Introduction to Lattice Energy

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Experimental Method of Lattice Energy

20150107

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I N D E X

Serial Name of the Topics Page

01 Introduction of Lattice Energy 02

02 Factors Affecting Lattice Energy of a Crystal 03

03 Attraction & Repulsion in a Crystal 04

04 The Relationship between Lattice Energies and Physical Properties

05

05 Determination of Lattice Energy 06

06 The Born-Haber Cycle 07

07 Born-Haber Cycle: Explanation with an Examples

Born-Haber Cycle for Sodium Chloride Born-Haber Cycle for Cesium Fluoride Born-Haber Cycle for Lithium Fluoride

08-13

08 Importance of Born-Haber Cycle 13

09 Applications of Lattice Energy Determination 13

10 Problem 14

Tables

Table 1 Selected Enthalpies of Sublimation at 298 K 9

Table 2 Selected Bond Dissociation Enthalpies at 298 K 10

Table 3 Summary of Reactions in the Born–Haber Cycle for the Formation of CsF(s)

12

Table 4 Representative Calculated Lattice Energies 13

Table 5 Experimental & Theoretical Lattice Energies in Kcal per Mole

14

Reference of the Topics Studied

01. Textbook of Physical Chemistry by Samuel Glasstone 02. Solid State Chemistry and its Applications by Anthony R West 03. General Chemistry (9th Edition) by Darrell D. Ebbing & Steven D. Gammon 04. Essentials of Physical Chemistry by Arun Bahl, B.S. Bahl & G.D. Tuli 05. A Textbook of Physical Chemistry by K. K. Sharma 06. Inorganic Chemistry (2nd Edition) by Catherine E. Housecroft 07. Inorganic Chemistry by J.D. Lee 08. Wikipedia.org 09. Chemwiki.com

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Introduction of Lattice Energy

Lattice energy, also called lattice enthalpy, of an ionic solid/crystal is a measure of the strength of bonds in that ionic compound. The positive and negative ion in an ionic crystal are held together by electrostatic forces. The energy released when a crystalline solid forms from ions is related to the lattice energy of the solid.

Lattice Energy is defined as:

The decrease in energy accompanying the process of bringing the ions, when separated from each other by an infinite distance, to the positions they occupy in the stable lattice.

The lattice energy is the change in energy that occurs when an ionic solid is separated into isolated ions in the gas phase.

Another definition of the lattice energy is to regard it as the energy of formation of a given quantity of the crystal, e.g., 1 mole, from “gaseous ions”. That is, from separated ions free from any solvent. In this way it is seen that the lattice energy is the same in magnitude as, but opposite in sign to, the energy of dissociation of the crystal. It follows, therefore, that the greater the lattice energy, numerically, the greater the energy required to break up the crystal into its constituent ions.

Greater the lattice energy, greater the strength of ionic bond. Lattice energy works against the solution process, so an ionic solid with relatively large lattice energy is usually insoluble.

After the formation of cations and anions separately, they combine to form ionic compound.

In this process, energy is released.

Generally, two elements bond ionically if the ionization energy of one is sufficiently small and the electron affinity of the other is sufficiently large and negative. This situation exists between a reactive metal (which has low ionization energy) and a reactive nonmetal (which has large negative electron affinity). In general, bonding between a metal and a nonmetal is ionic. This also explains why ionic bonding normally results in a solid rather than in ion-pair molecules.

A quantitative measure of the stability of any ionic solid is given by its lattice energy or lattice enthalpy (L.E.).

Lattice Energy is used to explain the stability of ionic solids. Some might expect such an ordered structure to be less stable because the entropy of the system would be low. However, the crystalline structure allows each ion to interact with multiple oppositely charge ions, which causes a highly favorable change in the enthalpy of the system. A lot of energy is released as the oppositely charged ions interact. It is this that causes ionic solids to have such high melting and boiling points. Some require such high temperatures that they decompose before they can reach a melting and/or boiling point.

