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From what I understand, it sounds like the Gibbs Free Energy change of a reversible reaction at equilibrium is zero. However, since I know that Gibbs Free energy change depends on temperature, does this not imply that equilibrium can only ever be reached at one very specific temperature? This doesn't sound right, as I know that equilibrium can be established at lots of different reaction temperatures, although the position of equilibrium changes with temperature of course. I asked three chemistry teachers at my school, who all seem just as baffled as I am about this.

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    $\begingroup$ At each specific temperature, there is a relationship between the concentrations of the products and reactants at equilibrium such that the Gibbs Free Energy change is zero when the concentrations satisfy this relationship. The relationship involves the equilibrium constant for the reaction, which is a function of temperature. $\endgroup$ – Chet Miller Mar 27 at 15:33
  • $\begingroup$ A change in temperature may put the system in a state that is not equilibrium. At this point, the reversible system will readjust to be in equilibrium at which point, the free energy change between reactants and products will be zero again. $\endgroup$ – Zhe Mar 27 at 15:42
  • $\begingroup$ @Zhe in order to get the free energy change between reactants and products to become zero again, either the temperature, the enthalpy change of the reaction, or the entropy change of the reaction must change, since ΔG = ΔH - TΔS. Which of these changes when the system readjusts to be in equilibrium again? $\endgroup$ – Purple dragon unicorn Mar 27 at 15:49
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    $\begingroup$ $\Delta G^0$ is a function of temperature, as are $\Delta H^0$ and $\Delta S^0$. $\endgroup$ – Chet Miller Mar 27 at 15:52
  • $\begingroup$ @ChetMiller My chemistry teacher told me that ΔH and ΔS only negligibly vary with temperature - hence why it's possible to answer questions such as "At what temperature does this reaction become feasible?" using very simple linear algebra. $\endgroup$ – Purple dragon unicorn Mar 27 at 16:03
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it sounds like the Gibbs Free Energy change of a reversible reaction at equilibrium is zero

Yes, that is true. The Gibbs free energy is zero when the reaction has reached an equilibrium, i.e. the reaction quotient Q is equal to the equilibrium constant K. You can express the Gibbs free energy in terms of the standard Gibbs free energy and the reaction quotient:

$$\Delta G = \Delta G^\circ + R T \ln(Q) = 0$$

No matter what the value of $\Delta G^\circ$, there is always a matching $Q$ that will result in a $\Delta G$ of zero. This is why at any temperature, the reaction will be able to reach equilibrium.

However, since I know that Gibbs Free energy change depends on temperature, does this not imply that equilibrium can only ever be reached at one very specific temperature?

As explained above, as long as $Q$ is able to change, there is a state of equilibrium for any value of $\Delta G^\circ$, i.e. at any temperature. However, if all of the reactants and products are pure liquids or solids, the expression for $Q$ is simply 1, and so $\Delta G$ does not change when the reaction proceeds forward or backward. In those cases, $\Delta G^\circ$ has to be zero in order to attain equilibrium, and that only happens at a specific temperature.

There are some electrochemical reactions where all reactant and product species are solids. This is great for making a battery because the voltage won't drop as you discharge the battery. Another more familiar example is the process of ice melting:

$$\ce{H2O(s) <=> H2O(l)}$$

The equilibrium constant expression and the reaction quotient for this process is simply 1. There is only one temperature at normal pressure where ice and liquid water exist side by side, and this temperature is called the normal melting point of water. At a temperature higher than that, water is all liquid, and at a temperature lower than that, it is all ice.

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  • $\begingroup$ Thanks. Do ΔH and ΔS depend on Q, then? $\endgroup$ – Purple dragon unicorn Mar 27 at 19:37
  • $\begingroup$ $\Delta S$ does. $\endgroup$ – Karsten Theis Mar 27 at 19:41
  • $\begingroup$ Thanks, @KarstenTheis . So, I think I now understand: as the reaction progresses, the relative concentrations of reactants and products changes, so ΔS changes which causes ΔG to change, until ΔG becomes zero at which point equilibrium is established. In terms of the standard entropy change, then, does that assume that the concentrations of all reactants and all products are equal (since under standard conditions, concentrations are all 1M?) And is that why the standard entropy change has only one value even though actually entropy change is affected by Q? $\endgroup$ – Purple dragon unicorn Mar 27 at 19:52
  • $\begingroup$ Yes, that’s it exactly. $\endgroup$ – Karsten Theis Mar 27 at 20:39
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Karsten Theis Mar 28 at 19:52
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In general, any expression for $G$ will by necessity have a dependence on $T$, if only through the explicit incorporation of temperature in $G = H - TS$.

Looking at the Calphad (Calculation of Phase Diagram) community, they go further. Referring to A.T. Dinsdale, "SGTE Data for Pure Elements", CALPHAD 15(4) 317-425 (1991), they state, for the purposes of defining $G$ for the elements:

The Gibbs energy is represented as a power series in terms of temperature $T$ in the form: $ G = a + bT + cT \ln(T) + \Sigma d_{n}T^{n}$

From the definition of $G$, $S$, and $H$ one can then get:

$S = -b-c-c\ln(T)-\Sigma nd_{n}T^{n-1}$ and

$H = a-cT-\Sigma (n-1)d_{n}T^{n}$

So, one sees, given the chosen representation for $G$, that both $H$ and $S$ should depend on $T$ unless $G$ can be represented in a very simple form (only $a$ and $b$ being non-zero). Since Dinsdale uses a more complex power series, one can rest assured that $G$, even for the elements, tends to be rich in temperature dependence.

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  • $\begingroup$ In which case, does this mean that equilibrium can only happen at one very specific temperature? Since the only thing that affects G is the temperature, but ΔG is zero at equilibrium? And therefore the power series of temperature must be zero at equilibrium? $\endgroup$ – Purple dragon unicorn Mar 27 at 17:42
  • $\begingroup$ @Purpledragonunicorn - no, it just means the equilibrium point (see comments to your question) will shift. For example, the solubility of salt in water is a function of temperature. That comes from the free energies changing with temperature. There is still an equilibrium point, it just changes. $\endgroup$ – Jon Custer Mar 27 at 17:53
  • $\begingroup$ Surely @JonCuster if you change the temperature, that changes ΔG such that ΔG is no longer zero. Now that ΔG is not zero, it will be negative for the forwards reaction and positive for the reverse reaction (or the other way around) meaning that only the forwards reaction is feasible, the backwards reaction is infeasible. And therefore, surely only the forward reaction will proceed, and not the backwards reaction, so it is therefore no longer in equilibrium? Or does something change after that to put it back into equilibrium? $\endgroup$ – Purple dragon unicorn Mar 27 at 17:58

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