In the Classroom
How Thermodynamic Data and Equilibrium Constants Changed When the Standard-State Pressure Became 1 Bar Richard S. Treptow* Department of Chemistry and Physics, Chicago State University, Chicago, IL 60628
Historically, the defined pressure for the standard state, i.e., the standard-state pressure, has been 1 standard atmosphere (101,325 Pa) and most existing data use this pressure. With the growing use of SI units continued use of the atmosphere is inconvenient and in some countries now illegal. It is recommended that thermodynamic data should be reported for a defined standard-state pressure of 10 5 Pa which is equal to 1 bar. Physical Chemistry Division Commission on Thermodynamics International Union of Pure and Applied Chemistry, 1982
In 1982 the IUPAC Commission on Thermodynamics proposed that the pressure used to define the standard state of a substance be changed from 1 atm to 1 bar (1). The new standard was favored because 1 bar is exactly 100,000 pascals, and the pascal is the SI unit for pressure. Any change in standardstate pressure would necessitate adjustments in the large volume of thermodynamic data published in the literature. The changes required in this case would be small, however, because the two pressures are nearly equal (1 atm = 1.01325 bar). Most thermochemists immediately accepted the new convention, and major compilations of thermodynamic data published since 1982 use the 1-bar standard state (2–7). Chemical educators, on the other hand, have been slow to abandon the old standard. Most general chemistry textbooks still define standard-state pressure as 1 atm. For consistency the thermodynamic data published in these texts are typically taken from the older literature. The reluctance to change is understandable. Although it is not an SI unit, the atmosphere is popular nonetheless. Students easily grasp its meaning, since it is based on a familiar physical entity. Furthermore, 1 atm continues to be used to define STP, the so-called standard temperature and pressure of gases. The IUPAC commission itself specified that its recommendation was not intended to apply to other “standard pressures”. For example, the normal boiling point of a liquid would continue to be defined as the temperature at which the vapor pressure of the liquid is 1 atm. This paper addresses the current dilemma that two different standard-state pressures are in common use. It asks how the choice of pressure affects thermodynamic data, equilibrium constants, and the results of equilibrium calculations. Finally, it concludes with some suggestions to chemical educators on how to manage the current state of affairs. A more advanced discussion of the 1-bar standardstate pressure appeared in this Journal shortly after it was first proposed (8). Effect on Thermodynamic Data The thermodynamic properties of a substance of most interest to a chemist are its enthalpy, entropy, and free energy. Let us discuss how each of these depends upon pressure. *Email:
[email protected].
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Enthalpy, Entropy, and Free Energy of Substances Figure 1 illustrates a system that contains a solid, a liquid, and a gas at each of the two pressures commonly used to define the standard state. The system has a movable lid, which adjusts to maintain its internal pressure. On the left the pressure is 1 atm, and on the right it is 1 bar. The slightly lower pressure of 1 bar is accomplished by hanging a small weight from a string so that it exerts a slight upward force on the lid. Each of the three substances can be said to be in its standard state if we further specify that the solid and liquid are pure substances and that the gas is ideal. Throughout our discussions the temperature of the system will be held constant. How does the difference in pressure affect the thermodynamic properties of the substances? An examination of data for solids and liquids reveals that their enthalpies, entropies, and free energies change only very slightly with pressure (2, 8). When the pressure changes from 1 atm to 1 bar these quantities change by amounts that are always much smaller than the uncertainty in the measurements themselves. Hence, for any condensed phase we can approximate H bar = H atm S
bar
G bar
(2)
G atm
(3)
=S =
(1)
atm
where, for example, the symbol H bar is the enthalpy of the substance at 1 bar. The gas in our system requires greater attention. Recall that enthalpy is defined as H = E + PV. Any change in the enthalpy of a substance resulting from a change in its pressure is ∆H = ∆E + ∆(PV ). For an ideal gas at fixed temperature both E and PV are constant. Thus, ∆H = 0 for the pressure change. In other words, the enthalpy of the gas is independent of pressure, and eq 1 applies to gases as well as solids and liquids. The entropy of the gas, on the other hand, is affected by pressure. When the pressure of the gas changes from P1 to P2 its entropy change is ∆S = nR ln(P1/P2). For our system P1 = 1 atm = 1.01325 bar and P2 = 1 bar. Therefore S bar = S atm + nR ln 1.01325
(4)
We see that the entropy, or disorder, of a gas is slightly greater at the lower pressure. This should be expected, since the gas occupies a slightly greater volume. Our knowledge of ∆H and ∆S when the pressure of the gas changes can be combined with the relationship ∆G = ∆H – T∆S to determine the difference in free energy. The result is G bar = G atm – nRT ln 1.01325
(5)
Hence, the free energy of a gas, or its ability to do useful work, is slightly less at the lower pressure. Table 1 summarizes the differences in the enthalpy, entropy, and free energy of substances at the two pressures.
