In the Classroom
A Mechanical Analogue for Chemical Potential, Extent of Reaction, and the Gibbs Energy Samuel V. Glass and Roger L. DeKock Department of Chemistry and Biochemistry, Calvin College, Grand Rapids, MI 49546-4388
Abstract concepts are often introduced to students with an analogy or a physical model that they understand from ordinary experience. Chemical equilibrium is one such concept for which numerous analogies and models have been published in this Journal (1–3). The majority of these analogies provide the student with a model that elucidates the dynamic nature of chemical equilibrium and the quantitative relationship between reactants and products. These models are generally aimed at the secondary school student. But what about the university student who is studying equilibrium in greater depth? As the concepts become more abstract, a physical analogy would be helpful. Chemical potential (4–7) is a foundational concept in the thermodynamics of phase stability and transitions, the mixing of gases, the colligative properties of solutions, and chemical equilibrium. Hence it is an important concept for students to understand. Before we present our analogy, we provide a short review of the chemical potential as it is currently found in textbooks. The abstract nature of this concept is readily apparent. The chemical potential is formally defined as follows (8):
µi = ∂G ∂n i
T , p, n j
This equation has been given several interpretations. “The chemical potential of a component of a phase is thus the amount by which the capacity of the phase for doing work (other than work of expansion) is increased per unit amount of substance added, for an infinitesimal addition at constant temperature and pressure” (9). Another textbook states, “µi may be thought of as the increase in the free energy of the system when one mole of component i is added to an infinitely large quantity of the mixture so that it does not significantly change the overall composition” (10). Our mechanical analogue provides insight into the concepts of chemical potential, µ, extent of reaction, ξ (6, 11– 13), and reaction Gibbs energy, ∆rG (6, 12, 13). The analogy relates the one-dimensional mechanical equilibrium of a rigid block between two Hooke’s law springs (14) and the chemical equilibrium of two perfect gases (6, 13). The concept of chemical potential usually is not introduced until a course in physical chemistry. We have built a physical model of our mechanical analogue (see Diagram 4) out of ordinary materials. This has been useful for us to introduce concepts of chemical equilibrium to first-year and upperclass students. We first review the standard treatment for the equilibrium between two perfect gases. Then we present our analogue. Finally, we summarize the close parallelisms between the two. Chemistry: Two Perfect Gases Consider the equilibrium1 A(g)
B(g)
Assume that total pressure p and temperature T are con190
stant and that the gases A and B behave according to the perfect gas equation, pV = nRT, where p = pA + pB. A and B are confined to volume V. We define ξ to be the extent of reaction, which is 0.0 mol when only A is present and 1.0 mol when only B is present. Start with 1.0 mol of pure A; as ξ increases, we have (1 – ξ) mol of A and ξ mol of B. nA = (1 – ξ ') n = ξ'
nA = 1 nB = 0
nA = 0 nB = 1
B
0
ξ'
ξ
1
Diagram 1
The partial pressures2 are: pA = (1 – ξ) p; pB = ξp. We define the positive direction of the reaction to correspond with A → B (i.e., to the right in Diagram 1). If an infinitesimal amount dξ mol of A is converted into B at constant T and p then dnA is negative and dnB is positive: dξ = dnB = { dnA. We now introduce the chemical potential. µ i0 is the standard chemical potential of component i (i.e., pure i at 1 bar). µi is the chemical potential of i under nonstandard conditions.
µi = µ0i + RT ln
pi p0
where p0 = 1 bar. The Gibbs energy function is: G = nAµA + n Bµ B
(1)
G = (1 – ξ ) µ A + ξ µB
(2)
or The change in Gibbs energy for a change in composition can be expressed as dG = µ AdnA + µ Bdn B = { µAd ξ + µBd ξ This becomes
∂G ∂ξ
= µA – µB
(3)
T, p
The reaction will proceed until G reaches a minimum value.
