Learning Objectives
By the end of this section, you will be able to:
- State and explain the second and third laws of thermodynamics
- Calculate entropy changes for phase transitions and chemical reactions under standard conditions
The Second Law of Thermodynamics
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 (Δ
S > 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:
ΔSuniv=ΔSsys+ΔSsurr
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:
- 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:
ΔSsys=Tsys−qrevandΔSsurr=Tsurrqrev
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.
- 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:
ΔSsys=TsysqrevandΔSsurr=Tsurr−qrev
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.
- The temperature difference between the objects is infinitesimally small, Tsys ≈ Tsurr, 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 1.
Table 1. The Second Law of Thermodynamics |
ΔSuniv > 0 |
spontaneous |
ΔSuniv < 0 |
nonspontaneous (spontaneous in opposite direction) |
ΔSuniv = 0 |
reversible (system is at equilibrium) |
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. As a result,
qsurr is a good approximation of
qrev, and the second law may be stated as the following:
ΔSuniv=ΔSsys+ΔSsurr=ΔSsys+Tqsurr
We may use this equation to predict the spontaneity of a process as illustrated in Example 1.
Example 1: Will Ice Spontaneously Melt?
The entropy change for the process is
H2O(s)⟶H2O(l)
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?
Show Answer
We can assess the spontaneity of the process by calculating the entropy change of the universe. If Δ
Suniv is positive, then the process is spontaneous. At both temperatures, Δ
Ssys = 22.1 J/K and
qsurr = −6.00 kJ.
At −10.00 °C (263.15 K), the following is true:
ΔSuniv=ΔSsys+ΔSsurr=ΔSsys+Tqsurr=22.1 J/K+263.15 K−6.00×103J=−0.7J/K
Suniv < 0, so melting is nonspontaneous (
not spontaneous) at −10.0 °C.
At 10.00 °C (283.15 K), the following is true:
ΔSuniv==ΔSsys+Tqsurr22.1J/K+283.15 K−6.00×103J=+0.9 J/K
Suniv > 0, so melting
is spontaneous at 10.00 °C.
Check Your Learning
Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of
Suniv?
Show Answer
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.
The Third Law of Thermodynamics
The previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal (
W = 1). According to the Boltzmann equation, the entropy of this system is zero.
S=klnW=kln(1)=0
This limiting condition for a system’s entropy represents the
third law of thermodynamics:
the entropy of a pure, perfect crystalline substance at 0 K is zero.
We can make careful calorimetric measurements to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions.
Standard entropies are given the label
S298∘ for values determined for one mole of substance at a pressure of 1 bar and a temperature of 298 K. The
standard entropy change (ΔS°) for any process may be computed from the standard entropies of its reactant and product species like the following:
ΔS∘=∑νS_298∘(products)−∑νS_298∘(reactants)
Here, ν represents stoichiometric coefficients in the balanced equation representing the process. For example, Δ
S° for the following reaction at room temperature
mA+nB⟶xC+yD is computed as the following:
=[xS298∘(C)+yS298∘(D)]−[mS298∘(A)+nS298∘(B)]
Table 2 lists some standard entropies at 298.15 K. You can find additional standard entropies in
Standard Thermodynamic Properties for Selected Substances.
Table 2. Standard Entropies (at 298.15 K, 1 atm) |
Substance |
S298∘ (J mol−1 K−1) |
carbon |
C(s, graphite) |
5.740 |
C(s, diamond) |
2.38 |
CO(g) |
197.7 |
CO2(g) |
213.8 |
CH4(g) |
186.3 |
C2H4(g) |
219.5 |
C2H6(g) |
229.5 |
CH3OH(l) |
126.8 |
C2H5OH(l) |
160.7 |
hydrogen |
H2(g) |
130.57 |
H(g) |
114.6 |
H2O(g) |
188.71 |
H2O(l) |
69.91 |
HCI(g) |
186.8 |
H2S(g) |
205.7 |
oxygen |
O2(g) |
205.03 |
Example 2: Determination of ΔS°
Calculate the standard entropy change for the following process:
H2O(g)⟶H2O(l)
Show Answer
The value of the standard entropy change at room temperature,
ΔS298∘ , is the difference between the standard entropy of the product, H
2O(
l), and the standard entropy of the reactant, H
2O(
g).
