(so ignoring any vibrational modes). We now calculate the energy required in pulling the static ions apart to infinity. (while preserving the complex ...

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Research: Science and Education

Thermodynamics of the Relationship between Lattice Energy and Lattice Enthalpy H. Donald B. Jenkins Department of Chemistry, University of Warwick, Coventry, West Midlands CV4 7AL United Kingdom; [email protected]

Lattice (potential) energy is the energy associated with the process in which a crystalline solid lattice, MpXq is converted into its constituent gaseous ions, pMq+(g) and qXp− (g): Mp Xq(s)

q+ p− pM (g) + qX (g)

(1)

When lattice energy, UPOT, is incorporated and made part of a Born–Fajans–Haber cycle (which is essentially an enthalpybased thermochemical cycle) it needs to be converted into a lattice enthalpy term. This lattice enthalpy, ∆HL, involves correction of the UPOT term by an appropriate number of RT terms. This comes about because enthalpy, H, and (internal) energy, U, are related as (2)

H = U + PV

Sometimes the correction term referred to above is (oversimplistically and mistakenly) assumed1 to be just the ∆nRT term of eq 4 alone, where ∆n is the number of gaseous ions produced (i.e., for process in eq 1, ∆n would be p + q). However as the discussion here shows, additional factors need be considered in the thermodynamics of this overall process. Although the magnitude of RT is approximately 2.5 kJ mol᎑1 and the required correction can often be relatively small in magnitude in some cases, when lattices are large and complex [as is the case for minerals, e.g., hydroxyapatite Ca10(PO4)6(OH)2] or where inorganic polyatomic ions exist in the lattice, (e.g., CaSO4), the situation needs careful evaluation. The lattice energy of the two compounds above (which are taken as examples) correspond to the processes Ca10(PO4)6(OH)2(s)

and hence ∆H = ∆ U + P ∆V + V ∆ P

For a process at constant pressure, P, in which gaseous ions (assumed to behave ideally) are generated, the relationship between ∆H and ∆U can be written (4)

∆H = ∆U + ∆n RT

(5) 10 Ca2 +(g) + 6 PO43 −(g) + 2 OH −(g)

(3)

Ca2 +(g) + SO42−(g)

CaSO4(s)

(6)

These processes, whereby the solid is disrupted to form gaseous ions, can be viewed as a potential energy curve drawn as a function of the ion separation (Figure 1).

10 Ca2ⴙ(g) ⴙ 6 PO43ⴚ(g) ⴙ 2 OHⴚ(g) true thermodynamic state

† 3 3 2 18 RT ⴙ 6 RT ⴙ 2 RT 2 2 2

10 Ca2ⴙ(g) ⴙ 6 PO43ⴚ(g) ⴙ 2 OHⴚ(g) stationary (hypothetical) state

∆U

10U [Ca2ⴙ(g)] ⴙ 6U [PO43ⴚ(g)] ⴙ 2U [OHⴚ(g)]

UPOT

Ca10(PO4)6(OH)2(s) true thermodynamic state

UACOUS

Ca10(PO4)6(OH)2(s) bottom of PE well (stationary ions on lattice points) absolute zero of 0 energy

U [Ca10(PO4)6(OH)2(s)]

Figure 1. Potential energy curve representing lattice energy for hydroxyapatite, Ca10(PO4)6(OH)2 (†assumes ions in the gas and crystal have equally excited degrees of freedom, so the vibrational RT terms are ignored here).

