role of entropy in gibbs free energy intuition

Question

I have looked at various derivations of the Gibbs free energy equation and the underlying definition of the Gibbs free energy.

However, I have been unable to attain direct insight/intuition over the significance of entropy in the equation.

Specifically, let's take the case of 2 gases reacting to form a solid (the reaction, overall, releases energy as more stable bonds are formed than the ones present before the reaction). Clearly, the entropy change for the system will be negative as the solidification strips the atoms of much of their "freedom".

However, what does it mean (in terms of this scenario) that if the temperature were high enough, the reaction would not occur? How is it that the entropy change will determine the spontaneity of the reaction, especially at high temperatures?

I really am looking for a verbal answer please (rather than some mathematical derivations, which often, at least to the novice (me), lack illuminative powers).

Thank you!

Answer 1

I'm currently taking my second thermodynamics course in my university's chemical engineering curriculum.

You might have some luck if you visualize entropy as not just a quantitative measurement of the randomness of a system or the amount of spatial and energetic states available to it; you can kind of think of entropy as the amount of energy that a system loses to frictional dispersion on an atomic scale.

For example, you can think of the large entropy change associated with rapidly decreasing the pressure of a gas by allowing it to expand through a valve to atmospheric conditions as a type of "frictional" response to the process being carried out quickly. If you were to take the same high pressure gas and expand it slowly, hence preventing large gradients in state variables across the volume of the system and keeping the system in equilibrium for the duration of the expansion process, the generation of entropy would be negligible.

Liken this to friction between your socks and a rug. If you slide your feet over the rug slowly, there won't be much heat generation. However, get up to a sprint then come to a sliding stop, and you'll find that things get a little hotter.

I might not have explained this very well, so certainly let me know if I'm wrong! But I think this might help you to better conceptualize your problem.

Answer 2

In your question you wrote that " How is it that the entropy change will determine the spontaneity of the reaction, especially at high temperatures." but that is not quite right. You are missing the boundary conditions. Entropy is maximized for an isolated systems as you remove the internal constraints and let it evolve spontaneously. If instead your system is interacting with an environment with which it can exchange heat and pressure work, that is "TdS" and "pdV" and if the environment is at constant temperature and constant pressure throughout the energy exchange (this is the "boundary condition") then the system reaches its equilibrium state with the environment when the system's Gibbs potential reaches its minimum at the prescribed T and p. While the system moves from one of its states to the final equilibrium state with its environment both its internal pressure and temperature may take all kinds of values different from that of the environment but then eventually will rest at those when the Gibbs potential is minimized. If the environment is finite so it has finite internal energy and entropy then the joint equilibrium of the system and its environment is achieved when their joint entropy is maximized as the two together form an isolated system.

Answer 3

As pjg notes, it's always about the increase in total entropy.

The Gibbs free energy is useful in situations where the initial and final temperature and pressure do not change. (Imagine the system is enclosed in a large thermal reservoir at temperature $T$, with everything at ambient pressure.) Confronting the math:

$$ G_{system} = H_{system} - TS_{system} $$

where the subscripts indicate that these are system quantities, and $T$ is the (fixed) common temperature of system and reservoir.

Under constant pressure conditions, the change in enthalpy $\Delta H_{system}$ is just the heat transferred from the reservoir to the system, so $-\Delta H_{system}/T$ is the change in reservoir entropy. For an exothermic reaction like your example, where the system generates heat, $\Delta H_{system} < 0$, and the reservoir entropy increases.

The total entropy change $\Delta S_{total}$ is the sum of the changes in the system and reservoir entropies:

$$ \Delta S_{tot} = \Delta S_{system} + \Delta S_{reservoir} = \Delta S_{system} - \frac{\Delta H_{system}}{T} $$

Note that

$$ \Delta G_{system} = - T \Delta S_{total} $$

and the Gibbs free energy decreases precisely when the total entropy increases.

In your specific example, the reaction "goes" when the increase in reservoir entropy exceeds the magnitude of the reduction in system entropy, so the total entropy increases (and the Gibbs free energy of the system decreases).

Note that, to the extent the heat released $- \Delta H_{system}$ is independent of temperature, the increase in reservoir entropy is reduced at higher temperatures (because of the $T$ in the denominator), so at a high enough temperature, the change in total entropy would become negative, and the reaction would stop.

Answer 4

For every irreversible process (like your chemical reaction) the entropy of the universe increases. Your entropy decrease during crystallization is offset by entropy increase associated with the heat transfer out into the surroundings. Heat only flows from cold to hot. So, if the surroundings get too hot, then the exothermic heat of crystallization can't escape and the crystallization can't occur.