Gibbs Free energy of a reversible reaction

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.




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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> 
> 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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