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To compute, a priori, any lattice energy we begin with standard crystal structure data for the solid. This defines the position of each atom in the lattice in the form of fixed atomic coordinates referred to a static reference frame. We assume that the ions in the solid are stationary on their lattice points (so ignoring any vibrational modes). We now calculate the energy required in pulling the static ions apart to infinity (while preserving the complex anions intact) in the gaseous state. The potential energy curve for Ca10(PO4)6(OH)2 is sketched in Figure 1. On the diagram is shown the zero-point energy, UACOUS, as well as the absolute zero of energy, from which we can define the total internal energy of the crystal, U[Ca10(PO4)6(OH)2(s)], and the total internal energies of the product ions, U[Ca2+(g)], U[PO43−(g)], and U[OH−(g)]. The total internal energy change for the lattice energy step can be written as 2+

3−

∆U = 10U Ca (g) + 6U PO4 (g) + 2U OH −(g) − U Ca10(PO4)6(OH)2(s)

(7)

Also in the figure, we distinguish between the energy states of the crystal when the ions in the crystal lattice are at rest (stationary ions)—as is implied by the static coordinates provided by the crystal structure data—and the true thermodynamic state of the solid crystal. The latter, as a result of its possession of acoustic (zero-point) energy, so permits the ions to oscillate about their lattice points. The separated (product) ions at infinity are shown in Figure 1. When at rest they execute no translational movement, rotation, or vibration. This represents an hypothetical (stationary) state. In their true thermodynamic state these ions will have translational and rotational energy and also possess vibrational modes. The latter are the more difficult to quantify, especially within a crystal. We therefore make the assumption that the ions are equally excited when in the lattice as they are in the gaseous state and thus the vibrational contribution to ∆U will approximately cancel for “reactants” and “products” involved in the lattice energy step and can therefore be disregarded. This leaves then the translational and rotational contributions for the ions at infinite separation to be considered. We assume for these energies, classical values of (3兾2)RT per ion (translational) and 0, (2兾2)RT, or (3兾2)RT per ion for their rotational energy depending on whether the particular ions in question are monatomic, diatomic, or polyatomic, respectively.

By combining eqs 9 and 4 for Ca10(PO4)6(OH)2, the lattice enthalpy, ∆HL, can be written (10) ∆H L = U POT + 2RT CaSO4 Lattice For CaSO4, the energy gap between the stationary and true thermodynamic states of the ions [Ca2+, (3兾2)RT translation; SO42−, (3兾2)RT translation + (3兾2)RT rotation] is (9兾2)RT (11 kJ mol᎑1, similarly not insignificant). Thus the following relationships can be seen:

How the lattice enthalpy ∆HL and lattice energy, UPOT are related can be established by combining eqs 12 and 4 for CaSO4, leading to 1 ∆H L = U POT + RT (13) 2 General Salt Lattice Generalizing for a general cation (anion), M+ (X−), labelled i, the individual translational–rotational energy gap between static and true thermodynamic states, for each ion contribution will be equal to (ci兾2)RT where ci = 3 for monatomic ions, 5 for linear polyatomic, and 6 for polyatomic ions. This total energy gap can then be found as the sum over all the ions present in the lattice p +q

=

i

If UACOUS is assumed3 to be 3RT per ion (of which there are 18, i.e., ∆n = 18) then

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ci RT = 2

•

c M+

p

cX−

+ q

2

2

RT

(14)

so that ∆U = UPOT + p

c M+ 2

+ q

cX− 2

RT − UACOUS (15)

Substituting into eq 4 ∆H L = UPOT + p

Vol. 82 No. 6 June 2005

c M+ 2

+ q

c X− 2

RT

− UACOUS + ( p + q ) RT

(9)

∆U = UPOT − 16 RT

ci RT 2

Thus for the binary salt MpXq, having p cations Mq+ and q anions Xp−

Ca10(PO4)6(OH)2 Lattice

(8)

∑ i

∑

∆U = UPOT + 38RT − UACOUS

(11)

If UACOUS is assumed3 to be 3RT per ion (of which there are two, i.e., ∆n = 2) then 3 ∆U = U POT − RT (12) 2

p +q

In the case of Ca10(PO4)6(OH)2 , the energy gap between the stationary and true thermodynamic states of the ions [Ca2+, (3兾2)RT translation; PO43−, (3兾2)RT translation + (3兾2)RT rotation; OH−, (3兾2)RT translation + (2兾2)RT rotation], is a total of 38RT ( ≈ 95 kJ mol᎑1, an important term). From Figure 1