Equilibrium At equilibrium,
∆rG = ∂G ∂ξ
=0 T, p
Combining this with eq 3, we obtain µA = µB. Solution for ξ at equilibrium: 0
∆rG = µ0B – µ0A = {RT ln
pB pA
Journal of Chemical Education • Vol. 75 No. 2 February 1998 • JChemEd.chem.wisc.edu
eq
In the Classroom where
Physics: Rigid Block and Two Springs
pB pA
= Kp = eq
ξeq 1 – ξeq
Thus we obtain
ξeq 1 – ξeq
= e{
µ 0B – µ 0A / RT
(4)
Comments 1. We can define a standard state of pressure p0 for all gases. Then µ i0 depends on T but not p.
Consider two Hooke’s law springs A and B attached to a rigid block of unit length. The other ends of the springs are attached to two pins separated by a distance of lA0 + lB0, where lA 0 and lB0 are fixed as the lengths of the springs in their relaxed state and lA 0 = lB0. In Diagram 4 we label relevant lengths.3 Assume the surface is level and frictional. Springs A and B have force constants kA and kB and behave according to F i = ki(li0 – li ) where li is the variable length of spring i and Fi is the force that spring i exerts on the block. Springs A and B are confined by the block and the pins. The length li will vary depending upon the horizontal position of the block. We define ci to be (li0 – li ), which corresponds to the change in length of spring i as a result of compression.
2. µi = µ i0 only at standard conditions (pure i at 1 bar). As ξ increases at constant total pressure, while the partial pressures change, the chemical potential of component i changes:
µi = µ0i + RT ln
1
A
l
lB0
A0
pi p0
lA
µA = µ0A + RT ln 1 – ξ + RT ln
p p0
(5)
p p0
cA
lB
cB
Diagram 4
3. We can plot the dependence of µA and µB on ξ:
µB = µ0B + RT ln ξ + RT ln
B
(6)
The point at which µA = µB corresponds to equilibrium.
We define x to be the distance from the midpoint of the two pins to the right edge of the block. Then (1 – x) corresponds to the distance from the midpoint of the two pins to the left edge of the block. We shall be concerned with values of x between 0 and 1, where the midpoint of the pins is contained by the block. Notice the values of cA , cB, and x for each of the configurations. A
µA
B
a) cA = 1
cB = 0
x = 0
µB
b) 0
ξ
ξeq
c = 0 A
1
c)
Diagram 2
c = 1– x’ A
4. We can plot the dependence of the Gibbs energy G on ξ: 0
G = µ0A + ξ ∆rG + RT ln
c = 1 B
x = 1
p + RT 1 – ξ ln 1 – ξ + ξ ln ξ p0
The point at which G is a minimum, or (∂G/∂ξ)T, p = 0, corresponds to equilibrium.
G
x = x'
cB = x '
Diagram 5
We define the positive direction in Diagrams 4 and 5 to be to the right. For an infinitesimal change dx in position of the block, dcA is negative and dcB is positive: dx = dcB = { dcA . We now introduce the spring forces. The force that spring A exerts on the block is FA = k Ac A = kA(1 – x)
(7)
FB = kBcB = kBx
(8)
Likewise,
0
ξ
ξ eq
1
Diagram 3
5. To solve for ξ eq we need to know only the ratio (µB0 – µA0)/RT (eq 4). That is, we need to know the relative standard chemical potentials of A and B and the temperature.
FA and F B will be considered positive, with FA to the right and FB to the left. We define the system to be the block, the two springs, and the two pins. The potential energy function is:4 2 V = 1 kA 1 – x + 1 kB x2 2 2
(9)
or
JChemEd.chem.wisc.edu • Vol. 75 No. 2 February 1998 • Journal of Chemical Education
191
In the Classroom
V = 1 1 – x F A + xF B 2
(10)
The change in potential energy for a change in position of the block can be expressed as dV = kAcAdcA + kBcBdcB = { F Adx + FBdx This becomes
dV = F – F B A dx
(11)
The block will come to rest at a position of minimum V.
