3
$\begingroup$

What does it mean when there is a negative delta $S$ $(\Delta S < 0)$? One person asked if this implied negative entropy, but I don't see how this could be possible.

Does this imply a negative change in entropy? If the former, how can negative entropy exist?

So what does $\Delta S\lt0$ mean?

$\endgroup$
4
  • 3
    $\begingroup$ entropy is just S. $\endgroup$ Commented Mar 15, 2017 at 6:04
  • 2
    $\begingroup$ Freezing or condensing water is a good example of a process where $\Delta S < 0$. $\endgroup$
    – Zhe
    Commented Mar 15, 2017 at 15:36
  • 1
    $\begingroup$ Delta means change, so negative delta mean decreasing of a value. An sure, yes, there are plenty of processes like that. $\endgroup$
    – Greg
    Commented Mar 18, 2017 at 6:10
  • 1
    $\begingroup$ Please note that while this question has garnered many views it is potentially ambiguous. I think you meant, what does $\Delta S <0$ mean but one answer has interpreted this as, what does $-\Delta S$ mean. $\endgroup$
    – Buck Thorn
    Commented Apr 20 at 7:57

3 Answers 3

4
$\begingroup$

Negative delta S ($\Delta S <0$) is a decrease in entropy in regard to the system.

For physical processes the entropy of the universe still goes up but within the confines of the system being studied entropy decreases.

One example is a freezer with a cup of liquid water in it. The freezer will utilize the electrical energy coming in to pump heat from the water until it becomes a solid (ice). At which point the entropy of the system (the contents of the freezer) decreases, however the electrical energy needed to be produced to power the freezer such as coal (burning a solid to a gas) and heat was wasted by the freezer in the process both of which create larger amounts of entropy than was reduced in the system by the freezer.

For chemical processes entropy can be a great driver of many reactions but it is not absolute. A system's favorability to release energy (enthalpy) competes with entropy. For example, an electron of hydrogen may have higher entropy if it drifts from the core proton but the electrostatic forces (and quantum mechanics) energetically keep it bound to the atom. For isobaric processes, you much determine the change in Gibbs free energy for the reaction to know which way it is driven. For isochoric processes, you must determine the Helmholtz free energy to know which way a reaction is driven.

One example is the oxidation of iron in air. When the oxygen is in the gas state it has higher entropy but the energy of bonding with iron is so great that at normal pressures oxygen goes from the gas phase and the iron rusts ("enthalpy wins") as delta G is negative.

Now we must consider statistical thermodynamics, this process is pressure dependent. At normal atmospheric pressure, the forward rate of oxygen entering the gas phase is the same as the reverse process. If the iron oxide were held in a sufficiently high vacuum the reverse process would occur and the iron oxide would reduce back to iron like many asteroids ("entropy wins"; note: thermodynamics is equilibrium after an infinite time). Accounting for pressure modifies the Gibbs free energy equation to: $$\Delta G = \Delta G^\circ -RT \ln(P) = \Delta H^\circ -T\Delta S^\circ -RT \ln(P)$$

One thing to note is that for chemical reactions the entropy and enthalpy values are for a standard temperature (such as $298\ \text{K}$). For a spontaneous system with $\Delta S^\circ < 0$ enthalpy must be negative, this heat in reality is absorbed by the system or the environment and produces entropy according to: $$ \int \mathrm dS \equiv \int \frac{C_v}{T}\,\mathrm dT \equiv \int \frac{\mathrm dQ}{T} $$

This in and of itself produces some entropy in the universe though may not net above zero as the bonding energy is still the major driving force.

$\endgroup$
4
  • $\begingroup$ Hmm, but $-\Delta S$ does not mean negative $S$ change, but negatively taken $\Delta S$. Similarly, $-T$ does not mean negative absolute temperature $T$, but negatively taken (as for subtraction) positive $T$. $\endgroup$
    – Poutnik
    Commented Mar 29, 2019 at 7:16
  • $\begingroup$ There are at least two wrong things here: "One thing to note is that for chemical reactions the entropy and enthalpy values are for a standard temperature (such as 298 K). For a spontaneous system with ΔS∘<0 enthalpy must be negative, this heat in reality is absorbed by the system or the environment". $\Delta S$ (and enthalpy) can be defined for non-standard conditions. However tables have been constructed for standard conditions for convenience. This generally means p=constant=1 bar (formerly it was 1 atm), but T must be declared (298K is common). $\endgroup$
    – Buck Thorn
    Commented Apr 20 at 7:44
  • $\begingroup$ And yes, enthalpy has to be negative (necessarily implying exothermicity) for a spontaneous process if $\Delta S$ is negative. $\endgroup$
    – Buck Thorn
    Commented Apr 20 at 17:42
  • $\begingroup$ When you write "For physical processes the entropy of the universe still goes up" do you mean spontaneous processes? $\endgroup$
    – Buck Thorn
    Commented Apr 22 at 6:00
1
$\begingroup$

The value of $\Delta S$ is defined as a difference between two entropies $S_1$ and $S_2$. You expect entropies $S_1$ and $S_2$ both to be positive, but the difference is either positive or negative, depending which of $S_1$ and $S_2$ is larger.

You have the same situation for temperature expressed using the Kelvin scale. All temperatures are positive, but a temperature change can either be an increase in temperature ($\Delta T$ is positive) or a decrease in temperature ($\Delta T$ is negative).

$\endgroup$
0
0
$\begingroup$

The original question (before being edited by Karsten) is an incredibly convoluted question, or rather like a purposely crafted tangle of two or even three questions:

(1) the title and the last sentence can be interpreted to ask the same thing, what does it mean when we say $\Delta S<0$. It means the change in S of a system is negative, which is equivalent to saying that the entropy of the system has decreased. A basic example is the response of an ideal gas when its temperature is reduced isochorically (at constant volume). Since $dU = -pdV + TdS$, $\left(\frac{\partial S}{\partial U}\right)_V = \frac{1}{T}$. For an ideal gas U is a monotonously increasing function of T (and T only), so reducing T reduces U: $$dT<0 \implies dU<0 $$ and $$\frac{dU}{T}<0 \implies dS<0 $$

(2) The first sentence (originally it read "What does it mean when there is a negative delta S (−ΔS)?") otoh asks what does it mean when we place a negative sign (negating) $\Delta S$. That is a mathematical operation with no bearing on the physical meaning of a change in S (as another answer explains).

(3) "One person asked if this implied negative entropy, but I don't see how this could be possible." Here you don't explain why you don't see how this could be possible, even if you are correct - the negative sign does not imply anything about the sign of the change in entropy).

(4) "If the former, how can negative entropy exist?" Finally here you ask something completely different, and you are correct, the change in entropy is negative. Entropy itself cannot be negative. The entropy of a system is always greater or equal to zero. Only changes in entropy can be negative. Perhaps that is what you meant all along.

$\endgroup$

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.