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I often see the equation $\Delta S_\text{universe}$ = $\Delta S_\text{surroundings}$ + $\Delta S_\text{system}$, where $\Delta S_\text{surroundings}$ can be re-expressed in terms of the enthalpy change of the system as

$$\Delta S_\text{surroundings} = \dfrac{-\Delta H_\text{system}}{T}$$.

This leaves me wondering why $\Delta S_\text{system}$ cannot also be re-expressed (or at least is never written) as a term with some combination of $\Delta H$ and $T$ (Maybe as $$\Delta S_\text{system} \overset{?}{=} \frac{- \Delta H_\text{surroundings}}{T}$$ although this would make $\Delta S_\text{universe}$ equal $0$).

Instead, $\Delta S_\text{system}$ is usually given as a standard value, seemingly independent of temperature. Is there a reason why we can only express the change in entropy of the surroundings, but not the change in entropy of the system, as related to the change in enthalpy and the temperature of a reaction?

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If the system experiences a change at constant pressure, the change in enthalpy of the system $\Delta H_{system}$ is equal to the heat Q transferred from the surroundings to the system. So the heat received by the surroundings is $-\Delta H_{system}$. Typically, the surroundings is regarded as an ideal constant temperature reservoir, and the reservoir is considered as being "internally reversible." So, if T is the temperature of the reservoir, the change in entropy of the surroundings is $\Delta S = -\frac{\Delta H_{system}}{T}$. The system is typically not regarded as experiencing an internally reversible change, so the same kind of approach can not be applied to the system.

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  • $\begingroup$ Could you specify what you mean by "internally reversible?" $\endgroup$ – user28295 Jun 8 '16 at 19:37
  • $\begingroup$ It's a term used in Fundamentals of Engineering Thermodynamics by Moran et al. I particularly like their approach in this book. "A reversible process is one for which no irreversibilities are present within the system or its surroundings. An internally reversible process is one for which there are no irreversibilities within the system. Irreversibiities may be located within the surroundings, however. Using the internally reversible process concept we refine the definition of the thermal reservoir ...we assume no internal irreversibilites present within a thermal reservoir." $\endgroup$ – Chet Miller Jun 8 '16 at 21:18
  • $\begingroup$ What would be a case where irreversibilities only occur in the surroundings (but not the system)? $\endgroup$ – user28295 Jun 8 '16 at 21:20
  • $\begingroup$ Covered in section 5.3.4 of the book. "If there were a spatial variation in temperature, say, there would be a tendency for a spontaneous energy transfer by conduction to occur within the system in the direction of decreasing temperature." Another example would be if a spontaneous chemical reaction is occurring. Other examples are mixing and diffusion. "For reversibility (within the system), however, no spontaneous processes can be present. From these considerations it can be concluded that the internally reversible process consists of a (continuous) series of equilibrium states. $\endgroup$ – Chet Miller Jun 8 '16 at 21:30
  • $\begingroup$ Oh. I didn't answer your question. Suppose you have a process for expanding an ideal gas reversibly, but, instead of gradually removing tiny weights from a piston, you control the motion of the piston by hand. In that case, your body is part of the surroundings, and there are a huge number of irreversibilities occurring within your body as you gradually decrease the force on the piston. So the gas receives the same reversible expansion in both cases, but, in the first case, the surroundings are handled reversibly and in the second case, they are not. $\endgroup$ – Chet Miller Jun 8 '16 at 21:36
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One of the purposes of defining a system and its surrounding in thermodynamics is to separate the interesting stuff from the boring stuff.

$\Delta S_\text{surroundings}$ can be re-expressed as $\dfrac{-\Delta H_\text{system}}{T}$

The best boring surrounding is a large homogeneous phase with the same temperature as the system, for example a stirred water bath. (If the temperature of the system changes, you have to quickly swap the bath to match the temperature.) In this specific scenario, the entropy change of the surrounding will be equal to the heat $-q$ transferred into the surrounding divided by the temperature. If the process in the system is happening at constant pressure, the system is closed and there is no non-PV work (no electrical, mechanical, etc. work, just expansion work), then the heat $q$ transferred out of the system will be equal to the change in enthalpy of the system, $\Delta H$.

So there are a lot of disclaimers to make before you can set $\Delta S_\text{surroundings} = \dfrac{-\Delta H_\text{system}}{T}$.

[...] why $\Delta S_\text{system}$ cannot also be re-expressed (or at least is never written) as a term with some combination of $\Delta H$ and $T$

Remember we want the interesting stuff to happen inside the system. In general, there is no way of predicting the entropy change in the system from other thermodynamic values unless the system is as boring as the surrounding (for example when the system is a block of hot metal and the surrounding a block of cold metal). Of course, if you put the interesting stuff in the surrounding and keep the system boring, $$\Delta S_\text{system} = \dfrac{-\Delta H_\text{surroundings}}{T}$$ would be appropriate.

Is ΔS of a system related to temperature and change in enthalpy?

In general, it is not. Entropy and enthalpy change are not coupled. They can both be negative, both positive, one negative and the other positive and vice versa. This shows you that there is no fixed relationship between entropy and enthalpy.

