In the quest to identify a property that may reliably predict the spontaneity of a process, a promising candidate has been identified: entropy. Processes that involve an increase in entropy of the system (ΔS > 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:
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:
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 12.2.
The Second Law of Thermodynamics | |
---|---|
ΔSuniv > 0 | spontaneous |
ΔSuniv < 0 | nonspontaneous (spontaneous in opposite direction) |
ΔSuniv = 0 | at equilibrium |
Will Ice Spontaneously Melt? The entropy change for the process
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:
Suniv < 0, so melting is nonspontaneous (not spontaneous) at −10.0 °C.
At 10.00 °C (283.15 K), the following is true:
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?
Entropy is a state function, so ΔSfreezing = −ΔSmelting = −22.1 J/K and qsurr = +6.00 kJ. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.
One of the challenges of using the ΔSsystem + ΔSsurr>0 to determine the spontaneity of a process is that it requires measurements of the entropy change for the surroundings (ie. ΔSsurr term) which is often inconvenient to measure. An alternative equation can be derived which is based on only the system's properties as shown below.
Condition for spontaneity:
ΔSsurr = qrev/T. But, 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:
The first law requires that qsurr = −qsys, and at constant pressure qsys = ΔHsys. So, this expression may be rewritten as:
Multiplying both sides of this equation by −T, and rearranging yields the following:
For simplicity’s sake, the subscript “sys” will be omitted henceforth. Also, we introduce a new thermodynamic property, ΔG to represent ΔH−TΔS. Therefore, In a spontaneous process:
The free energy change (ΔG= ΔHsys−TΔSsys) is therefore a reliable indicator of the spontaneity of a process. It is directly related to the previously identified spontaneity indicator, ΔSuniv as ΔG=−TΔSuniv. Table 12.3 summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.
Relation between Process Spontaneity and Signs of Thermodynamic Properties | ||
---|---|---|
ΔSuniv > 0 | ΔG < 0 | spontaneous |
ΔSuniv < 0 | ΔG > 0 | nonspontaneous |
ΔSuniv = 0 | ΔG = 0 | at equilibrium |
Gibb's Free energy is a thermodynamic property that can be interpreted as the energy available to do work. It is mathematically expressed as:
Under standard conditions (1atm and 25°C):
Free energy is a state function, and at constant temperature and pressure, the Standard free energy change (ΔG) may be expressed as the following:
where ν represents the stoichiometric coefficients, ΔGf
Using Standard Free Energies of Formation to Calculate ΔG°consider the decomposition of yellow mercury(II) oxide.
Calculate the standard free energy change at room temperature, using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?
Solution The required data are available in Appendix G and are shown here.
Compound | |||
---|---|---|---|
HgO (s, yellow) | −58.43 | −90.46 | 71.13 |
Hg(l) | 0 | 0 | 75.9 |
O2(g) | 0 | 0 | 205.2 |
(a) Using free energies of formation:
(b) Using enthalpies and entropies of formation:
Both ways to calculate the standard free energy change at 25 °C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (not spontaneous) at room temperature.
Check Your Learning Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?
(a) 140.8 kJ/mol, nonspontaneous
(b) 141.5 kJ/mol, nonspontaneous
As was previously demonstrated in this chapter’s section on entropy, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:
The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:
These four scenarios are summarized in Figure 12.11.
Predicting the Temperature Dependence of Spontaneity The incomplete combustion of carbon is described by the following equation:
How does the spontaneity of this process depend upon temperature?
Solution Combustion processes are exothermic (ΔH < 0). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, ΔS > 0). The reaction is therefore spontaneous (ΔG < 0) at all temperatures.
Check Your Learning Popular chemical hand warmers generate heat by the air-oxidation of iron:
How does the spontaneity of this process depend upon temperature?
ΔH and ΔS are negative; the reaction is spontaneous at low temperatures.
The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, Suniv > 0. If ΔSuniv < 0, the process is nonspontaneous, and if ΔSuniv = 0, the system is at equilibrium.
Gibbs free energy (G) is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A negative value for ΔG indicates a spontaneous process; a positive ΔG indicates a nonspontaneous process; and a ΔG of zero indicates that the system is at equilibrium. A number of approaches to the computation of free energy changes are possible.