9 RT − UACOUS 2

∆U = U POT +

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and since2 UACOUS ≈ 3( p + q)RT ∆H L = UPOT +

p

c M+ 2

+ q

c X− 2

RT

− 3 ( p + q ) RT + ( p + q ) RT

∆H L = UPOT +

p

c M+ c − − 2 + q X − 2 2 2

(17)

RT (18)

Complex Lattices and Minerals Containing More than Two Types of Ions Equation 18 is the key equation that is readily extendable to more general complex lattices in the form ions in formula unit

∆H L = UPOT +

∑ si i

UACOUS = 3ni 1 +

ci − 2 RT 2

(19)

where si is the number of ions of type i in the formula unit and ci is defined as above according as whether ion i is monatomic (ci = 3), linear polyatomic (ci = 5), or polyatomic (ci = 6), respectively. Conclusion The thermodynamics involved in incorporating the lattice energy, UPOT into a Born–Fajans–Haber enthalpy cycle is often oversimplified (and has never been explicitly documented). For complex and large lattices this can produce significant errors. The lattice enthalpy, ∆HL, is shown, for a general lattice, to be related to the lattice energy, UPOT by the formula: ions in formula unit

∆H L = UPOT +

∑ si i

for all lattices. Frequently ∆HL and UPOT are assumed to be synonymous. For the NaCl lattice, possessing monatomic ions, eq 18 reveals that ∆HL = UPOT − RT while for CaCl2: ∆HL = UPOT − (3兾2)RT, yet the first undergraduate text reached off the shelf simply equates ∆HL and UPOT without distinction for these two cases. This is clearly of minor consequence. For larger lattices, as will be illustrated, the distinction becomes increasingly important however and needs to be accounted for. 2. The author declines to mention specific texts. 3. We can, of course, calculate UACOUS for an individual crystal lattice either from a knowledge of heat capacities as a function of temperature and from the zero-point energy of the lattice or by means of either the Einstein (1–4) or Debye (5) theory of heat capacities. If θE and θD are the Einstein and Debye characteristic temperatures, respectively, then on the basis of the Einstein model of the crystal containing ni ions:

ci − 2 RT 2

(20)

1 θE 12 T

2

−

1 θE 720 T

4

+ ... RT (22)

The corresponding expression using the Debye model takes the form: UACOUS = 3ni 1 +

1 θD 20 T

2

−

1 θD 1680 T

4

+ ... RT (23)

At room temperature all the alkali halides with the exception of LiF, NaF, KF, and LiCl are above their Debye temperature (6) and the error in putting:

UACOU US = 3ni RT ≈ 15 kJ/mol

(24)

is less than 5% (or 0.6 kJ mol᎑1). This is of course an approximation and still dependent on classical ideas about the equipartition of energy. For salts where this approximation is invalid one has to rely on computing the zero-point energy and adding H298 − H0 from tabulated information. For most work, however, it is satisfactory to assume that UACOUS ≈ 3RT per ion.

Dedication This article is dedicated to Poppy, our grandchild. Literature Cited

Notes 1. Innumerable textbooks2 assume the relationship between lattice energy and lattice enthalpy to be that of eq 4 with the change in energy brought about by the process shown in eq 1, UPOT replacing the internal energy change ∆U :

(21)

∆H L = UPOT + ∆ nR T

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1. 2. 3. 4. 5. 6.

Einstein, A. Ann. Physik. 1907, 22, 180. Einstein, A. Ann. Physik. 1907, 22, 800. Einstein, A. Ann. Physik. 1911, 34, 170. Einstein, A. Ann. Physik. 1911, 34, 590. Debye, P. Ann. Physik. 1912, 39, 789. Münster, A. Statistical Thermodynamics; Hall, V., Scritan, T., Scientific Translation Service; Springer-Verlag: Berlin, 1974.

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