There is one special case where entropy change and enthalpy change of the system are related - when the change in Gibbs energy is zero. The change in Gibbs energy is expressed in terms of entropy and enthalpy changes as (in the following, the subscript "system" is omitted because all the quantities refer to the system):

$$\Delta G = \Delta H - T \Delta S$$

When $\Delta G$ is zero (for example when a substance is at its melting temperature and both solid and liquid phase are present), the following relationship holds:

$$ \Delta S = \frac{\Delta H}{T}$$

Notice that there is no negative sign. This relationship can be used to estimate the melting point of a substance. Notice also that for chemical reactions at equilibrium, $\Delta_r G^\circ$ is not equal to zero, but $\Delta_r G$ is.

So while at equilibrium,

$$\Delta_r S = \Delta_r S^\circ + R \ln K = \frac{\Delta_r H}{T}$$

is correct, you can't calculate $\Delta_r S^\circ$ from $\Delta_r H^\circ$, i.e.

$$ \Delta_r S^\circ \neq \frac{\Delta_r H^\circ}{T}.$$

The only exception is when $K$ happens to be one, as it is by definition for melting when the substance is at the melting point.

$ΔS_\text{system}$ is usually given as a standard value, seemingly independent of temperature

Both $ΔS_\text{system}$ and $ΔH_\text{system}$ are dependent on temperature, and the heat capacity is used to describe the temperature-dependency. If you are evaluating the temperature-dependency of $ΔG_\text{system}$, however, the largest dependency comes from having the temperature directly in the defining equation:

$$\Delta G = \Delta H - T \Delta S$$

So even if entropy and enthalpy were temperature-independent (again, they are not), $\Delta G$ would still depend significantly on temperature. Oddly enough, a rigorous treatment of the temperature-dependence of $\Delta G$ gives the result that you would get from differentiating the defining equation with respect to $T$:

$$\frac{d \Delta G}{dT} = - \Delta S$$

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I probably don't have the full answer to your question, but I may have some insight. I like this topic, because it's the central idea in Chemistry, but textbooks care a lot about the math of it (which is easy), but not as much about explaining the ideas behind it in detail.

Since the reaction is assumed to take place at constant temperature, the kinetic-rotational energy of the particles of the reacting system does NOT increase of decrease following the reaction. If it were increased or decreased, the temperature would have changed.

Where does the heat released in the reaction come from (in an exothermic reaction)? The heat that is released in the reaction does NOT come from a decrease in the speed, rotation, or vibration of the particles, but from changes in electrical potential. So that's why heat can get released - bonds form with lower electrical potential energy and the difference in electrical potential is released - while the speed, vibration, rotation of the system stays the same - which is why temperature is constant.

So you see, the amount of heat released cannot be used to easily predict how the entropy of the system changed. The entropy change in the system is the result of the fact that when the system reacts, while it has the same amount of kinetic-vibrational energy, that energy is distributed in a way in which either more of it is in a form that cannot be used to do work (increase in entropy) or in a way that more of it is in a form that can be used to do work (more concentrated kinetic-vibrational energy and lower entropy).

By the way, notice that the entropy of the system is not about the electrical potential stored in the system (the source of enthalpy changes in a reaction), but is a fraction of the kinetic-vibrational energy of the particles (the higher percentage of that energy, whose amount in the system does not change when the reaction occurs in constant temperature, is distributed in a way that can NOT be used for work, the higher the entropy).

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  • $\begingroup$ Is it possible to use the fact that since ΔHsurroundings = ΔHsystem and Tsurroundings = Tsystem, if ΔSsystem were to equal -ΔHsurroundings/T, then from the equation ΔSuniverse = ΔSsurroundings + ΔSsystem, ΔSuniverse would always be 0, in order to show that -ΔHsurroundings/T CANNOT equal ΔSsystem? $\endgroup$ – user28295 Jun 8 '16 at 5:54
  • $\begingroup$ "If ΔSsystem were to equal -ΔHsurroundings/T" - I believe that you can't relate these in a straightforward way as in principle you could, for example, have a reaction which releases heat (changes in electrical potential), while experience no opposite change in the entropy of the system (no more concentrated kinetic-vibrational motion in the products compared to the reactants). $\endgroup$ – Daniel Jun 8 '16 at 5:59
  • $\begingroup$ Entropy and energy have different dimensions, so the entropy can't be "a fraction of the kinetic-vibrational energy". $\endgroup$ – Karsten Theis Apr 25 '19 at 19:12
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I don't know if you have done some statistical mechanics, but maybe this is related to counting of states. First of all, we express the entropy change of the surroundings in terms of enthalpy change and temperature of the system because we simply can define our system in most cases and so can circumvent and measures quantities that defines a system (enthalpy change, for example, can be measured in a calorimetry expt.) then the entropy change of the surrounding can be inferred because the that of the universe is zero (only for reversible processes).

Second, for the counting I mentioned earlier, Boltzmann defines entropy as $$S = k\ln W$$ $W$ represents the count of possible states of the system/surrounding. But the number of states of the system multiplied by those of the surroundings equal the states of the universe. So if you know two quantities (here, system and universe) through knowing their entropies, you can get the remaining. Sure, this doesn't prevent you from doing what you suggested, which is expressing system entropy change as surroundings' enthalpy change and temperature, but you can't define them specifically for the surroundings and don't have a readily available tool to measure them.

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  • $\begingroup$ We don't know the entropy change of the universe (nothing the OP said indicates a process that is at equilibrium). Even if it were at equilibrium, we still could not express system entropy via the surrounding's enthalpy change unless your system is very simple (no reactions, and enthalpy equal to heat transfer). $\endgroup$ – Karsten Theis Apr 25 '19 at 19:19

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