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Factors Affecting Lattice Energy of a Crystal

The value of lattice energy depends upon the following two factors:

(a) Size of the ions In order to have the greater force of attraction between the cations and anions their size should be small as the force of attraction is inversely proportional to the square of the distance between them. Because the ionic radii of the cations decrease in the order K+ > Na+ > Li+ for a given halide ion, the lattice energy decreases smoothly from Li+ to K+. Conversely, for a given alkali metal ion, the fluoride salt always has the highest lattice energy and the iodide salt the lowest.

Figure 1: A Plot of Lattice Energy versus the Identity of the Halide for the Lithium, Sodium, and Potassium Halides

Lattice Energies of Alkali Metals Halides (kJ/mol)

F- Cl- Br- I- Li+ 1036 853 807 757Na+ 923 787 747 704K+ 821 715 682 649Rb+ 785 689 660 630Cs+ 740 659 631 604

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Because lattice energy is inversely related to the inter-nuclear distance, it is also inversely proportional to the size of the ions. This effect is illustrated in figure, which shows that lattice energy decreases for the series LiX, NaX, and KX as the radius of X− increases. Because r0 in is the sum of the ionic radii of the cation and the anion (r0 = r+ + r−), r0 increases as the cation becomes larger in the series, so the magnitude of U decreases. A similar effect is seen when the anion becomes larger in a series of compounds with the same cation.

(b) Charge on Ions

Greater the charge on ions greater will be the force of attraction between them and therefore, greater will be the strength of the ionic bond. Because the lattice energy depends on the product of the charges of the ions, a salt having a metal cation with a +2 charge (M2+) and a nonmetal anion with a −2 charge (X2−) will have a lattice energy four times greater than one with M+ and X−, assuming the ions are of comparable size (and have similar inter-nuclear distances). For example, the calculated value of U for NaF is 910 kJ/mol, whereas U for MgO (containing Mg2+ and O2− ions) is 3795 kJ/mol.

Lattice energies are highest for substances with small, highly charged ions.

Attraction & Repulsion in a Crystal

The stability of an ionic crystal depends on the balancing of at least three forces:

a. The electrostatic, or Coulomb, forces between the ions which give a resultant attraction falling off with the square of the distance

b. Van der Waals forces of attraction diminishing according to the seventh power of the distance. And,

c. Interatomic repulsive forces, falling off still more rapidly with distance.

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The result of the attractive and repulsive forces is to lead to an equilibrium position of minimum potential energy, i.e., of greatest stability, as depicted in the figure below:

Figure 2: Potential Energy of Crystal Lattices (Shown for NaCl)

The r0, corresponds to the minimum, represents the equilibrium value of the ionic separation in the stable crystal lattice. At all temperatures there will, however, be some thermal oscillation about the equilibrium positions; in some cases rotation also occurs.

The Relationship between Lattice Energies and Physical Properties

The magnitude of the forces that hold an ionic substance together has a dramatic effect on many of its properties. The melting point, for example, is the temperature at which the individual ions have enough kinetic energy to overcome the attractive forces that hold them in place. At the melting point, the ions can move freely, and the substance becomes a liquid. Thus melting points vary with lattice energies for ionic substances that have similar structures. The melting points of the sodium halides (Figure 3), for example, decrease smoothly from NaF to NaI, following the same trend as seen for their lattice energies (Figure 1). Similarly, the melting point of MgO is 2825°C, compared with 996°C for NaF, reflecting the higher lattice energies associated with higher charges on the ions. In fact, because of its high melting point, MgO is used as an electrical insulator in heating elements for electric stoves.

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Figure 3: A Plot of Melting Point versus the Identity of the Halide for the Sodium Halides. The melting points follow the same trend as

the magnitude of the lattice energies in Figure 1

The hardness of ionic materials—that is, their resistance to scratching or abrasion—is also related to their lattice energies. Hardness is directly related to how tightly the ions are held together electrostatically, which, as we saw, is also reflected in the lattice energy. As an example, MgO is harder than NaF, which is consistent with its higher lattice energy.