ΔS298∘=S298∘(H2O(l))−S298∘(H2O(g))=(70.0 Jmol−1K−1)−(188.8 Jmol−1K−1)=−118.8Jmol−1K−1
The value for
ΔS298∘ is negative, as expected for this phase transition (condensation), which the previous section discussed.
Check Your Learning
Calculate the standard entropy change for the following process:
H2(g)+C2H4(g)⟶C2H6(g)
Show Answer
−120.6 J mol−1 K−1
Example 3: Determination of ΔS°
Calculate the standard entropy change for the combustion of methanol, CH
3OH:
2CH3OH(l)+3O2(g)⟶2CO2(g)+4H2O(l)
Show Answer
The value of the standard entropy change is equal to the difference between the standard entropies of the products and the entropies of the reactants scaled by their stoichiometric coefficients.
ΔS∘=ΔS298∘===∑νS298∘(products)−∑νS298∘(reactants)[2S298∘(CO2(g))+4S298∘(H2O(l))]−[2S298∘(CH3OH(l))+3S298∘(O2(g))]{[213.8+4×70.0]−[2(126.8)+3(205.03)]}=−371.6J/mol⋅ K
Check Your Learning
Calculate the standard entropy change for the following reaction:
Ca(OH)2(s)⟶CaO(s)+H2O(l)
Show Answer
24.7 J/mol·K
Key Concepts and Summary
The second law of thermodynamics states that a spontaneous process increases the entropy of the universe,
Suniv > 0. If Δ
Suniv < 0, the process is nonspontaneous, and if Δ
Suniv = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process.
Key Equations
ΔS∘=ΔS_298∘=∑νS_298∘(products)−∑νS_298∘(reactants)
ΔS=Tqrev
- ΔSuniv = ΔSsys + ΔSsurr
ΔSuniv=ΔSsys+ΔSsurr=ΔSsys+Tqsurr
Exercises
- What is the difference between ΔS, ΔS°, and
ΔS298∘ for a chemical change?
- Calculate
ΔS298∘ for the following changes.
SnCl4(l)⟶SnCl4(g)
CS2(g)⟶CS2(l)
Cu(s)⟶Cu(g)
H2O(l)⟶H2O(g)
2H2(g)+O2(g)⟶2H2O(l)
2HCl(g)+Pb(s)⟶PbCl2(s)+H2(g)
Zn(s)+CuSO4(s)⟶Cu(s)+ZnSO4(s)
- Determine the entropy change for the combustion of liquid ethanol, C2H5OH, under standard state conditions to give gaseous carbon dioxide and liquid water.
- Determine the entropy change for the combustion of gaseous propane, C3H8, under standard state conditions to give gaseous carbon dioxide and water.
- “Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is
Fe2O3(s)+2Al(s)⟶Al2O3(s)+2Fe(s) .
- Using the relevant
S298∘ values listed in Standard Thermodynamic Properties for Selected Substances, calculate S298∘ for the following changes:
N2(g)+3H2(g)⟶2NH3(g)
N2(g)+25O2(g)⟶N2O5(g)
- From the following information, determine
ΔS298∘ for the following:
N(g)+O(g)⟶NO(g);ΔS298∘=?
N2(g)+O2(g)⟶2NO(g);ΔS298∘=24.8 J/K
N2(g)⟶2N(g);ΔS298∘=115.0 J/K
O2(g)⟶2O(g);ΔS298∘=117.0 J/K
- By calculating ΔSuniv at each temperature, determine if the melting of 1 mole of NaCl(s) is spontaneous at 500 °C and at 700 °C.
SNaCl(s)∘=72.11mol⋅ KJSNaCl(l)∘=95.06mol⋅ KJΔHfusion∘=27.95 kJ/mol
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
Show Selected Answers
2.
