$$\mathrm dG < V\,\mathrm dp - S\,\mathrm dT$$
For a process at constant pressure and temperature ($\mathrm dp = \mathrm dT = 0$), and in the absence of any non-pV work, $\mathrm dG < 0$. Since energy cannot be created or destroyed (and Gibbs free energy is a measure of energy) where does the energy go when this spontaneous change occurs? Likewise for internal energy: $\mathrm dU = T\,\mathrm dS - p\,\mathrm dV$?
Also, why is this even useful? I understand that it's easy to hold temperature and pressure constant in a lab but we only seem to measure enthalpy anyway using $\mathrm dH = C_p\,\mathrm dT$. If it's so easy to keep temperature and pressure constant yet so hard to keep entropy constant then why do we measure enthalpy (which decreases for a spontaneous change when entropy and pressure are held constant) rather than Gibbs free energy. I suspect I am barking up the wrong tree and I am hoping that you'll be able to put me right; the only thought that I had was that the fact that Gibbs free energy muct decrease for a spontaneous change would help to predict whether a spontaneous change would occur when done in a lab - given that $p$ and $T$ will be constant; is this right?
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) is technically incorrect: there is a guide on meta, see e.g. chemistry.meta.stackexchange.com/q/443/16683 $\endgroup$