In addition to determining melting point and hardness, lattice energies affect the solubilities of ionic substances in water. In general, the higher the lattice energy, the less soluble a compound is in water. For example, the solubility of NaF in water at 25°C is 4.13 g/100 mL, but under the same conditions, the solubility of MgO is only 0.65 mg/100 mL, meaning that it is essentially insoluble.

High lattice energies lead to hard, insoluble compounds with high melting points.

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Determination of Lattice Energy

Ionic crystals may be regarded as regular 3D arrays of point charges. The forces that hold them together are entirely electrostatic and may be calculated by summing all the electrostatic repulsions and attractions in the crystal.

Lattice energy can be calculated from Coulomb’s law or an experimental value may be obtained from thermodynamic data.

The Born-Haber Cycle

Direct experimental determination of the lattice energy of an ionic solid is difficult and not reliable. However, this quantity can be indirectly determined from experiment by means of a thermochemical “cycle” originated by Max Born and Fritz Haber in 1919 and now called the Born–Haber cycle. The reasoning is based on Hess’s law of constant heat summation.

There are several important concept to understand before the Born-Haber Cycle can be applied to determine the lattice energy of an ionic solid: Ionization Energy, Electron Affinity, Dissociation Energy, Sublimation Energy, Heat of Formation, and Hess's Law.

Ionization Energy is the energy required to remove an electron from a neutral atom or an ion. This process always requires an input of energy, and thus will always have a positive value. In general, ionization energy increases across the periodic table from left to right, and decreases from top to bottom. There are some exceptions, usually due to the stability of half-filled and completely filled orbitals.

Electron Affinity is the energy released when an electron is added to a neutral atom or an ion. Usually, energy released would have a negative value, but due to the definition of electron affinity, it is written as a positive value in most tables. Therefore, when used in calculating the lattice energy, we must remember to subtract the electron affinity, not add it. In general, electron affinity increases from left to right across the periodic table and decreases from top to bottom.

Dissociation energy is the energy required to break apart a compound. The dissociation of a compound is always an endothermic process, meaning it will always require an input of energy. Therefore, the change in energy is always positive. The magnitude of the dissociation energy depends on the electronegativity of the atoms involved.

Sublimation energy is the energy required to cause a change of phase from solid to gas, bypassing the liquid phase. This is an input of energy, and thus has a positive value. It may also be referred to as the energy of atomization.

The heat of formation is the change in energy when forming a compound from its elements. This may be positive or negative, depending on the atoms involved and how they interact.

Hess's Law states that the overall change in energy of a process can be determined by breaking the process down into steps, then adding the changes in energy of each step. The Born-Haber Cycle is essentially Hess's Law applied to an ionic solid.

The Born-Haber Cycle can be expressed by the general equation:

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Heat of formation = Heat of atomization + Dissociation energy + (sum of Ionization energies) + (sum of Electron affinities) + Lattice energy

Note: In this general equation, the electron affinity is added. However, when plugging in a value, determine whether energy is released (exothermic reaction) or absorbed (endothermic reaction) for each electron affinity. If energy is released, put a negative sign in front of the value; if energy is absorbed, the value should be positive.

Rearrangement to solve for lattice energy gives the equation:

Lattice energy = Heat of formation - Heat of atomization - Dissociation energy - (sum of Ionization energies) - (sum of Electron Affinities)

General Born-Haber Cycle for the Formation of Metal Halide

Born-Haber Cycle: Explanation with an Examples

Born-Haber Cycle for Sodium Chloride

The Born-Haber Cycle is based on the first law of thermodynamics. This cycle involves the formation of an ionic compound (crystal lattice) from the reaction of a metal (often a group I or II element) with a non-metal.

To understand the concept of ‘Born-Haber Cycle’ let us consider the formation of NaCl from its constituents.