Journal of Chemical Education • Vol. 76 No. 2 February 1999 • JChemEd.chem.wisc.edu
In the Classroom
∆ H, ∆ S, and ∆ G for Reactions The principles developed above can be applied to determine the effect of pressure on the thermodynamic properties of reactions. Consider, for example, the vaporization of water: H2O(ᐉ) → H2O(g)
Figure 1. A solid, liquid, and gas at each of the pressures commonly used to define the standard state.
Figure 2 shows this simple reaction occurring at each of the standard pressures. We can calculate ∆Hvap , ∆Svap , and ∆G vap at either pressure by use of appropriate data from the literature.1 The first line of Table 2 shows the results obtained using data based upon the 1-atm standard and at a temperature of 298.15 K. Let us next convert these values to the 1-bar standard. ∆H vap does not depend on pressure, since the enthalpy of both the reactant and product are taken to be pressure independent. On the other hand, ∆Svap will increase by nR ln 1.01325 at the new pressure, because the entropy of the H2O(g) produced is greater by this amount. This increase is 0.1094 J/K for one mole of gas, since R = 8.3143 J/K ⭈ mol. Similarly, ∆Gvap will decrease by nRT ln 1.01325, which is 0.03263 kJ for one mole of gas at 298.15 K. The second line of Table 2 lists the 1-bar values for ∆Hvap , ∆Svap, and ∆Gvap. Incidentally, since ∆G vap > 0 at either pressure the results affirm that the vaporization of water is nonspontaneous at 298.15 K. The example above illustrates the procedure for converting the thermodynamic properties of a reaction from one standardstate pressure to the other. The adjustment only requires knowledge of the amount of gas produced or consumed. The conversion formulas are ∆H bar = ∆H atm
(6)
∆S bar = ∆S atm + ∆nR ln 1.01325
(7)
∆G bar = ∆G atm – ∆nRT ln 1.01325
(8)
where ∆n is the change in the number of moles of gas in the reaction. If a reaction results in an increase in the amount of gas, its entropy change will be slightly greater and its free energy change slightly less at 1 bar.
Figure 2. The reaction H2O(ᐉ) → H2O(g) at each of the pressures commonly used to define the standard state.
Table 1. Effect of Pressure on the Enthalpy, Entropy, and Free Energy of Substances Substance
H b a r – H a tm
Solid or liquid
0
Ideal gas
0
S b a r – S a tm
G b a r – G a tm
0
0
nR ln 1.01325 ᎑ nRT ln 1.01325
Table 2. Thermodynamic Properties for the Reaction H2 O (ᐍ ᐍ) → H2 O(g) at 298.15 K for Pressures Used To Define the Standard State ∆Hv a p / k J ∆Sv a p /J K᎑1 ∆Gv a p / k J Pressure 1 atm
44.012
118.81
8.590
1 bar
44.012
118.92
8.557
Converting Thermodynamic Data in Tables Textbooks typically present tables of thermodynamic properties of substances. The properties most commonly tabulated are ∆Hf°, S°, and ∆Gf°, where the subscript indicates the formation reaction for the substance and the superscript implies one or the other standard state. When such tables are constructed or when values are taken from them, it may be necessary to convert data from one standard-state pressure to the other. The conversions can be accomplished by use of eqs 2, 4, 6, and 8. Table 3 lists thermodynamic properties for a few selected substances based upon both pressures.2 The table gives the value of ∆nf , the change in the number of moles of gas in the formation reaction. The properties affected by the choice of pressure are S° for all gases and ∆Gf° for all substances whose ∆n f is not zero. Even in these cases the effects are small. Effect on Equilibrium Constants Perhaps the greatest value of thermochemical data is that they can be used to determine equilibrium constants for reactions. How does the choice of standard-state pressure
JChemEd.chem.wisc.edu • Vol. 76 No. 2 February 1999 • Journal of Chemical Education
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In the Classroom Table 3. Thermodynamic Data for Selected Substances at 298.15 K for Pressures Used To Define the Standard State ∆Hf ° /k J mol ᎑1 ∆G f° / k J mol ᎑1 S ° /J K᎑1 mol ᎑1 ∆nf /mol Substance a tm bar a t m b a r ∆Hf or ∆Hf ∆G f a t m ∆G f b a r S S H 2 O(ᐉ)
᎑285.830
69.91
69.91
᎑237.178
᎑237.129
᎑1.5
H 2 O(g)
᎑241.818
188.716
188.825
᎑228.588
᎑228.572
᎑0.5
0
191.50
191.61
0
0
0
0
130.575
130.684
192.34
192.45
N 2 (g) H 2 (g)
᎑46.11
NH3 (g) I 2 (g)
62.438
260.58
260.69
H I(g)
26.48
206.485
206.594
0
0
0
᎑16.48
᎑16.45
᎑1.0
19.360
19.327
1.0
1.72
1.70
0.5
Na 2 SO4 (s)
᎑1387.08
149.58
149.58
᎑1270.23
᎑1270.16
᎑2.0
Na2 SO4 ⴢ10 H2 O(s)
᎑4327.26
592.0
592.0
᎑3647.40
᎑3646.85
᎑17.0
affect these constants and the partial pressures of substances at equilibrium calculated from them? Consider the equilibrium between liquid and gaseous water at 298.15 K as illustrated in Figure 3 and represented by the equation H2O(ᐉ)
H2O(g)
We will examine both its thermodynamic and practical equilibrium constants.
