# Understanding the derivation of Gibbs Free Energy

In the equation: $$ΔG=ΔH−TΔS$$

I am having trouble understanding my professor's derivation of it from $$ΔS_{\mathrm{univ}}=ΔS_{\mathrm{sys}}+ΔS_{\mathrm{surroundings}}$$

where he says $$ΔS_{\mathrm{surroundings}}= \frac{-ΔH_{\mathrm{sys}}}{T}$$ which I have learned is only true for a reversible process. How is it possible that this heat can be reversed? And why can't we say that the change of entropy for the system is also equal to the equation. I understand that it would mean the reaction is at equilibrium, but would it not be since heat is reversible? $$ΔS_{\mathrm{sys}}= \frac{ΔH_{\mathrm{sys}}}{T}$$

• It's a simplification.
– Zhe
Nov 15, 2017 at 12:55
• Could you give a more in-depth explanation of how this simplification makes sense conceptually? Nov 18, 2017 at 23:45
• Well, you take $\approx$ and treat it like $=$. I'm not sure what you're asking...
– Zhe
Nov 19, 2017 at 18:29

The surroundings are treated as a constant volume reservoir, so the first law of thermodynamics, $$\mathrm{d}U_\mathrm{surroundings} = \mathrm{d}q_\mathrm{surroundings} + \mathrm{d}w_\mathrm{surroundings},$$ yields $$\mathrm{d}U_\mathrm{surroundings} = \mathrm{d}q_\mathrm{surroundings} \text{, or } \Delta U_\mathrm{surroundings} = q_\mathrm{surroundings}.$$ Notice that $q_\mathrm{surroundings}$ is equal to a state function, so it behaves as one, that is, it is independent whether the process is reversible or not. If the process is reversible $$q_\mathrm{rev,surroundings} = \Delta H_\mathrm{surroundings}, \tag1\label{eq:rev}$$ if it is not reversible $$q_\mathrm{surroundings} = \Delta H_\mathrm{surroundings}, \tag2\label{eq:irrev}$$ from \eqref{eq:rev} and \eqref{eq:irrev} follows $$q_\mathrm{surroundings} = q_\mathrm{rev,surroundings}.$$ Therefore $$\Delta S_\mathrm{surroundings} = \frac{q_\mathrm{rev,surroundings}}{T},$$ in particular $$q_\mathrm{rev,surroundings} = -q_\mathrm{sys},$$ then $$\Delta S_\mathrm{surroundings} = \frac{-q_\mathrm{sys}}{T}.$$ If the process occurs at constant pressure $$q_\mathrm{sys} = \Delta H_\mathrm{sys},$$ therefore $$\Delta S_\mathrm{surroundings} = \frac{-\Delta H_\mathrm{sys}}{T}.$$

You are right. Only when the process is reversible we have:

$$dS = \frac{dQ} {T}$$

For an irreversible process we have instead:

$$dS > \frac{dQ} {T}$$

This is the Clausius inequality and when replacing $$dQ$$ with its expression derived from the first law ($$dU = dQ - PdV$$; assuming only PV work) we obtain a general criterion for spontaneous change:

$$dU + PdV - TdS < 0$$

The Gibbs energy is the special case of this general criterion when $$P$$ and $$T$$ are held constant. In such case:

$$dU + PdV - TdS <0 \implies d(U+PV-TS) < 0 \implies d(H-TS)<0 \implies dG<0$$

Other expressions (such as the Helmholtz energy) are derived from the same general criterion.