3
$\begingroup$

To the best of my understanding, the quantity of Gibbs free energy of an entity depends on its concentration in the system. Equilibrium is reached when ∆G = 0, or when the Gibbs free energies of all the entities are equal to each other.

If Gibbs free energy is dependent on concentration, then at standard conditions, the concentrations of all entities is 1M—this is fixed by convention. How then can you have a ∆Gº value? All the concentrations are fixed and there is no change in concentration nor energy. Is ∆Gº then just a rudimentary concept for those who do not know calculus? Should it be better written as dGº?

In a similar vein, though slightly unrelated, I would also want to know if enthalpy is related to concentration, similar to how Gibbs free energy is. Why or why not? If enthalpy is also concentration dependent, then does ∆Hº exist?

$\endgroup$
  • 1
    $\begingroup$ Yes, $\Delta G^0$ is defined for standard conditions, including 1M concentrations of everything. It is very real and has a lot of practical consequences. Whether or not the system is stable, is irrelevant for the purposes of this definition. $\endgroup$ – Ivan Neretin Jun 17 '16 at 13:13
  • $\begingroup$ @IvanNeretin Could you explain why ∆Gº is defined when there seems to be no change in energy of the system at standard conditions? $\endgroup$ – lightweaver Jun 17 '16 at 13:25
  • $\begingroup$ OK, I got it. Say, we have a reaction $\ce{A + B->C + D}$. Now, $\Delta G^0$ of this reaction is the energy of the change (real or imaginary, this is not important) from one system ($\ce{A + B}$ in standard state) to another system ($\ce{C + D}$ in their standard state). $\endgroup$ – Ivan Neretin Jun 17 '16 at 13:30
1
$\begingroup$

It exists in the sense that it is a well defined concept, but it does not exist as something that is directly physically observed.

The IUPAC Gold Book defines standard reaction quantities as follows:

Infinitesimal changes in thermodynamic functions with extent of reaction divided by the infinitesimal increase in the extent when all the reactants and products are in their standard states. For the quantity X they should be denoted by $Δ_r X °$, but usually only Δ X ° is used. For specific types of reactions the subscript r is replaced by: f for formation, c for combustion, a for atomization and superscript ‡ for activation.

So though delta is capitalized, it still represents an infinitesimal concept.

Also, the requirement that all the substances are in their standard states requires that they not be mixed together, and for gases and solutes requires fictional ideal states, so it is not directly observed.

Particularly, standard state is defined as:

State of a system chosen as standard for reference by convention. Three standard states are recognized: For a gas phase it is the (hypothetical) state of the pure substance in the gaseous phase at the standard pressure p = p°, assuming ideal behaviour. For a pure phase, or a mixture, or a solvent in the liquid or solid state it is the state of the pure substance in the liquid or solid phase at the standard pressure p = p°. For a solute in solution it is the (hypothetical) state of solute at the standard molality , standard pressure or standard concentration and exhibiting infinitely dilute solution behaviour. For a pure substance the concept of standard state applies to the substance in a well defined state of aggregation at a well defined but arbitrarily chosen standard pressure.

$\endgroup$
  • $\begingroup$ $\mathrm G$ itself is not a physical observable, either. $\endgroup$ – hBy2Py Jun 17 '16 at 13:58
  • $\begingroup$ @Brian I agree, but I'm trying to say that beyond that, to the extent that you can observe an equilibrium constant and corresponding delta G, it isn't going to be the standard equilibrium constant and standard delta G, because that involves the concept of the each substance being in a pure state and/or a fictitious state. $\endgroup$ – DavePhD Jun 17 '16 at 14:06
  • 1
    $\begingroup$ Oh, I'm not disagreeing with your answer at all! Just noting that nothing about any $\mathrm G$ or $\Delta\mathrm G$ is experimentally measurable, standard state or no. $\endgroup$ – hBy2Py Jun 17 '16 at 15:58
1
$\begingroup$

You don't measure $\Delta G^\circ$ directly. You measure concentrations, which gives you the equilibrium constant, which gives you $\Delta G^\circ$.

You can measure the enthalpy change by, for example, measuring the heat released or absorbed during the reaction. The heat exchanged will depend on the concentrations used but when the heat is converted to $\Delta H$ (using the heat capacity) it is typically converted to units of energy per mole, which effectively is the 1 M reference in this particular example, making it the standard enthalpy change ($\Delta H^\circ$).

If there is a volume change during the reaction the heat exchanged will also depend on the pressure and if the pressure wasn't 1 bar then the $\Delta H$ value needs to be corrected before it can be called ($\Delta H^\circ$). One usually assumes that the volume change is negligible if neither product nor reactants are gasses.

$\endgroup$
0
$\begingroup$

I think I understood what you were asking, and the people here just explained to you what $\Delta G$ is.

$\Delta G^\circ$ is defined as the $\Delta G$ in the natural conditions, at $298\ \mathrm K$ and with every concentration at $1\ \mathrm{mol/l}$. But think of any reaction. Are they at equilibrium when the concentration of reactants is equal to that of products? Probably not. For example, a strong acid dissolving in water has a very high equilibrium constant, meaning that it reaches equilibrium when there is very little reactant and a lot of product. Another example is a very insoluble salt dissolving in water. The equilibrium is reached when there is mostly reactant, but a tiny tiny fraction of it will dissolve. In both cases, when all the concentrations are $1\ \mathrm{mol/l}$, the system will be trying like crazy to turn the reactants into products, or vice-versa. Since:

$$\Delta G=\Delta G^\circ+RT\ln Q$$

in equilibrium:

$$0=\Delta G^\circ+RT\ln K$$

$$\Delta G^\circ=-RT\ln K$$

If $K$ (the equilibrium constant) is high (the case of the strong acid), $\Delta G^\circ$ will be very negative; meaning that when all concentrations are equal, there is a very strong tendency to form products. If $K$ is small (much smaller than 1, the case of the insoluble salt), $\Delta G^\circ$ will be very positive, meaning that when all concentrations are equal, there is a strong tendency to form reactants back.

$\endgroup$
  • 1
    $\begingroup$ Note that the definition of standard state makes no reference to fixed temperature. It is possible to have various standard states of a substance as the reference temperature varies. However, the most favoured value for the reference temperature is indeed 298.15 K. $\endgroup$ – Loong Jul 6 '16 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.