The Thermodynamic Equilibrium Constant The thermodynamic equilibrium constant can be calculated from the standard-state free energy change of the reaction. For example, if the standard pressure is 1 atm, we have the relationship ∆G atm = ᎑RT ln K atm
(9)
The value of ∆G atm in Table 2 for the vaporization of water can be used to give K atm = 3.127 × 10᎑2. The thermodynamic equilibrium constant is expressed in terms of activities: K atm = aH2 O(g) /aH2O(ᐉ)
(10)
If the H2O(g) present in the equilibrium system is taken to be an ideal gas, its activity is the dimensionless ratio PH2O(g) /1 atm, where PH2O(g) is the vapor pressure of water. The H2O(ᐉ) is present as a pure condensed phase. Hence, it is in its standardstate and its activity is 1 by definition. Thus K atm = PH2O(g) /1 atm
The Practical Equilibrium Constant Chemists define practical equilibrium constants, which are somewhat more convenient for calculating partial pressures or concentrations of species at equilibrium.3 These constants are generally not dimensionless, since they have the units of the standard-state incorporated into them. For the equilibrium under discussion the practical constant is defined as KP = PH2O(g)
where the symbol KP indicates the constant is based upon pressure units. Equations 11 and 13 combine to give KP = K atm ⭈ 1 atm
KP = K bar ⭈ 1 bar
(12) ᎑2
214
(15)
which gives KP = 3.169 × 10᎑2 bar. In effect, KP is either of the thermodynamic constants with the appropriate pressure unit affixed to it. The two KP values are identical, except for the obvious fact that they are expressed in different units. Not surprisingly, either can be used to calculate the vapor pressure of water cited above.
From this equation we can calculate the vapor pressure of water at 298.15 K. The result is PH2O(g) = 3.127 × 10 ᎑2 atm = 23.8 torr. If the standard pressure is 1 bar, a similar treatment leads to K bar = 3.169 × 10 ᎑2 and the relationship from which we obtain PH2 O(g) = 3.169 × 10 bar = 23.8 torr. Although K atm and K bar are numerically different, the vapor pressures calculated from them are identical. Equations 11 and 12 show that each constant is merely the dimensionless ratio of the vapor pressure to the standard-state pressure. Vapor pressure is, of course, the directly measurable property.
(14)
which yields KP = 3.127 × 10᎑2 atm. Similarly, eqs 12 and 13 give
(11)
K bar = P H2 O(g) /1 bar
(13)
Figure 3. The equilibrium H2O(ᐉ)
H 2O(g) at 298.15 K.
Journal of Chemical Education • Vol. 76 No. 2 February 1999 • JChemEd.chem.wisc.edu
In the Classroom
Table 4. Equilibrium Constants for Selected Reactions at 298.15 K Thermodynamic Constants
Reaction H 2 O(ᐉ )
K H2 O(g)
N 2 (g) ⫹ 3H2 (g) H 2 (g) ⫹ I2 (g)
2NH3 (g) Na 2 SO4 (s) ⫹ 10H2 O(g)
KP
3.169 × 10 ᎑2
5.96 × 10
5.81 × 10
5
1.012 × 10
᎑1 6
Table 4 summarizes the four equilibrium constants for the vaporization of water. The table also presents three other reactions and their constants. The value of ∆n is listed for each reaction. Note the relationship between ∆n and the unit on the practical equilibrium constant.