To obtain the lattice energy of NaCl, one think of solid sodium chloride being formed from the elements by two different routes, as shown in Figure below.

Process – I

In one route, NaCl(s) is formed directly from the elements, Na(s) and1 2 Cl g . The enthalpy

change for this is∆H , which is 411kJ per mole of NaCl.

Na (s) + 1/2 Cl2 (g) NaCl (s) + Hf0

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Figure 4: Born–Haber cycle for NaCl

Process – II

The second route consists of five steps:

1. Sublimation of Sodium Metallic sodium is vaporized to a gas of sodium atoms. (Sublimation is the transformation of a solid to a gas.) The enthalpy change for this process, measured experimentally, is ∆ 108 per mole of sodium. Enthalpy of sublimation is always positive because energy is required to sublime a solid.

Table 1: Selected Enthalpies of Sublimation at 298 K

Substance ΔHsub (kJ/mol)Li 159.3Na 107.5K 89.0Rb 80.9 Cs 76.5Be 324.0Mg 147.1Ca 177.8Sr 164.4 Ba 180.0

2. Dissociation of chlorine

Dissociation of ½ mole of Cl2 (g) into 1 mole of separate gaseous chlorine, Cl(g) atoms. It is the enthalpy of atomization of chlorine. The enthalpy change for this equals the Cl– Cl bond dissociation energy, which is 240 kJ per mole of bonds, or 120 kJ per mole of Cl atoms. Therefore, ∆ 120 / The ΔH for this reaction, too, is always positive because energy is required to dissociate any stable diatomic molecule into the component atoms.

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Table 2: Selected Bond Dissociation Enthalpies at 298 K

Substance D (kJ/mol)H2(g) 436.0 N2(g) 945.3 O2(g) 498.4 F2(g) 158.8 Cl2(g) 242.6 Br2(g) 192.8 I2(g) 151.1

3. Ionization of sodium

Sodium atoms are ionized to Na+ ions. The enthalpy change is essentially the first ionization energy of atomic sodium, which equals ∆ 496 per mole of Na. Energy is needed to ionize any neutral atom. Hence, regardless of the compound, the enthalpy change for this portion of the Born–Haber cycle is always positive.

4. Formation of chloride ion The electrons from the ionization of sodium atoms are transferred to chlorine atoms. The enthalpy change for this is the electron affinity of atomic chlorine, which equals ∆ 349 per mole of Cl atoms. Electron affinities can be positive, negative, or zero. In this case, ΔH is negative because of the highly negative electron affinity of chlorine.

5. Formation of NaCl(s) from ions The ions Na+ and Cl- formed in Steps 3 and 4 combine to give solid sodium chloride. Because this process is just the reverse of the one corresponding to the lattice energy (breaking the solid into ions), the enthalpy change is the negative of the lattice energy. If U be the lattice energy, the enthalpy change for Step 5 is∆ . Because it is the reverse of the equation used to define lattice energy and U is defined to be a positive number, ΔH5 is always negative, as it should be in a step that forms bonds.

According to Hess’s law:

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In summing the equations, we have canceled terms that appear on both the left and right sides of the arrows. The final equation is simply the formation reaction for NaCl(s). Adding the enthalpy changes, we find that the enthalpy change for this formation reaction is375 . But the enthalpy of formation has been determined calorimetrically and equals 411 . Equating these two values, we get:

∆

,

,

,

Therefore, the lattice enthalpy of the formation of NaCl is 786kJ/mol.

