# Does Gibbs free energy of formation always equal zero for elements in standard state?

I was looking through the appendix in my chemistry textbook when I noticed that every compound that had a $\Delta H_f$ of zero also had a $\Delta G_f$ of zero. So of course the compounds that have a $\Delta H_f$ of zero are the elements in their standard states but why does that make their $\Delta G_f$ zero? Does every compound that has a $\Delta H_f$ of zero also have a $\Delta G_f$ of zero? What exactly is $\Delta G_f$?

• Do you know about relationship between $\Delta G$ and $\Delta H$? – Freddy Nov 23 '15 at 8:28
• Yeah $\Delta G$ = $\Delta H$ - T $\Delta S$ where T is the temperature in Kelvins and S is entropy. – Lubed Up Slug Nov 23 '15 at 8:41
• so from that relation, $\Delta H = 0$ does not always mean $\Delta G$ is not equal to 0. – Freddy Nov 23 '15 at 8:54
• Well isn't that a little different from what I am describing? That is just $\Delta H$ for a reaction or physical process not the $\Delta H_f$ of a compound right? – Lubed Up Slug Nov 23 '15 at 9:01
• If you look more closely through your textbook, you will find it pointed out that the heats of formation and free energies of formation of all elements in their natural states at 298 and 1 bar are taken to be zero by convention. This in no way limits your ability to use these tables to determine the changes in enthalpy and free energy for arbitrary chemical reactions involving compounds comprised of these elements. – Chet Miller Nov 23 '15 at 12:54

Does Gibbs free energy of formation always equal zero for elements in standard state?

No. Elements occur in different allotropes.

$\Delta H_f^o$ and $\Delta G_f^o$ are defined to be zero at 298K, 1 bar for the lowest energy allotrope, with the exception that the values for white phosphorous are defined to be zero even though it is not the lowest energy allotrope.

The other allotropes do not have zero $\Delta H_f^o$ and $\Delta G_f^o$ even though they are still elements.

So of course the compounds that have a $\Delta H_f$ of zero are the elements in their standard states but why does that make their $\Delta G_f$ zero?

$\Delta H_f = 0$ does not imply $\Delta G_f = 0$

The two values are independently defined to be zero for a particular allotrope as explained above.

Does every compound that has a $\Delta H_f$ of zero also have a $\Delta G_f$ of zero?

If a compound (other than by definition) happens to have $\Delta H_f = 0$, then $\Delta G_f$ could be any value, it is not necessarily zero.

What exactly is $\Delta G_f$

It is the change in Gibbs energy upon forming a compound in its standard state from its constituent elements in their standard states at 298K.

• For allotropes of elements which have both $\Delta G_f$ and $\Delta H_f$=0, is $\Delta S_f$ also zero? I have never seen this hold. – zed111 Sep 20 '16 at 5:21

If you look more closely through your textbook, you will find it pointed out that the heats of formation and free energies of formation of all elements in their natural states at 298 and 1 bar are taken to be zero by convention. This in no way limits your ability to use these tables to determine the changes in enthalpy and free energy for arbitrary chemical reactions involving compounds comprised of these elements.