After doing some research and reading I found some problems which I will try to state as clearly as possible.

The definition of Gibbs Free Energy says "the greatest amount of mechanical work which can be obtained from a given quantity of a certain substance in a given initial state, without increasing its total volume or allowing heat to pass to or from external bodies, except such as at the close of the processes are left in their initial condition. (I have taken the definition from the wikipedia). The equation of the Gibbs energy $$ G= H -TS$$ is also clear. But the way I understand it is this: the maximum non-expansion work , like transporting an electron, breaking a chemical bond, moving real life things etc. which can be obtained from a system . These things made sense during the study of Second Law of thermodynamics but as I moved to Chemical Equilibrium these concepts began to tremble (for me).
In elementary classes it is said that Equilibrium is a state where the composition of reactants and products do not change over time, but in higher chemistry classes it is said that equilibrium is a condition corresponds to $$ \Delta G = 0$$ So, my first question is how do these two concepts mean same thing? The next thing which causes problem is chemical potentials. If we adhere to the formal meaning of potential i.e. something which is stored and can be used when proper conditions are met, so chemcial potential would mean the potential of substance to react and again this is related to Gibbs energy whose definition I gave above. So, how do chemical potential and Gibbs energy can have any relation?
A question which is off topic over here but I want to mention it, why do we bother so much about standard things like $$ \mu_\mathrm A = \mu_\mathrm A^\circ + RT \ln(p_\mathrm A)$$ why we wanted to express it in that standard (that little circle) form ?
I want to make myself clear that the conception of Gibbs energy was quite clear to me in the context of thermodynamics, we simply meant it to be the work which can be extracted from a substance, but it all-pervading use has made me to doubt myself, just like in mathematics the number $\mathrm e$ appears in odd places. Even if Gibbs energy (according to the understanding that gave above in bold) is appearing mathematically then also it would be having some physical meaning because Thermodynamics and Equilibrium are natural sciences and not the mathematics.

Thank you, any help will be much appreciated.

  • $\begingroup$ Your equation gives the chemical potential of a species in a gas mixture at a partial pressure of $p_A$ bars in the mixture. In this equation, $\mu^o_A$ is the molar free energy of the pure species at 1 bar and at the same temperature. $\endgroup$ Oct 8, 2019 at 13:00
  • $\begingroup$ @ChetMiller Your such a small comment has solved my such a big problem. I’m really grateful to you. Can you please tell me one more thing: what is the meaning of chemical potential the way you understand it? $\endgroup$ Oct 8, 2019 at 14:17
  • $\begingroup$ I understand it as the partial derivative of the total Gibbs free energy with respect to the number of moles of the given species holding the number of moles of all other species (and T and P) constant. $\endgroup$ Oct 8, 2019 at 15:15
  • $\begingroup$ @ChetMiller Okay, means chemical potential means that how much the Gibbs energy of a system gonna change if change the number of moles of a particular substance. Am I right? $\endgroup$ Oct 8, 2019 at 16:10
  • $\begingroup$ Yes, you are right. Chemical potential is the partial derivative of the Gibbs energy per molar amount of the compound. $\endgroup$
    – Poutnik
    Oct 8, 2019 at 16:54

1 Answer 1


Your first question:

Before the second law was understood it used to be thought that the maximum amount of work that could be extracted from a reaction was $-\Delta H$ but many experiments showed that this was not the case. It is true that, according to the first law, (the law of conservation of energy), that external work done must be equal to the loss of energy of the system, unless some heat is given to or taken from the surroundings. This was first understood by Gibbs. In an isothermal reaction working under reversible conditions, the heat absorbed from the surroundings is $T\Delta S$, if this is positive then the work done will be even greater than the heat of reaction and so heat will be taken from the thermostat.

Second question:

The $\Delta G$ you mention is actually the gradient of the free energy with extent of reaction $(\partial G/\partial \xi)_{T,p}$ at constant temperature and pressure. This differential is sometimes written as $\Delta G'$. The equation in general is $\Delta G'=\Delta G^\text{o}+RT\ln(Q)$ where $Q$ is the ratio of partial pressures, for example, in a gas phase reaction. At equilibrium a plot of $G$ vs $\xi$ reaches a minimum and the gradient is zero, i.e. $\Delta G'=0$ then $\Delta G^\text{o}=-RT\ln(K_p)$ where $K_p$ is the equilibrium constant and is used instead of $Q$ at equilibrium. This reaction tells us how the position of equilibrium can be determined in terms of standard state free energies of reactants and products at 1 atm pressure.

(The extent of reaction is zero when only reactants are present and is one when one mole of reactants have changed to product)

  • $\begingroup$ Thank you, your answer is very nice. Can you please write something on chemical potential too? $\endgroup$ Oct 7, 2019 at 16:55
  • $\begingroup$ @porphyrin When we say that at "equilibrium" it reaches a minimum do we mean to the equilbrium state of both system and surroundings? I understand that $\Delta G'$ must be zero because the entropy (of the isolated system, system + surroundings) should stop increase at equilibrium. Is it that correct or the condition for $\Delta G'=0$ comes from the fact that at chemical equilibrium the chemical potentials are the same? $\endgroup$
    – ado sar
    Feb 2, 2021 at 23:56
  • $\begingroup$ yes system and surroundings so chemical potential the same. $\endgroup$
    – porphyrin
    Feb 3, 2021 at 9:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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