Born-Haber Cycle for Cesium Fluoride

Let’s use the Born–Haber cycle to determine the lattice energy of CsF(s). CsF is a nearly ideal ionic compound because Cs is the least electronegative element that is not radioactive and F is the most electronegative element. To construct a thermochemical cycle for the formation of CsF, we need to know its enthalpy of formation, ΔHf, which is defined by the following chemical reaction:

→

Figure 5: The Born–Haber Cycle Illustrating the Enthalpy Changes Involved in the Formation of Solid Cesium Fluoride from Its

Elements

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The following steps are involved in the formation CsF from its constituents:

1. Sublimation of Cs

Cs s → Cs g

Enthalpyofsublimation, ΔH ΔH 76.5kJ/mol

2. Ionization of Cs Cs g → Cs g

EnthalpyofIonization, ΔH ΔH 375.7kJ/mol 3. Dissociation of F

12F g → F g

Enthalpyofdissociation, ΔH ΔH 79.4kJ/mol 4. Formation of Fluoride

F g → F g Electronaffinityoffluorine, ΔH ΔH 328.2kJ/mol

5. Formation of CsF(s) Cs g F g → CsF s

Enthalpyofformation, ΔH ΔH 553.5kJ/mol

According to Hess’s law:

ΔHf = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5

Or, −ΔH5 = ΔH1 + ΔH2 + ΔH3 + ΔH4 − ΔHf

Or, U = 76.5 kJ + 375.7 kJ/ + 79.4 kJ + (−328.2 kJ) − (−553.5 kJ)

= 756.9 kJ/mol

Table 3: Summary of Reactions in the Born–Haber Cycle for the Formation of CsF(s)

Reaction Enthalpy Change (kJ/mol) (1) Cs(s) → Cs(g) ΔHsub = 76.5 (2) Cs(g) → Cs + (g) + e− I1 = 375.7 (3) ½F2(g) → F(g) ½D = 79.4 (4) F(g) + e− → F−(g) EA = −328.2 (5) Cs + (g) + F−(g) → CsF(s) −U = −756.9 Cs(s) + ½F2(g) → CsF(s) ΔHf = −553.5

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Born-Haber Cycle for Lithium Fluoride

Importance of Born-Haber Cycle

1. It is useful to determine the electron affinities which are otherwise difficult to determine by other means.

2. It is useful in analyzing and correlating the stability of various ionic solids. 3. It also provides an explanation for why most metals fail to form stable ionic compounds

in low valence states, such as AlO, MgCl.

Applications of Lattice Energy Determination

1. Estimation of electron affinities One of the most valuable applications of the lattice energy is to calculate the electron affinities of halogen atoms, i.e., work required to remove an electron from a halogen ion, since these can only be determined directly with some difficulty. Use is made of the Born-Haber Cycle.

2. Fluoride affinities 3. Estimation of standard enthalpies of formation and disproportionation

Table 4: Representative Calculated Lattice Energies

Substance U (kJ/mol)NaI 682CaI2 1971MgI2 2293

NaOH 887 Na2O 2481

NaNO3 755Ca3(PO4)2 10,602

CaCO3 2804

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Table 5: Experimental & Theoretical Lattice Energies in Kcal per Mole

The discrepancy between the experimental lattice energy and that calculated on the assumption of purely ionic linkages increases as the valence partakes increasingly of covalent character.

Problem

Draw a Born-Haber cycle for the formation of RbCl to determine the lattice enthalpy of the compound using the following data:

∆H 431kJ

∆H 86kJ

∆H 122kJ

∆H 408kJ

∆H 349kJ

Solution:

Born-Haber Cycle for the formation of RbCl:

Figure 6: Born-Haber Cycle for RbCl

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Given,

∆H ∆H 431kJ

∆H ∆H 86kJ

∆H ∆H 122kJ

∆H ∆H 408kJ

∆H ∆H 349kJ

And, Lattice Energy (L.E.), ΔH5 = U =?

According to Hess’s law:

ΔHf = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5

Or, −ΔH5 = ΔH1 + ΔH2 + ΔH3 + ΔH4 − ΔHf

Or, U = 86 kJ/mol + 122 kJ/mol + 408 kJ/mol + (−349 kJ/mole) − (−431 kJ/mol)

= 698 kJ/mol

Therefore, Lattice energy for the formation of RbCl = 698 kJ/mol.