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Usually change is indicated as the difference between some initial and final state. However, for ΔG there seem to be multiple definitions and I am a bit confused. For example, at equilibrium, ΔG = 0 but I am confused as to what the initial and final state between which this 'change' is measured since equilibrium is just a single point. Could someone please explain exactly what this change is measured for? I am only in high school so I haven't learnt more complex thermodynamics so I would appreciate it if it's possible to include a relatively basic explanation.

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Planckton seems to forget that the Gibbs energy is a value that changes with concentration. Let's consider a reaction $\ce{A + B <=> C + D}$. In their pure state, the Gibbs energies of formation of $\ce{A, B, C}$ and $\ce{D}$ have the values $\pu{G°(A), G°(B), G°(C), G°(D)}$, resspectively.

Let's start from $1$ mole of the pure substances $\ce{A}$ and $\ce{B}$, without any $\ce{C}$ and $\ce{D}$. The reactants are not yet mixed. Their total free energies are $\ce{G°(A) + G°(B)}$. In a first step, these substances must be mixed, even without any reaction. This simple mixture changes their Gibbs energies, because this mixture is spontaneous and not reversible. In the mixture, before the reaction starts, the Gibbs energies of $\pu{A}$ and $\pu{B}$ are $$\pu{G(A)_{t=0}} < \pu{G°(A)}$$ $$\pu{G(B)_{t=0}} < \pu{G°(B)}$$. At this point, there is no $\ce{C}$ and no $\ce{D}$ in the mixture, and their Gibbs energies are zero. $$\ce{G(C)_{t=0} = G(D)_{t=o} = 0}$$ At the beginning, before the reaction starts, the variation of Gibbs energy due to the reaction is $$\ce{\Delta G_r,_{t=0} = G(C)_{t=0} + G(D)_{t=o} - G(A)_{t=0} - G(B)_{t=0} = - G(A)_{t=0} - G(B)_{t=0} < 0}$$ As $\ce{\Delta G_r}$ is negative, the reaction can proceed. So the reaction starts, and both [$\ce{A}$] and [$\ce{B}$] decreases. Simultaneously, [$\ce{C}$] and [$\ce{D}$], being zero in the beginning, are becoming finite. But as time in going on, the reaction proceeds and $\ce{\Delta G_r}$ becomes : $$\ce{\Delta G_r = G(C)_t + G(D)_t - G(A)_t - G(B)_t }$$ As a consequence, $\ce{\Delta G_r }$ is still negative, but slightly increasing with time, as the concentrations and the free energies of $\pu{C}$ and $\pu{D}$ are increasing, and $\pu{A}$ and $\pu{B}$ are decreasing. After a rather long time, $$\ce{G(C)_{t=∞} + G(D)_{t=∞} = G(A)_{t=∞} - G(B)_{t=∞}}$$ At this point, $\ce{\Delta G_r = 0}$ The reaction is at an equilibrium.

The most important result is the fact that the equilibrium constant of the reaction $\pu{K}$ depends on the values of the Gibbs energies in their pure state, and not in the mixture. This constant depends on the $\pu{G°}$ values, according to : $$\ce{K_r = \frac{[C][D]}{[A][B]} = G°(C) + G°(D) - G°(a) - G°(B) = \Delta G°_r}$$

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  • $\begingroup$ Thank you for you response. My confusion seems to arise from whether to interpret ΔG as the change in Gibb's energy of the complete reaction vs. the 'instantaneous' diffierence of the gibb's energy between the reactants and products at any given point in the reaction. For example, when calculating ΔG=ΔH−TΔS is this for the whole reaction at a given temeprature or is this at a single point in the reaction? $\endgroup$
    – planckton
    Aug 15, 2021 at 21:23
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    $\begingroup$ @planckton. ΔG is valid for both the whole reaction and at a single point in the reaction. The only difference is the index ° which yields an important change. Without this index (°), ΔG defines any point during the reaction. With this index (°), ΔG° only characterizes the difference between the pure reactants $\pu{A + B}$ and the pure final products $\pu{C + D}$. $\endgroup$
    – Maurice
    Aug 16, 2021 at 15:03
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Enthalpy, entropy and Gibbs energy are extensive properties and also state functions.

Any difference can also be defined as:

Reactants$\ce{->}$Products; $\Delta_r H = \Sigma H_{\text{Products}} - \Sigma H_{\text{Reactants}}$

In this definition, there is no initial and final states but rather all terms are in same state.

Similarly, the change in Gibbs energy and entropy can be defined as:

$\Delta_r G = \Sigma G_{\text{Products}} - \Sigma G_{\text{Reactants}}$

$\Delta_r S = \Sigma S_{\text{Products}} - \Sigma S_{\text{Reactants}}$

Gibbs energy is defined as, $G=H-TS$. Absolute value of Gibbs energy, i.e., $G_{\text {initial}}$ and $G_{\text {final}}$ cannot be calculated but change in Gibbs energy can be calculated as:

$\Delta_r G=\Delta_r H-T\cdot \Delta_r S$, at constant temperature and pressure

Another way of thinking is by writing the differential form:

$\mathrm dG=\mathrm dH-T\mathrm dS-S\mathrm dT$


More general case:

$G$ is defined in terms of $H$ and $S$, absolute value of both cannot be calculated.

In the equation, $\Delta G_{\text{system}}=\Delta H_{\text{system}}-T\cdot \Delta S_{\text{system}}=-T\cdot \Delta S_{\text {total}}$, the change in Gibbs energy is equal to the product of change in total entropy (system + surroundings) from initial to final state and the constant temperature.

Also note that, $\Delta H_{\text{system}}$ is the total energy change and $T\cdot \Delta S_{\text{system}}$ is the unavailable energy. Therefore, $\Delta G_{\text{system}}$ is the available energy.

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  • $\begingroup$ @BuckThorn Doesn't that change in composition is reconciled when we are calculating the difference between products and reactants? $\endgroup$
    – Apurvium
    Aug 15, 2021 at 7:02
  • $\begingroup$ You are right, you are defining the Gibbs free energy of reaction which is a property of a given fixed state. But I think this might be improved by explaining more clearly what makes this special compared to some other (discrete) Gibbs free energy change. $\endgroup$
    – Buck Thorn
    Aug 15, 2021 at 8:50
  • $\begingroup$ No, G=H-TS is a general definition. $\Delta G$ can be defined for any thermodynamic process. The difference between the reaction free energy change and the general free energy is similar to the difference between a force and a potential. The free energy change is a partial molar quantity (or rather a difference in partial molar quantities or chemical potentials). $\endgroup$
    – Buck Thorn
    Aug 15, 2021 at 10:27

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