## Why is deltaG of formation 0 for diatomic molecules?

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

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

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

sofiakavanaugh
Posts: 58
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

### Why is deltaG of formation 0 for diatomic molecules?

Why is deltaG(formation) zero for diatomic molecules again? and same with enthalpy values? (But not for entropy values?)

Vincent Tse 1K
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am
Been upvoted: 1 time

### Re: Why is deltaG of formation 0 for diatomic molecules?

Heat of formation of a certain element is defined to be 0 for that element in its standard state (which is usually the state it is found in naturally, the diatomic one).

I suppose ΔG has a similar definition and entropy doesn't, but I'm not too entirely sure about those.

Jason Liu 1C
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am

### Re: Why is deltaG of formation 0 for diatomic molecules?

ΔS isn't 0 for diatomic molecules because there is always entropy. S is only 0 when T = 0K.

Anna Li 2E
Posts: 21
Joined: Sat Jul 22, 2017 3:00 am

### Re: Why is deltaG of formation 0 for diatomic molecules?

The heat of formation is 0 only for certain diatomic molecules. This occurs for the elements where their natural/most common state is a diatomic. These elements are H2, N2, F2, O2, L2, Cl2, Br2.

This is due to the fact that standard enthalpy of formation is defined as enthalpy change of a substance when it is formed from the most stable form of itself in standard condition (P=1 bar and T=25C). As these diatomics are already in their most stable form, their standard enthalpy of formation is zero. This is also true for Gibbs free energy (where G=0) as each diatomic molecule is in its standard state.