Tamera Scott 1G
Posts: 65
Joined: Fri Sep 28, 2018 12:27 am

### ΔG= ΔH -TΔS

Should you convert all the units into kJ/mol in the beginning of the problem or can you wait until the end when solving for ΔG with ΔG= ΔH -TΔS?

Camille Marangi 2E
Posts: 60
Joined: Fri Sep 28, 2018 12:26 am

### Re: ΔG= ΔH -TΔS

As long as both the enthalpy and entropy are joules/mol, you can wait until after calculations to convert to KJ/mol.

Vincent Li 4L
Posts: 48
Joined: Fri Sep 28, 2018 12:19 am

### Re: ΔG= ΔH -TΔS

In most cases, if you have to calculate change in entropy and change in enthalpy separately before using the Gibbs Free Equation, entropy tends to be in J/mol whereas enthalpy tends to be in kJ/mol. When calculating each of the constituent thermodynamic values, it doesn't really matter if it's in J/mol or kJ/mol as long as it's consistent. When using the actual delta G equation, it is important, however, to either convert both to kJ/mol or J/mol before getting the final answer.

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