We now define two more energies. The definition of the Helmholtz energy, A, is
A;U2TS (4.5)
The infinitesimal dA is therefore equal to
dA 5dU2T dS2S dT which becomes, for a reversible process,
dA5 2S dT2p dV
where we have used the definition of dU and the entropy for a reversible process as substitutions. Parallel to the above conclusions regarding dU and dH, their natural variables, and spontaneity, we state that the natural variables of A are T and V, and that for an isothermal, isochoric process,
(dA)T,V#0 (4.6)
is sufficient to ensure the spontaneity of a process. Again, the “equal to” part of the sign applies to processes that occur reversibly. This definition has some application, because some chemical and physical processes do occur under conditions of con- stant volume (for example, bomb calorimetry).
We also define the Gibbs energy, G, as
G;H2TS (4.7)
The infinitesimal dG is
dG5dH2T dS2S dT
Substituting for the definition of dH and again assuming a reversible change, we get dG5 2S dT1V dp
This equation implies certain natural variables, namely, T and p, such that the following spontaneity condition is
(dG)T,p#0 (4.8)
This is the spontaneity condition we have been looking for! We therefore make the following, perhaps premature, statements. For a system, under conditions of con- stant pressure and temperature, for any process,
If DG,0: the process is spontaneous
If DG.0: the process is not spontaneous (4.9) If DG50: the system is at equilibrium
Because G (and A) are state functions, these statements reflect the fact that
edG5 DG, not G.
The state functions U, H, A, and G are the only independent energy quantities that can be defined using p, V, T, and S. It is important to note that the only type of work we are considering at this point is pressure-volume work. If other forms of work are performed, then they must be included in the definition of dU. (Usually, they appear as dwnon-pV. We will consider one type of non-pV work in a later chapter.)
Furthermore, it must be understood that the condition DG,0 defines only spontaneity, not speed. A reaction may be thermodynamically favorable but might proceed at a snail’s pace. For example, the reaction
2H2 (g)1O2 (g) h 2H2O (,)
has a very negative DG. However, hydrogen gas and oxygen gas can coexist in an isolated system for millions of years before all of the reactant gas has converted
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4.3 | The Gibbs Energy and the Helmholtz Energy 105
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into liquid water. At this point, we cannot address the speed of the reaction. We can address only whether it can occur spontaneously.
The Helmholtz energy is named after the German physician and physicist Hermann Ludwig Ferdinand von Helmholtz (Figure 4.1). He is known for the first detailed, specific enunciation of the first law of thermodynamics in 1847. The Gibbs energy is named for Josiah Willard Gibbs, an American mathematical physicist (Figure 4.2).
In the 1870s, Gibbs took the principles of thermodynamics and applied them mathematically to chemical reactions. In doing so, Gibbs established that the thermo- dynamics of heat engines was also applicable to chemistry.
The usefulness of the Helmholtz energy, A, can be demonstrated by starting with the first law:
dU5dq1dw Because dS$dq/T, we can rewrite the equation above as
dU2T dS#dw
If dT 50 (that is, for an isothermal change), this can be written as d(U2TS)#dw
Because the quantity inside the parentheses is the definition of A, we can substitute:
dA#dw which we integrate to get
DA#w (4.10)
This says that the isothermal change in Helmholtz energy is less than or, for reversible changes, equal to the work done by the system on the surroundings.
Because work done by the system has a negative value, equation 4.10 means that the DA of an isothermal process is the maximum amount of work a system can do on the surroundings. The connection between work and the Helmholtz energy is the reason that Helmholtz energy is represented by A. It comes from the German word Arbeit, meaning “work.”
A similar expression can be derived for the Gibbs energy, but using a slightly dif- ferent understanding of work. So far, we have always discussed work as pV work, work performed by expanding gases against external pressures. This is not the only kind of work. Suppose we define a sort of work that is non-pV work. We can write the first law of thermodynamics as
dU5dq1dwpV1dwnon-pV
Making the same substitution for dS$dq/T, and also substituting for the defini- tion of pV work, we get
dU1p dV2T dS#dwnon-pV
If temperature and pressure are constant (the crucial requirements for a useful G state function), then we can rewrite the differential as
d(U1pV2TS)#dwnon-pV
U1pV is the definition of H. Substituting:
d(H2TS)#dwnon-pV
Also, H 2 TS is the definition of G:
dG#dwnon-pV which we can integrate to get
DG#wnon-pV (4.11)
F i g u r e 4 . 1 He r m an n Lu dw i g Ferdinand von Helmholtz (1821–1894), German physicist and physiologist.
In addition to studying various aspects of physiology including sight and hearing, Helmholtz made important contributions to the study of energy. He was one of the first people to clearly enunciate what became the first law of thermodynamics.
© Corbis
Figure 4.2 Josiah Willard Gibbs (1839–1903), American physicist. Gibbs applied the mathematics of thermody- namics to chemical reactions in a rigorous fashion, thereby extending the applicabil- ity of thermodynamics from engines to chemistry. However, his work was so much ahead of his contemporaries that it took almost 20 years for his contributions to be recognized.
