5
$\begingroup$

Another statement of second law of thermodynamics can be formulated in terms of system properties and not properties + surroundings (isolated system). For a closed system at constant temperature and pressure:

$$dG_{T,p,n_i} \leq 0$$

Are $n_i$ constant during a process? We know that during a chemical reaction Gibbs free energy take its minimum value at equilibrium. But during the reaction the number of molecules $n_i$ change. So would it better the inequality to take the form of:

$$dG_{T,p} \leq 0$$

?

Interpretation of the differential In mathematics the differential of a function $f(x)$ can be interpreted as the change in $f(x)$ for an infinitesimal change in the argument $x$, that is:

$$df(x)= f(x+dx)-f(x)$$

I have trouble in the interpretation of $dG \leq 0$. When I was introduced in thermodynamics I always thought of the spontaneity inequalities in the following manner.

Suppose that the system is at equilibrium state (state $1$) at time $t$. Something happens (e.g. a chemical reaction takes place) so the system will reach a new equilibrium state (state $2$). In that case $G$ can be thought only as function of $ξ$ (the extent of reaction). Because the Gibbs energy must take its minimum value at equilibrium I interpreted the $dG \leq 0$ as:

$$dG(t)= G(t+dt)- G(t) \leq 0$$

In other words Gibbs free energy decreases until it takes its minimum value. And because we can plot $G$ as function of $ξ$ the reaction will proceed in the direction where $G$ is minimized as shown in the following figure (the left figure):

enter image description here

Although the above interpretation seems intuitive I have some problems.

$1$st Problem With the above interpretation is not clear if the equality should be interpreted as that the function have a constant value over an inteval or that it has reached a minimum value. For example if we were monitoring $G$ over time and we noticed that $G$ has a constant value would this imply that the system have reached equilibrium? Look at the following figure:

                                                    

In the mid region of the function (where it is constant) the differential is zero. So how $dG \leq 0$ should be interpreted? In general we can't find a function $φ$ so that $G=φ(t)$ and this should be expected as Thermodynamics doesn't deal with how the state of the system varies with time.

$2$nd Problem In the first scheme I have drawn an alternative path (right figure) that the system could take in order to minimize $G$. In that case there is an abrupt change in the extent of the reaction but still the systems keeps decreasing its $G$. To be honest I haven't seen any reaction reaching equilibrium in that way but I can't find a reason why it should not behave this way. It should be noted that the line representing the path depicts the change in $ξ$ in an abrupt manner and it should not be confused with a continuous change (think it like a $5$ step process).

In summary I want to make clear how to interpret the differential expression of the second law. Is a smooth incease of entropy (of surroundings + system) over time assumed?

$\endgroup$
4
  • 2
    $\begingroup$ Usually, it is the derivative of G with respect to reaction conversion is equal to zero at equilibrium. For multiple reactions, it would be minimum of G with respect to all species perturbations under constraint of conservation of atomic species. $\endgroup$ Jun 7 at 17:58
  • $\begingroup$ Point 2 it is interesting. I do not think anything goes exactly like that but I think to see a similarity with oscillating reactions. I also would like to know how to discard it by reasons/principle rather than by feeling. $\endgroup$
    – Alchimista
    Jun 8 at 8:26
  • $\begingroup$ Yet, the trick of thermodynamics is to swap the real dynamics under the carpet.... At the end what you are following are state functions, and their construction does require smooth continuous (and slow) changes. But, then, you have state functions. So between left & right the observation at equilibrium would be the same. $\endgroup$
    – Alchimista
    Jun 8 at 8:36
  • $\begingroup$ I can't understand why the question is closed. $\endgroup$
    – Anton
    Jun 26 at 10:45