## enthalpy of formation vs standard reaction enthalpy

$\Delta G^{\circ} = -nFE_{cell}^{\circ}$

inlovewithchemistry
Posts: 104
Joined: Fri Sep 28, 2018 12:19 am

### enthalpy of formation vs standard reaction enthalpy

when comparing 4J5 and 4J7, why can you use the enthalpy of formation value for 4J5, but are required to calculate the enthalpy of the reaction for 4J7?
is it because there are 2H2O2 in 4J7? how can both be plugged into the formula for delta G?

Philip Lee 1L
Posts: 30
Joined: Fri Sep 28, 2018 12:21 am

### Re: enthalpy of formation vs standard reaction enthalpy

This is because the reactions in 4J5 are formation reactions while the reactions in 4J7 are not formation reactions.

A formation reaction is a reaction where a species is synthesized from its elements in their most stable forms. For example, the formation reaction of NH3(g) in 4J5 (1/2)N2(g) + (3/2)H2(g) --> NH3(g). Notice that there are fractional coefficients in the formation reaction. This is to ensure that the product of the formation reaction is exactly one mole of the species of interest.

For 4J7, the reactions are NOT formation reactions. However, the enthalpy of the reaction can be calculated with the enthalpies of formation of each species participating in the reaction.

The change in enthalpy value that is plugged into the formula for delta G must be the change in enthalpy for the entire reaction. For 4J5, the reactions that are being considered are formation reactions, so the enthalpy of formation represents the enthalpy change for the entire formation reaction, and the enthalpy of formation can be plugged into the formula for delta G. For 4J7, you cannot just plug enthalpies of formation into the formula for delta G because those do not represent the entire reaction. You must use enthalpies of formation to calculate the enthalpy of formation for the entire reaction and then plug the enthalpy of formation for the entire reaction into the formula for delta G.