Equilibrium At equilibrium, dV/dx = 0. Combining this with eq 11, we obtain F A = FB . The solution for x at equilibrium is kA(1 – xeq) = kBxeq or
x eq 1 – x eq
=
kA kB
(12)
Comments 1. We can define a standard state for all springs, compressed by length c0 and exerting force Fi0 = ki c0. Then Fi0 depends on ki but not ci. (Assuming an ideal spring, ki is independent of T and p.) 2. Fi = F i0 only at standard conditions (ci = c0). As x changes and the lengths of the springs change, the force exerted by spring i changes: Fi = F 0i + k i(ci – c0). 3. We can plot the dependence of FA and FB on x with eqs 7 and 8. The point at which F A = FB corresponds to equilibrium. kA
FA kB
FB
x eq
0
1
x Diagram 6
4. We can plot the dependence of the potential energy V on x with eq 9. The point at which V is a minimum, or dV/dx = 0, corresponds to equilibrium.
V
0
xeq
x
1
Diagram 7
5. To solve for xeq we need to know only the ratio kA /k B (eq 12). That is, we need to know the relative strengths of the two springs.
192
Force and Chemical Potential Chemical potential “can be regarded as providing the force which drives chemical systems to equilibrium” (10). Just as the two springs come to an equilibrium position at which their forces are balanced, so the two reacting gases come to an equilibrium at which their chemical potentials are equal. In other words, if at some point x′, F A > F B or dV/dx < 0, then the block will move towards spring B; likewise, if at some stage of the reaction ξ ′, µA > µB or (∂G/∂ ξ)T,p < 0, then the reaction A → B is spontaneous. The extent of reaction, ξ, is correlated to the composition of the reaction mixture, or the relative amounts of reactants A and products B. The chemical potentials µA and µB are dependent on ξ. Similarly, x is correlated to the configuration of the mechanical system, or the position of the block and the lengths of springs A and B. The spring forces FA and FB are dependent on x. It is worthwhile to point out that both force and chemical potential are gradients. All potentials are gradients, whether they be electrical potential, gravitational potential, coulombic potential, or chemical potential. All these potentials refer to the variation of energy with respect to variation in some quantity. For example, in the case of gravitational potential it is the variation in energy with respect to the distance from some point of reference. In the case of a Hooke’s law spring, it is the variation of the energy with respect to the amount of extension or compression from the relaxed state (the point of reference). In the case of the chemical potential the quantity being varied is the extent of reaction. Mechanical and Chemical Equilibrium In the study of mechanics, three types of equilibrium are distinguished (14, 15). Our two-spring system is an example of stable equilibrium. For values of x neighboring xeq, the net force is directed towards the equilibrium point so as to restore the system to equilibrium; V is a minimum at xeq. An example of unstable equilibrium is the teeter-totter or seesaw model (2), in which the forces experienced at points near the equilibrium point are directed away from the position of equilibrium.5 In other words, V is a maximum. An example of neutral equilibrium, in which all points near an equilibrium point are also equilibrium positions, is a sphere on a level surface experiencing no net force (15). V is constant in this case. Of these three, chemical equilibrium is analogous to stable equilibrium. We point out four differences between mechanical and chemical equilibrium. Probing these differences helps us to understand the chemical concepts at a deeper level. First, chemical equilibrium is dynamic: the reactions A → B and A ← B continue at equilibrium at a constant rate and the relative proportions of A and B remain constant. Our example of mechanical equilibrium, on the other hand, is static.6 Second, the two-spring system involves oscillations of the block, which are damped by frictional forces until equilibrium is reached. However, we could choose a surface whose frictional properties damp the body so that it does not oscillate at all. In fact, frictional forces can be used as a mechanical analogue for the slow rate at which some chemical reactions reach equilibrium. Chemical equilibrium in a closed system (such as we have been considering) does not involve oscillations during the approach to equilibrium. Chemicals don’t “overshoot” the equilibrium position. An open system, however, may involve oscillations (16). Third, while F A and F B vary directly with x, µA and µB vary logarithmically with ξ (eqs 5–8). Fourth, the Gibbs energy is