Converting Equilibrium Constants The previous section illustrates that a given reaction has only one equilibrium constant although it can be expressed in a variety of ways. The general relationship between K bar and K atm for any reaction is K bar = K atm ⭈ 1.01325 ∆n
Practical Constants
bar
3.127 × 10 ᎑2 6.171 × 10 2
2HI(g)
Na 2 SO4 ⴢ10 H2 O(s)
K
a tm
(16)
This equation can be used to convert thermodynamic equilibrium constants based on one standard-state pressure into the other. It can be applied, for example, to verify the data of Table 4. The equation affirms that the four constants are numerically equal for any reaction in which the number of moles of gas does not change. Summary and Recommendations The principal effect of the change in standard-state pressure from 1 atm to 1 bar is a slight increase the entropy of any gas in its defined standard state. The increase can be easily expressed and calculated, since gases are defined to be ideal in their standard state. The entropy increase causes a slight decrease in the free energy of the gas. The choice of standard-state pressure also affects the entropy change, free energy change, and thermodynamic equilibrium constant for any reaction involving a change the number of moles of gas. Practical equilibrium constants obtained by affixing units to thermodynamic equilibrium constants are equivalent for both pressures. Calculations of the partial pressures of gases at equilibrium give the same results regardless of the equilibrium constant used. This conclusion should be expected, because equilibrium conditions are not affected by our arbitrary choice of standard-state pressure. The current situation of two different standard-state pressures being in common use is unlikely to change in the near future. The appeal of the atmosphere as a unit suggests that the 1-atm standard will remain in introductory courses. Because the thermodynamic literature generally no longer reports data for this standard, general chemistry textbook authors will find it increasingly necessary to convert data from
5
3.127 × 10 ᎑2 atm
3.169 × 10 ᎑2 bar
5.96 × 10 atm
5.81 × 10 bar
5
6.171 × 10 2 1.155 × 10
᎑1 6
∆n
KP ᎑2
6.171 × 10 2 1.012 × 10
᎑1 6
5
᎑2
6.171 × 10 2
atm
10
1.155 × 10
᎑1 6
bar
1 ᎑2 0
10
10
1 bar to 1 atm. Admittedly, these conversions result in only small alterations in values. They should be made, nonetheless, in the interest of correctness. The 1-bar standard should surely be used in advanced courses where students are more likely to encounter the thermochemical literature and where they can fully appreciate the advantage of SI units. In the final analysis, it must be remembered that the choice of standard pressure is arbitrary, and either is acceptable if properly used. Notes 1. All quantities reported in this paper have been obtained directly or by calculation from data obtained from a single literature source (2). This was done for consistency and verifiability. 2. The substances included in Table 3 were chosen to provide the reader with data that can be used to verify the results of Tables 2 and 4. 3. Textbooks often do not distinguish thermodynamic and practical equilibrium constants. This practice leads to confusion in calculations of partial pressures or concentrations of species at equilibrium. If the dimensionless thermodynamic equilibrium constant is used for such a calculation, the answers obtained appear to have no units. The correct unit must then be added in seemingly arbitrary fashion. References 9–11 discuss the various equilibrium constants and their units.
Literature Cited 1. Cox, J. D. Pure Appl. Chem. 1982, 54, 1239. 2. Wagman, D. D.; Evans, W. H.; Parker, V. B.; Schumm, R. H.; Halow, I.; Bailey, S. M.; Churney, K. L.; Nuttall, R. L. J. Phys. Chem. Ref. Data 1982, 11(Suppl. 2). The NBS Tables of Chemical Thermodynamic Properties. 3. Chase, M. W., Jr.; Davies, C. A.; Downey, J. R., Jr.; Frurip, D. J.; McDonald, R. A.; Syverud, A. N. J. Phys. Chem. Ref. Data 1985, 14(Suppl. 1). JANAF Thermochemical Tables, 3rd ed. 4. Cox, J. D.; Wagman, D. D.; Medvedev, V. A. CODATA Key Values for Thermodynamics; Hemisphere: New York, 1989. 5. Thermodynamic Properties of Individual Substances, 4th ed.; Gurvich, L. V.; Veyts, I. V.; Alcock, C. B., Eds.; Hemisphere: New York, 1989. 6. Thermochemical Properties of Inorganic Substances, 2nd ed.; Knacke, O.; Kubaschewski, O.; Hesselmann, K., Eds.; Springer: Berlin, 1991. 7. Barin. I. Thermochemical Data of Pure Substances, 3rd ed.; VCH: Weinheim, 1995. 8. Freeman, R. D. J. Chem. Educ. 1985, 62, 681–686. 9. Tykodi, R. J. J. Chem. Educ. 1986, 63, 582–585. 10. Mills, I. M. J. Chem. Educ. 1989, 66, 887–889. 11. Mills, I. M. J. Chem. Educ. 1995, 72, 954–955.
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