Delta H of formation for diatomic molecules.

[wushuguy]

I just finished up my first class of chemistry, and have a question on something that I just can't figure out. The delta H of formation for diatomic molecules is 0. I just don't understand why this is. If there is a reaction, and O2 is one of the products, why is there no heat released when O2 forms? Doesn't each molecule of Oxygen go from a higher energy state to a lower energy state when O2 forms? Thanks in advance for any help.

 

[Fewmet]

I just finished up my first class of chemistry, and have a question on something that I just can't figure out. The delta H of formation for diatomic molecules is 0. I just don't understand why this is. If there is a reaction, and O2 is one of the products, why is there no heat released when O2 forms? Doesn't each molecule of Oxygen go from a higher energy state to a lower energy state when O2 forms? Thanks in advance for any help.

It's been a while since I have worked in this area, but my memory is that the naturally occurring elemental form is arbitrarily assigned. It has to do with the fact that you cannot measure the absolute heat content of a substance, so we work with relative sales. The zero of that relative scale is the naturally occurring (and therefore most stable) form of the substance.

 

[SpectraCat]

I just finished up my first class of chemistry, and have a question on something that I just can't figure out. The delta H of formation for diatomic molecules is 0. I just don't understand why this is. If there is a reaction, and O2 is one of the products, why is there no heat released when O2 forms? Doesn't each molecule of Oxygen go from a higher energy state to a lower energy state when O2 forms? Thanks in advance for any help.

There are several issues with your post:

First, the heat of formation is NOT zero for all diatomics, it is only zero for elements where the thermodynamic standard state is a hononuclear diatomic, like O2, N2 or H2. Furthermore, it is *defined* as zero, because the heat of formation of *any* element in its thermodynamic standard state is defined as zero. For a heteronuclear diatomic molecule like HF or CO, the heat of formation is definitely NOT zero.

Second, the heat of formation does NOT refer to just any old reaction where O2 is "formed" as a product. It has a very specific definition ... the heat of formation is *defined* as the heat released (or consumed) when a chemical species is formed from its component elements in their thermodynamic standard states.

I suggest you review this material in your book ... it is particularly important to make sure that you understand the definition of the thermodynamic standard state.

EDIT: Also, Fewmet's explanation of the need for defining a consistent scale against which to measure relative thermodynamic changes during chemical reactions is correct. The choice was made to set the heats of formations of elements in their standard states equal to zero, to give a reference point against which other thermodynamic changes could be measured. The absolute energy of a chemical compound (or atom) is harder to define, and chemical reactions always represent changes in bonding between atoms, keeping the total number of atoms conserved, so only a relative measure of the energy (or other thermodynamic variable) is required.

 

[wushuguy]

Wow. I was not aware of how unclear my understanding of this was. Thank you both for setting me straight.

 

[ZealScience]

Because ΔH formation is defined by formation of a compound from it's elemental form. For example the ΔH formation is the enthalpy change of the reaction of O2 + C → CO2. From the same definition ΔH formation of O2 is enthalpy change of O2 → O2, aparently 0.

If you are looking for heat released by forming O2, then it's called "ΔH of atomization" which is the enthalpy change of 2O → O2, but it is not the ΔH formation. Thus no need for taking care of it.

 

[Zeppos10]

If you like, you can also use absolute enthalpy values. To do so you have to reason as follows: 1. start with hydrogen. 2. assume an ideal gas consisting of H-atoms at stp. 3. From the kinetic theory H=(5/2)RT. 4. Using the H of atomization (see ZealScience #5)you can calculate the Habs for H2 = H-molecule. 5. Likewise for other elements.
But now you have to distinguish 'absolute enthalpy'from 'enthalpy of formation'.
All reaction enthalpies are not affected ofcourse.
[cf Spencer ,1996, J Chem Edu 73:631-6]