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Enthalpy Change of Formation?

I understand that the enthalpy change of formation is the formation of 1 mole of a compound from its constituent elements, with all substances in their standard states at 1 atmosphere (1 atm or 101.3 kPa).

The equation used to calculate the enthalpy of formation of water is:

H2(g)+1/2 O2(g)----------------> H2O(l)

I understand this equation as all the substances are in the standard states.

However I don't understand the equation:

H2(g)+1/2 O2(g)----------------> H2O(g)

As H2O isn't a gas in it's standard state.

Basically I don't understand how you can calculate the enthalpy change of formation for water vapour when H2O isn't a gas in it's standard state, yet the definition states that all substances must be in their standard states.

Can anyone please help me?

Reply 1

Freypops

Could you not work out the enthalpy change evaporating h2o and then Hess' law would state that the total enthalpy change should be the same?

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Reply 2

oversizedcarrot

Original post by _paul

I understand thatthe enthalpy change of formation is the formation of 1 mole of a compound from its constituent elements, with all substances in their standard states at 1 atmosphere (1 atm or 101.3 kPa).

The equation used to calculate the enthalpy of formation of water is:

H2(g)+1/2 O2(g)----------------> H2O(l)

I understand this equation as all the substances are in the standard states.

However I don't understand the equation:

H2(g)+1/2 O2(g)----------------> H2O(g)

As H2O isn't a gas in it's standard state.

Basically I don't understand how you can calculate the enthalpy change of formation for water vapour when H2O isn't a gas in it's standard state, yet the definition states that all substances must be in their standard states.

Can anyone please help me?

The change in state for the H2O from liquid to gas in the reaction doesn't matter, it'll just be the enthalpy of formation of gaseous H2O instead of liquid H2O that is factored into the equation you are trying to work out. Hess's law states that the enthalpy change of a reaction is independent of the route taken. Therefore we can use the enthalpy of formation to construct a Hess's law triangle to work out the enthalpy change for this reaction to form the product water from hydrogen gas and oxygen.

Say we create an overall reaction for the formation of water:
2H2(g) + O2(g) -----> 2H2O(g OR l)

Underneath that reaction we use enthalpy of formation to construct a hess's law triangle(imagine the central bold part connecting to the above reaction underlined).
So 2H2(g) + O2(g) <---------2H2(g) + O2(g) ------> 2H2O(g or l or s)
The part in bold is the enthalpy of formation reaction forming the products and reactants (or compounds in reaction). It's this bold part which is the enthalpy of formation which has to be in standard states or conditions, not the reactants and products underlined in the overall reaction. That's why the H2O can be in gas, liquid or even solid state.

By definition, the enthalpy of formation of H2 + O2 will be zero.
The enthalpy of formation of H2O will be different depending on which state it is in.

The enthalpy change of the reaction (ΔH°) = ΣΔH_f° (products) - ΣΔH_f° (reactants)