ΔS298∘ for each change is as follows:
ΔS298∘===∑νΔS298∘(products)−∑νΔS298∘(reactants)1ΔS298∘SnCl4(g)−1ΔS298SnCl4(l)[1 mol(366mol KJ)]−[1 mol(259mol KJ)]=107 J/K
ΔS298∘===∑νΔS298∘(products)−∑νΔS298∘(reactants)1ΔS298∘Cs2(l)−1ΔS298∘Cs2(g)[1 mol(151.3mol KJ)]−[1 mol(237.7mol KJ)]=−86.4 J/K
ΔS298∘===∑νΔS298∘(products)−∑νΔS298∘(reactants)1ΔS298∘Cu(g)−1ΔS298∘Cu(s)[1 mol(166.3mol KJ)]−[1 mol(33.15mol KJ)]=133.2 J/K
ΔS298∘===∑νΔS298∘(products)−∑νΔS298∘(reactants)1ΔS298∘H2O(g)−1ΔS298∘H2O(l)[1 mol(188.8mol KJ)]−[1 mol(70.0mol KJ)]=118.8 J/K
ΔS298∘===∑νΔS298∘(products)−∑νΔS298∘(reactants)2ΔS298∘H2O(l)−[1ΔS298∘O2(g)+2ΔS298∘H2(g)][2 mol(70.0mol KJ)]−[1 mol(205.2mol KJ)+2 mol(130.7mol KJ)]=−326.6 J/K
ΔS298∘===∑νΔS298∘(products)−∑νΔS298∘(reactants)[1ΔS298∘PbCl2(s)+1ΔS298∘H2(g)]−[1ΔS298∘Pb(s)+2ΔS298∘HCl(g)][1 mol(136.0mol KJ)+1 mol(130.7mol KJ)]−[1 mol(64.81mol KJ)+2 mol(186.9mol KJ)]=−171.9 J/K
ΔS298∘===∑νΔS298∘(products)−∑νΔS298∘(reactants)[1ΔS298∘Cu(s)+1ΔS298∘ZnSO4(s)]−[1ΔS298∘Zn(s)+1ΔS298∘CuSO4(s)][1 mol(33.15mol KJ)+1 mol(110.5mol KJ)]−[1 mol(41.6mol KJ)+1 mol(109.2mol KJ)]=−7.2 J/K
4. The reaction is
C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g)
ΔS298∘===∑νΔS298∘(products)−∑νΔS298∘(reactants)[3ΔS298∘CO2(g)+4ΔS298∘H2O(g)]−[1ΔS298∘C3H8(g)+5ΔS298∘O2(g)][3 mol(213.6mol KJ)+4 mol(69.91mol KJ)]−[1 mol(269.9mol KJ)+5 mol(205.03mol KJ)]=100.6 J/K
6.
ΔS298∘ for each change is as follows:
ΔSsys∘ΔSsys∘ΔSsys∘===∑νΔS298∘(products)−∑νΔS298∘(reactants)(2ΔS298∘NH3)−(1ΔS298∘N2+3ΔS298∘H2)[2 mol(192.8mol KJ)]−[1 mol(191.6mol KJ)+3 mol(130.7mol KJ)]=−198.1 J/K
ΔSsys∘ΔSsys∘ΔSsys∘===∑νΔS298∘(products)−∑νΔS298∘(reactants)(2ΔS298∘N2O5)−(1ΔS298∘N2+25×ΔS298∘O2)[1 mol(355.7mol KJ)]−[1 mol(191.6mol KJ)+25mol(205.2mol KJ)]=−348.9 J/K
8. The process is
NaCl(s)⟶NaCl(l) . At 500 °C, the following is true:
ΔSuniv=ΔSsys+Tqsurr=(95.06−72.11)mol⋅ KJ+500+273.15−27.95×103molJ=−13.2mol⋅ KJ
At 700 °C, the following is true:
ΔSuniv=ΔSsys+Tqsurr=(95.06−72.11)mol⋅ KJ+700+273.15−27.95×103molJ=−5.8mol⋅ KJ
As Δ
Suniv < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem.
Glossary
second law of thermodynamics: entropy of the universe increases for a spontaneous process
standard entropy (S°): entropy for a substance at 1 bar pressure; tabulated values are usually determined at 298.15 K and denoted
S298∘
standard entropy change (ΔS°): change in entropy for a reaction calculated using the standard entropies, usually at room temperature and denoted
ΔS298∘
third law of thermodynamics: entropy of a perfect crystal at absolute zero (0 K) is zero
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