Standard State of Compounds

[KDang_1D]

Our TA explained that substances in their standard state have a standard enthalpy of formation of 0.

In lecture, Dr. Lavelle did an example in which we calculated the standard enthalpy of the combustion of methane. He stated that gases are in their standard state at 1atm and pure liquids are always in their standard state. Given this, why did methane, H2O, and CO2 have non-zero values for their individual standard enthalpies of formation?

[Sebastian Lee 1L]

I think what your TA meant is that substances in their standard state in their most stable form have a standard enthalpy of formation of 0. The most stable form of an element is one that only contains that element in its natural state. For example, oxygen's most stable form is O2 gas so the enthalpy of formation for O2 is 0.

A substance must be in its standard state (ie a gas at 1 atm) in order for you to use standard enthalpies of formation in the first place. In the problem you give, H2O contains hydrogen and oxygen and is thus made from their stable forms hydrogen gas and oxygen gas. However, combining hydrogen gas and oxygen gas to create water has a certain enthalpy which is the enthalpy of formation given in a table. The standard enthalpy of formation of H2O is not 0 because hydrogen and oxygen do not exist as water in its most pure, stable form.

[BAlvarado_1L]

these compounds are not in their standard state, for example- C(s, graphite) + O2(g) → CO2(g)
C(s) and O2(g) are their standard states with standard enthalpies of formation=0