© Interfoto/Alamy
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106 Chapter 4 | Gibbs Energy and Chemical Potential
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That is, when non-pV work is performed, DG represents a limit. Again, because work performed by a system is negative, DG represents the maximum amount of non-pV work a system can perform on the surroundings. For a reversible proc- ess, the change in the Gibbs energy is equal to the non-pV work of the process.
Equation 4.11 will become important to us in Chapter 8, when we discuss electro- chemistry and electrical work.
The “T dS” term in the definition of dA and dG represents the amount of inter- nal energy or enthalpy that is tied up in random molecular energy states, and so is not available to perform work of any sort, whether it be pressure-volume work or non-pressure-volume work. (Recall that we previously defined heat as disordered en- ergy transfer while work is defined as ordered energy transfer.) It is also why A and G are sometimes referred to as the Helmholtz free energy and the Gibbs free energy, respectively (this terminology is commonly used for G but less common for A, but both are discouraged by IUPAC). It is the amount of energy that is “free” to do work.
This is what equations 4.10 and 4.11 imply.
Calculate the change in the Helmholtz energy for the reversible isothermal compression of 1 mole of an ideal gas from 100.0 L to 22.4 L. Assume that the temperature is 298 K.
Solution
The process described is the third step in a Carnot-type cycle. Because the process is reversible, the equality relationship DA 5 w applies. Therefore, we need to calculate the work for the process. The work is given by equation 2.7:
w5 2nRT ln Vf Vi
Substituting for the various values:
w5 2(1 mol)a8.314 J
mol#Kb(298 K) ln 22.4 L 100.0 L w53710 J
Because for this reversible process DA 5 w, we have DA 5 3710 J ExamplE 4.2
A battery performs 776 J of work on a circuit. What state function is related to this work, and what is the value of the state function?
Solution
Electrical work is a form of non-pV work. As such, it is related to DG. Thus, according to equation 4.11,
DG #2776 J for the process.
ExamplE 4.3
Work done by a system is negative, whether the work is pV or non-pV.
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Because many processes can be made to occur isothermally (or at least re- turned to their original temperatures), we can develop the following expressions for DA and DG:
A5U2TS
dA5dU2T dS 2S dT
dA5dU2T dS for an isothermal change or, integrating:
DA5 DU2T DS (4.12)
Similarly, for the Gibbs energy:
G5H2TS
dG5dH2T dS2S dT
dG5dH2T dS for an isothermal change We integrate to get
DG5 DH2T DS (4.13)
Both equations 4.12 and 4.13 are for isothermal changes in the system. They also allow us to calculate DA or DG if changes in other state functions are known.
Just as we can determine DU, DH, and DS for chemical processes using a Hess’s- law approach, we can also determine DG and DA values for chemical reactions using a products-minus-reactants scheme. Because DG is a more useful state function, we focus on that. We define Gibbs energies of formation DfG similarly to the enthalpies of formation, and tabulate those. If the DfG values are determined at standard thermo- dynamic conditions, we use the ° superscript and label them DfG°. We can then deter- mine the DG of a reaction, DrxnG, just like we did the enthalpies of reactions. However, with DG we have two ways to calculate the Gibbs energy change for a reaction.
We can use the DrxnG values and a products-minus-reactants approach, or we can use equation 4.13. The choice depends on the information given (or the information you are able to get). Ideally, both approaches should give you the same answer.
Note that the above paragraph implies that DfG for elements in their standard states is exactly zero. The same is true for DfA. This is because a formation reac- tion is defined as the formation of a chemical species from its constituent chemical elements in their standard states.
ExamplE 4.4
Determine DrxnG (25°C 5 298.15 K) for the following chemical reaction using the two methods mentioned above for determining DrxnG, and show that they yield the same answer. Assume standard conditions. Appendix 2 in the back of the book lists the various thermodynamic data.
2H2 (g) 1 O2 (g) h 2H2O (,) Solution
The following data were obtained from Appendix 2:
H2 (g) O2 (g) H2O (,)
DfH, kJ/mol 0 0 2285.83
S, J/mol#K 130.68 205.14 69.91
DfG, kJ/mol 0 0 2237.13
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108 Chapter 4 | Gibbs Energy and Chemical Potential
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We begin by calculating DrxnH:
DrxnH52(2285.83)2(2 # 011 # 0) DrxnH5 2571.66 kJ
Now, we calculate DrxnS:
DrxnS 5 2(69.91)2(2 # 130.681205.14)
The 2s are from the stoichiometry of the balanced chemical reaction. We get DrxnS 52326.68 J/K
In combining DrxnH and DrxnS, we need to make the units compatible. We convert DrxnS into kilojoule-containing units:
DrxnS 520.32668 kJ/K Using equation 4.13, we calculate DrxnG:
DG5 DH2T DS
DG5 2571.66 kJ2(298.15 K)(20.32668 kJ/K)
Notice that the K temperature units cancel in the second term. Both terms have the same units of kJ, and we get
DG 52474.26 kJ
using equation 4.13. Using the idea of products-minus-reactants, we use the DfG values from the table to get
DrxnG52(2237.13)2(2 # 010) kJ DrxnG5 2474.26 kJ
This shows that either way of evaluating DG is appropriate.
ExamplE 4.4 (continued)
This entropy change is reasonable when going from 3 moles of gas to 2 moles of liquid.
Convert to kJ units by dividing by 1000.
The method you use may depend on the data available.