# 13.5: Entropy Changes and Spontaneity


Learning Objectives

• State and explain the second and third laws of thermodynamics
• Calculate entropy changes for phase transitions and chemical reactions under standard conditions

## Connecting Entropy and Heat to Spontaneity

In the quest to identify a property that may reliably predict the spontaneity of a process, we have identified a very promising candidate: entropy. Processes that involve an increase in entropy of the system ($$ΔS_{sys} > 0$$) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:

$ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr} \label{$$\PageIndex{1}$$}$

To illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:

1. The objects are at different temperatures, and heat flows from the hotter to the cooler object. This is always observed to occur spontaneously. Designating the hotter object as the system and invoking the definition of entropy yields the following: $ΔS_\ce{sys}=\dfrac{−q_\ce{rev}}{T_\ce{sys}}$ and $ΔS_\ce{surr}=\dfrac{q_\ce{rev}}{T_\ce{surr}} \label{$$\PageIndex{2}$$}$The arithmetic signs of qrev denote the loss of heat by the system and the gain of heat by the surroundings. Since Tsys > Tsurr in this scenario, the magnitude of the entropy change for the surroundings will be greater than that for the system, and so the sum of ΔSsys and ΔSsurr will yield a positive value for ΔSuniv. This process involves an increase in the entropy of the universe.
2. The objects are at different temperatures, and heat flows from the cooler to the hotter object. This is never observed to occur spontaneously. Again designating the hotter object as the system and invoking the definition of entropy yields the following: $ΔS_\ce{sys}=\dfrac{q_\ce{rev}}{T_\ce{sys}}$ and $ΔS_\ce{surr}=\dfrac{−q_\ce{rev}}{T_\ce{surr}} \label{$$\PageIndex{3}$$}$ The arithmetic signs of qrev denote the gain of heat by the system and the loss of heat by the surroundings. The magnitude of the entropy change for the surroundings will again be greater than that for the system, but in this case, the signs of the heat changes will yield a negative value for ΔSuniv. This process involves a decrease in the entropy of the universe.
3. The temperature difference between the objects is infinitesimally small, $$T_{sys} ≈ T_{surr}$$, and so the heat flow is thermodynamically reversible. See the previous section’s discussion). In this case, the system and surroundings experience entropy changes that are equal in magnitude and therefore sum to yield a value of zero for ΔSuniv. This process involves no change in the entropy of the universe.

These results lead to a profound statement regarding the relation between entropy and spontaneity known as the second law of thermodynamics: all spontaneous changes cause an increase in the entropy of the universe. A summary of these three relations is provided in Table $$\PageIndex{1}$$.

 $$ΔS_{univ} > 0$$ spontaneous $$ΔS_{univ} < 0$$ nonspontaneous (spontaneous in opposite direction) $$ΔS_{univ} = 0$$ reversible (system is at equilibrium)

Definition: The Second Law of Thermodynamics

All spontaneous changes cause an increase in the entropy of the universe, i.e., $ΔS_\ce{univ} > 0.$

For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, $$q_{surr}$$ is a good approximation of qrev, and the second law may be stated as the following:

\begin{align} ΔS_\ce{univ} &=ΔS_\ce{sys}+ΔS_\ce{surr} \\[4pt] &=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \label{4} \end{align}

We may use this equation to predict the spontaneity of a process as illustrated in Example $$\PageIndex{1}$$.

Example $$\PageIndex{1}$$: Will Ice Spontaneously Melt?

The entropy change for the process

$\ce{H2O}(s)⟶\ce{H2O}(l) \nonumber$

is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C?

Solution

We can assess the spontaneity of the process by calculating the entropy change of the universe. If ΔSuniv is positive, then the process is spontaneous. At both temperatures, ΔSsys = 22.1 J/K and qsurr = −6.00 kJ.

At −10.00 °C (263.15 K), the following is true:

\begin{align*} ΔS_\ce{univ}&=ΔS_\ce{sys}+ΔS_\ce{surr} \\[4pt] &=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \\[4pt] &=\mathrm{22.1\: J/K+\dfrac{−6.00×10^3\:J}{263.15\: K}} \\[4pt] &=−0.7\:J/K \end{align*}

Suniv < 0, so melting is nonspontaneous (not spontaneous) at −10.0 °C.

At 10.00 °C (283.15 K), the following is true:

\begin{align*} ΔS_\ce{univ} &=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \\[4pt] &= 22.1 \:J/K + \dfrac{−6.00×10^3\:J}{283.15\: K} \\[4pt] &=+0.9\: J/K \end{align*}

$$\Delta S_{univ} > 0$$, so melting is spontaneous at 10.00 °C.

Exercise $$\PageIndex{1}$$

Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of $$\Delta S_{univ}$$?

Entropy is a state function, and freezing is the opposite of melting. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.

## Summary

The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, $$S_{univ} > 0$$. If $$ΔS_{univ} < 0$$, the process is nonspontaneous, and if $$ΔS_{univ} = 0$$, the system is at equilibrium.

## Key Equations

• $$ΔS^\circ=ΔS^\circ_{298}=∑νS^\circ_{298}(\ce{products})−∑νS^\circ_{298}(\ce{reactants})$$
• $$ΔS=\dfrac{q_\ce{rev}}{T}$$
• $$ΔS_{univ} = ΔS_{sys} + ΔS_{surr}$$
• $$ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T}$$