Journal of Chemical Education • Vol. 75 No. 2 February 1998 • JChemEd.chem.wisc.edu
In the Classroom Table 1. A Summary of the Strong Relationship between the Chemistry and the Mechanical Analogue Chemistry
Physics
V = 1 1 – x F A + xF B 2
G = 1 – ξ µA + ξ µB ∂G ∂ξ
= µB – µA
dV = F – F B A dx
T, p
∆rG = ∂G ∂ξ
dV = 0 at equilibrium dx
= 0 at equilibrium T, p
µA = µB at equilibrium ξeq 1 – ξeq µi = ∂G ∂n i
= e{
F A = F B at equilibrium x eq
µ 0B – µ 0A / RT
1 – x eq F i = ∂V ∂c i
T , p, n j
given (17) by G = H – TS. Chemical systems are driven toward equilibrium both by the tendency to minimize their enthalpy and by the tendency to maximize their entropy. We find no such parallel in mechanical potential energy. In conclusion, we believe that our analogy elucidates the concepts of the chemical potential, the extent of reaction, and the reaction Gibbs energy by comparison with the spring force, the position of the block, and the mechanical potential energy. We summarize the strong parallelisms of our results in Table 1. Notes 1. An example is the equilibrium between two isomers such as n-butane and isobutane (6 ). Note that such an equilibrium need not be reached rapidly at ambient temperature, and in this case it certainly is not. 2. These formulas are deceivingly simple. They should show that the quantity (1 – ξ ) and ξ are divided by the total number of moles in the system, i.e., 1 mol. That is, the term multiplying the pressure is a mole fraction. Also, the number “1” in (1 – ξ ) has units of moles as well. Of course, extent of reaction ξ has units of moles. 3. The example can be worked with vectors, but for the sake of simplicity and because vectors are not relevant to the analogy, we use scalars instead. 4. This potential energy function should not be mysterious to the student. In physics they have all studied a Hooke’s law spring, V = (1/2) kx 2 . We have defined the reference point for the potential energy, V = 0, such that each spring is isolated from the system in its relaxed state, where l i = l i 0.
=
kA kB
cj
5. In this example, the forces act parallel to a vertical axis, whereas the position of equilibrium is determined along the horizontal axis. 6. Chemical thermodynamics can be treated at the macroscopic level or at the molecular level. Some analogies are directed at the macroscopic level (e.g., the teeter-totter) and some at the molecular level (e.g., fish swimming between two connected tanks). Our analogue is meant to handle only the macroscopic aspect.
Literature Cited 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.
Laurita, W. J. Chem. Educ. 1990, 67, 598. Russell, J. M. J. Chem. Educ. 1988, 65, 871–872. Olney, D. J. J. Chem. Educ. 1988, 65, 696–697. Winn, J. S. Physical Chemistry; HarperCollins: New York, 1995; Chapters 5–7, 10. Atkins, P. W. Physical Chemistry, 5th ed.; Freeman: New York, 1994; Chapters 5–7, 9, 10. Smith, E. B. Basic Chemical Thermodynamics, 4th ed.; Oxford University Press: New York, 1993; pp 48–55. Denbigh, K. G. The Principles of Chemical Equilibrium, 4th ed.; Cambridge University Press: New York, 1981; pp 76–82. Winn, J. S. Op. cit., p 125. Denbigh, K. G. Op. cit., p 79. Smith, E. B. Op. cit., p 49. Winn, J. S. Op. cit., p 202. Treptow, R. S. J. Chem. Educ. 1996, 73, 51–54. Atkins, P. W. Op. cit., pp 272–276. Skinner, R. Mechanics; Blaisdell: Waltham, MA, 1969; p 339. Resnick, R.; Halliday, D.; Krane, K. S. Physics, 4th ed.; Wiley: New York, 1992; Vol. 1, pp 154–159, 303–304. Scott, S. K. Oscillations, Waves, and Chaos in Chemical Kinetics; Oxford University Press: New York, 1994; p 12. Winn, J. S. Op. cit., p 118.
JChemEd.chem.wisc.edu • Vol. 75 No. 2 February 1998 • Journal of Chemical Education
193
