I was looking at the enthalpy change for water-splitting reaction:
$$ \Delta H^o_R = [\Delta H^0_{H_2(g)} +\frac{1}{2}\Delta H^0_{O_2(g)}]-\Delta H^0_{H_2O(l)} = 285.83 kJ/mol$$
According to the book "Thermal physics"by Schroeder; at constant T and P; if there are no other forms of work on the system besides compression/expansion. Then, then $\Delta H^o_R = Q$ and if we have. However, when there are other forms of work being done we then have $\Delta Gº_R \leq W_{other}$ and $\Delta H^o_R = Q + W_{other}$. Does The value for the Gibbs free energy in this mean itcase is definitely not possible$ \Delta Gº_R = 237 kJ/mol $. We can relate $\Delta G$ and $\Delta H$ by $\Delta G =\Delta H -T\Delta S$ .
My confusion arises first from reading that the gibbs free energy is the work we need to changedrive the enthalpy of a reaction purely by compression/expansion, say electrical work ? Is. However, the enthalpy change inshows that the energy required could be done via heat and/or another form of work is higher than that of the Gibbs free energy? My guess so far is zerothat we can take some energy from the environment for "free", but what happens when enthalpy iswe drive this reaction only changed by heat? Ifvia heat and say electrical work are used, then G is the minimum worksuch that needs to be supplied by the electric work$\Delta H = Q$, would $\Delta G = 0 $ ?